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Parents often use SCERT Class 10 English Solutions Unit 4 Chapter 2 Trills and Thrills: Birdwatching in India Questions and Answers Activities Notes Pdf to assist their kids with homework.

Class 10 English Trills and Thrills: Birdwatching in India Question Answer Notes Kerala Syllabus

Std 10 English Trills and Thrills: Birdwatching in India Question Answer

10th Class English Trills and Thrills: Birdwatching in India Question Answer – Textual Questions And Answers

a. What is special about ornithology?
പക്ഷിശാസ്ത്രത്തിന്റെ പ്രത്യേകത എന്താണ്?
Answer:
Ornithology as a hobby or profession is full of adventures, rewards and disappointments. It is one of the most peaceful of pursuits and it has excitement and thrills.
ഒരു ഹോബി അല്ലെങ്കിൽ ഇഷ്ടപ്പെട്ട ഒരു കാര്യം എന്ന നിലയിൽ പക്ഷിശാസ്ത്രം സാഹ സികത, പ്രതിഫലം, നിരാശ എന്നിവയാൽ നിറ ഞ്ഞതാണ്. ഇത് ഏറ്റവും സമാധാനപരമായ പ്രവർത്തനങ്ങളിൽ ഒന്നാണ്, ഇതിന് ആവേ ശവും സന്തോഷവുമുണ്ട്.

b. Is there any difference between what we read in books about birds and their real life? Explain.
പക്ഷികളെക്കുറിച്ച് പുസ്തകങ്ങളിൽ വായി ക്കുന്നതും അവയുടെ യഥാർത്ഥ ജീവിതവും തമ്മിൽ എന്തെങ്കിലും വ്യത്യാസമുണ്ടോ? വിശദീകരിക്കുക.
Answer:
There are differences between what we read in books about birds and their real life. Salim Ali had grown up only on the traditional accounts of the nesting habits of birds. But while sitting hidden in a canvas hide, high up on a stepladder, a few yards away from the nests, he noticed some happenings which showed that in real life birds behaved differently and not as seen in the textbooks.
പക്ഷികളെക്കുറിച്ചുള്ള പുസ്തകങ്ങളിൽ വായി ക്കുന്നതും അവയുടെ യഥാർത്ഥ ജീവിതവും തമ്മിൽ വ്യത്യാസങ്ങളുണ്ട്. പക്ഷികളുടെ കൂടു കെട്ടൽ ശീലങ്ങളെക്കുറിച്ചുള്ള പരമ്പരാഗത വിവരണങ്ങൾ വായിച്ചാണ് സലിം അലി വളർന്നത്. എന്നാൽ കൂടുകളിൽ നിന്ന് ഏതാനും യാർഡുകൾ അകലെ. ഒരു സ്റ്റെപ് ഏണിയിൽ ഉയരത്തിൽ ഒരു ക്യാൻവാസ് മറവിൽ ഒളിച്ചിരുന്ന് പക്ഷികളെ നിരീക്ഷിക്കു മ്പോൾ യഥാർത്ഥ ജീവിതത്തിൽ പക്ഷികൾ പാഠപുസ്തകങ്ങളിൽ കാണുന്നതുപോലെയല്ല. വ്യത്യസ്തമായിട്ടാണ് പെരുമാറുന്നതെന്നു കാണിക്കുന്ന ചില അസാധാരണ സംഭവങ്ങൾ അദ്ദേഹം ശ്രദ്ധിച്ചു.

c. On what occasions can elephants be dangerous? Why?
ഏതൊക്കെ സന്ദർഭങ്ങളിൽ ആനകൾ അപ കടകാരികളാകാം? എന്തുകൊണ്ട്?
Answer:
Elephants can be dangerous if they are rogues or females with small calves.
It is a tusker, a male elephant, that leads the herd. He is often a rogue and he wants to show off his power. A female elephant becomes dangerous if she is with a calf as she wants to make sure that her calf is safe.
തെമ്മാടികളായ കൊമ്പനാനകളും ചെറിയ കുട്ടികളുള്ള പെൺ ആനകളും ആണെങ്കിൽ അവ അപകടകാരികളാകാം. കൂട്ടത്തെ നയിക്കു ന്നത് ഒരു ആൺ ആനയാണ്. അവൻ പല പ്പോഴും ഒരു തെമ്മാടിയാണ്. തന്റെ ശക്തി പ്രക ടിപ്പിക്കാൻ അവൻ ആഗ്രഹിക്കുന്നു. തന്റെ കുട്ടി സുരക്ഷിതമാണെന്ന് ഉറപ്പാക്കാൻ ആഗ്രഹിക്കു ന്നതിനാൽ ഒരു പെൺ ആന തന്റെ കുട്ടിയോ ടൊപ്പം ആണെങ്കിൽ അപകടകാരിയാകുന്നു.

d. What was the impact of the construction of the Parambikulam dam?
പറമ്പിക്കുളം അണക്കെട്ടിന്റെ നിർമ്മാണ ത്തിന്റെ അനന്തരഫലങ്ങൾ എന്തായിരുന്നു
Answer:
The construction of the Parambikulam dam meant the dismantling of the romantic. Cochin Forest tramway. The huge dam and reservoir submerged Parambikulam and all the beautiful places around it.
പറമ്പിക്കുളം അണക്കെട്ടിന്റെ നിർമ്മാണം കാരണം കൊച്ചി ഫോറസ്റ്റ് ട്രാംവേ പൊളിച്ചു മാറ്റേണ്ടി വന്നു. വലിയ അണക്കെട്ടും ജലസം ഭരണിയും പറമ്പിക്കുളത്തെയും ചുറ്റുമുള്ള എല്ലാ മനോഹരമായ സ്ഥലങ്ങളെയും വെള്ള ത്തിൽ മുക്കി.

e. What was the sight the forest guard pointed to?
ഫോറസ്റ്റ് ഗാർഡ് ചൂണ്ടിക്കാണിച്ച് കാഴ്ച എന്തായിരുന്നു?
Answer:
The sight the forest guard pointed to was a tusker coming down the same path from the opposite direction.
ഫോറസ്റ്റ് ഗാർഡ് ചൂണ്ടിക്കാണിച്ച് കാഴ്ച എതിർദിശയിൽ നിന്ന് അതേ വഴിയിലൂടെ വരുന്ന ഒരു കൊമ്പനാനയായിരുന്നു.

f. How did Salim Ali react when he saw the tusker?
കൊമ്പനാനയെ കണ്ടപ്പോൾ സലിം അലി എങ്ങനെ പ്രതികരിച്ചു
Answer:
When he saw the tusker, Salim turned to flee as fast as he could, motioning to Omar who was a bit behind him to do the same.
ആനയെ കണ്ടപ്പോൾ സലിം അലി കഴിയുന്നത വേഗത്തിൽ ഓടിപ്പോകാനായി തിരിഞ്ഞു. തന്റെ പിന്നിലാ യി രുന്ന ഒമറിനോട് അങ്ങനെ ചെയ്യാൻ ഞാൻ ആംഗ്യം കാണിച്ചു.

g. Do animals always attack when they encounter human beings? Explain.
മനുഷ്യരെ നേരിടുമ്പോൾ മൃഗങ്ങൾ എപ്പോഴും ആക്രമിക്കുമോ? വിശദീകരിക്കുക.
Answer:
Animals do not always attack when they encounter human beings. They attack only if they feel threatened or if, in the case of animals like lions and tigers, they are hungry.
മനുഷ്യരെ നേരിടുമ്പോൾ മൃഗങ്ങൾ എപ്പോഴും ആക്രമിക്കാറില്ല. ഭീഷണി തോന്നിയാലോ സിംഹങ്ങൾ, കടുവകൾ തുടങ്ങിയ മൃഗങ്ങളുടെ കാര്യത്തിൽ ആണെങ്കിൽ വിശന്നാലോ മാത്രമേ അവ മനുഷ്യരെ ആക്രമിക്കൂ.

h. What made Salim Ali’s hair stand on end?
സലിം അലിയുടെ രോമങ്ങൾ എണീറ്റ് നിൽക്കാൻ കാരണമെന്താണ്?
Answer:
Salim Ali was standing on the edge of a mountain to watch a yellow-naped Yuhina. For a better view, he took a step back. He heard a pebble rolling down. When he looked back he realised that if he had gone backwards for 2 inches more he would have fallen into the abyss. This thought made his hair stand on end.
മഞ്ഞ നിറത്തിലുള്ള തലമുടിയുള്ള യൂഹിനയെ കാണാൻ സലിം അലി ഒരു മലയുടെ അരികിൽ നിൽക്കുകയായിരുന്നു. മികച്ച കാഴ്ചയ്ക്കായി, അയാൾ ഒരു പടി പിന്നോട്ട് മാറി. ഒരു കല്ല് താഴേക്ക് ഉരുളുന്നത് അയാൾ കേട്ടു. തിരിഞ്ഞു നോക്കിയപ്പോൾ, 2 ഇഞ്ച് കൂടി പിന്നോട്ട് പോയി രുന്നെങ്കിൽ താൻ അഗാധത്തിലേക്ക് വീഴുമാ യിരുന്നെന്ന് അയാൾക്ക് മനസ്സിലായി. ഈ ചിന്ത അയാളെ ഭയപ്പെടുത്തി, അയാളുടെ രോമ ങ്ങൾ എണീറ്റ് നിൽക്കാൻ പ്രേരിപ്പിച്ചു.

10th Std English Trills and Thrills: Birdwatching in India Question Answer Notes – Let’s rewind

Question 1.
Why does the author say that the birds Go had not read the textbooks?
Answer:
The author says that the birds had not read the textbooks because what is found in the textbooks about the nesting habits of birds differs from their actual nesting habits. Birds don’t make their nests as prescribed in the textbooks.

Question 2.
You know what happened when Salim Ali and Omar went bird-watching in Parambikulam. Why do people engage in such adventurous but challenging activities?
Answer:
People engage in such adventurous but challenging activities because of the joy and satisfaction they get when they do things they want to do. Man by nature is curious. All human progress is the result of the adventurous and challenging activities undertaken by curious and brave people.

Question 3.
The sight of the Yellow-naped Yuhina enthralled Salim Ali that he forgot his surroundings. Narrate the incident.
Answer:
The incident took place when Salim Ali was on the Himalayan trail from Almora to the Lipu Lekh Pass on his way to Lake Manasarover and Mount Kailas in 1945. He was at a particularly narrow part of the trail with a thousand feet of vertical scarp one side and the roaring Kali River some 300 feet vertically down the other side. He then saw this Yellow-naped Yuhina. To have a closer look he took a step back. He heard a pebble tumbling down. When he looked back he realised that if he had gone back two inches further he would have fallen into an abyss. He was lucky!

10th English Trills and Thrills: Birdwatching in India Question Answer – Let’s recall and recreate:

Question 1.
The forest guard cautions Salim Ali and Omar of the possible dangers of exploring Parambikulam and also the etiquette to be followed while trekking in a forest. Write down the instructions that he might have given them.
Answer:
a) Always stay with me. Never wander off the marked trail, as it’s easy to get into some other troubles.
b) If you need a rest or feel unwell, inform me immediately.
c) As there is lot of grass around even on the trail, beware of snakes.
d) Smoking is strictly prohibited outside of designated areas. Forest fires spread quickly and are devastating.
e) Observe from a distance. Enjoy the wildlife, but never approach, feed, or touch any animal.
f) Do not collect any plants, flowers, rocks, or historical artefacts. Everything in the forest belongs here. You can only take pictures.
g) If you need to use the restroom, tell me. I will tell you where to go.
h) There are dangerous plants and insects. So don’t try to touch them.
i) The terrain is uneven. There may be small stones on the way. So walk with caution.
j) Make sure you have enough water to drink.

Question 2.
A memorial lecture is conducted to honour Induchoodan, the birdman of Kerala. A noted birdwatcher is invited to deliver the lecture. Draft the speech that he is likely to deliver. You may give instances from “Trills and Thrills”.
Answer:
Honoured guests, fellow nature enthusiasts, ladies and gentlemen,

It’s a great privilege for me to stand here today and deliver this memorial lecture in honour of a man who didn’t just observe birds, but taught an entire generation how to truly listen to them. Induchoodan’s real name was K. K. Neelakantan. He was born at Mongombu, a village in Alleppy district, in 1923. Although he was a professor of English literature, he chose to write his most important works in his mother tongue, Malayalam. This choice was profoundly significant. It was through his brilliant, accessible, and deeply personal writing that the world of birds was opened up to the Malayalees.

His greatest book was “Keralathile Pakshikal”. In that he describes 261 species of birds. He brought them to life with his vivid language and his own illustrations. He transformed bird-watching into a movement of the masses. He didn’t just teach us the names of birds; he taught us their habits, their songs, their joys, and their struggles.

Induchoodan understood that conservation could only succeed if people cared, and people could only care if they understood. His works, like Pakshikalum Manushyarum and Pullu Thottu Poonara Vare, blended rigorous scientific observation with the charm of a captivating narrative, winning him numerous awards and many admirers. “Pullu Thottu Poonara Vare” is an idiom which means from something small to something great.

He showed us that a true ornithologist isn’t just someone who collects data, but someone who connects with nature on a spiritual level. Beyond birds, Induchoodan was a fierce and uncompromising environmental crusader. He co-founded the Kerala Natural History Society (KNHS), providing a platform for like-minded individuals to pool their knowledge and actively engage in conservation. His death in 1992 at the age of 69 was great loss to not only the state, but the nation and humanity at large. He is remembered by posterity as the “Birdman of Kerala” just like Salim Ali is known as the “Birdman of India”.

Kerala will never forget Induchoodan!
Thank you!

Question 3.
(a) Read the observation diary of an ornithologist given on p. 136. Now, write down the features of a birdwatcher’s diary in the box given there.
Study the sample diary given below.

Answer:

b) You visit Parambikulam Tiger Reserve and spot the Broad-tailed Grass Warbler. Write a Birdwatcher’s diary.
Answer:
Birdwatcher’s Diary Entry

Date and time: Saturday, 25th September 2025
Location: Parambikulam Tiger Reserve, Kerala
Weather: Overcast morning, becoming sunny, slightly humid. Approx 22°C. Time: 08:30 09:00 AM

The morning started like any other at Parambikulam. The air was thick with the calls from different kinds of birds. I was focusing on a patch of tall, dense grassland adjacent to a small, marshy depression-the classic, slightly wet, high-altitude habitat that holds the promise of the elusive Broad- tailed grass-warblers.

For a long time, nothing seemed to happen. I was almost getting disappointed and I was planning to move away. But then I heard a distinct, slightly harsh ‘chirrup-chirrup’ call, unlike anything else I’d heard that morning. It was low.

I froze, lifting my binoculars to the spot. The bird was absolutely skulking. I waited for five minutes, and then, a small, dark bird came out in a short, clumsy, parachuting display flight. It was, so quick, but the marks were unmistakable. It was the Broad-tailed Grass Warbler!

The most striking feature was its heavy, fanned, and broad tail-dark brown, appearing almost black underneath, giving it a ‘rear-heavy’ appearance as it fluttered briefly against the sky. Soon it dropped back into the cover of the dense grass.

I managed to keep my lens fixed on the spot where it landed. It slipped through the grass with an almost unbelievable agility, looking on the ground for insects. I caught a final, excellent look as it perched for a split second on a dead grass stem, revealing the pale, buffy breast and the very small head.

This is a ‘Vulnerable’ species. Seeing it is a memory that I will cherish for long. It gave me great satisfaction.

Question 4.
You have learnt to create a web page in Class IX. Now, develop the content of a web page on the tourist attractions of your area highlighting the variety of birds there. You may also give details of migratory birds.
Welcome to My Village Thrikkakara – A Bird Watcher’s Paradise!
Answer:
Escape the city’s hustle and bustle and immerse yourself in the tranquillity of my village. Here, nature thrives, offering a peaceful retreat for every visitor. While our village is known for its cultural heritage and scenic beauty, its true highlight is the incredible variety of birds that call this place their home. When talking of cultural heritage I must mention “Thrikkarappan”, very much seen during Onam celebrations.

Thrikkakara is a unique habitat for birds with its perfect mix of lush bushes, small hills, extensive farmlands, and a lake. This diverse landscape attracts a wide range of bird species, from vibrant migratory birds to rare resident ones. You don’t need to be an expert bird watcher to enjoy the spectacle just a pair of binoculars and keen eyes will do!

Near the lake you can spot a variety of water – birds like Kingfishers, Herons, and Egrets, especially in the morning. In the winter, you might even see some migratory species that travel thousands of miles to visit this place.

In the bush areas you will see the Cuckoos and can hear their melodious songs. In the farmlands you can see peacocks strutting proudly, sparrows flitting among the crops, and kites soaring high in the sky.

Enjoy Local Cuisine: No visit is complete to my village without tasting our delicious local food. Enjoy traditional dishes made from fresh, locally sourced ingredients. You’ll find a range of flavours that will leave you wanting more!

Plan Your Trip: Our village is perfect for a day trip or a weekend getaway. The best time to visit for bird-watching is from October to March, when the weather is pleasant and migratory birds are plenty.

Thrikkakara is between Ernakulam and Aluva. You can easily reach the place by road. The nearest railway station is Edapally.

Come and experience the charm of our village a peaceful haven where nature and culture beautifully intertwine. We eagerly wait for you!

Question 5.
Read the information on Thattekkad Bird Sanctuary given on p. 137. Identify other bird sanctuaries in Kerala. Study the variety of birds that live there and the details of migratory birds too. Prepare a write-up on the bird wealth of the state and the need to preserve it.

Answer:

Sanctuary	Habitat to	Migratory birds
Kumarakom	Waterfowl, Cuckoo, Owl, Moorhen, Egret, Heron, Darter, Cormorant, Kites, Woodpecker, Kingfisher, Parrots, Skylarks	Siberian Crane, Teal, Pintail Duck, Marsh Harriers, Flycatchers, Gulls, Booted Eagle, Steppe Eagle, Black-tailed Godwit
Mangalavanam	Little cormorant, Brahminy kite, Black-crowned Night Heron, White- breasted Water hen, Indian Pond Heron, Little Egret, Cattle Egret	Common Redshank, Common Greenshank, Common Redshank, Marsh Sandpiper
Kadalundi	Kingfishers, Brahminy Kites, Malabar Hornbills, Woodpeckers, Darters, Egrets, Cormorants,	Terns, Gulls, Sandpipers, Plovers, Whimbrels, Redshanks, Greenshanks, Curlews, Blacktailed and Bar-tailed Godwits, Curlew Sandpipers, Sanderlings, Crab Plovers, Openbill Storks
Pathiramanal	Darter, Indian Shag, Heron, Different Kinds of Egrets, Cormorants, Stork-billed kingfisher, Watercock, Different Bitterns, Pheasant-tailed and Bonze-tailed Jacanas	Pintail Duck, Common Teal, Gulls, Whistling Ducks, Cotton Pygmy-Goose

Kerala is blessed with a variety of birds. Thus we have a bird wealth, supported by its diverse landscapes, including the Western Ghats, extensive backwaters, and lush coastal wetlands. Kerala is Home to over 500 species of birds and it is significant habitat for both resident and migratory populations. This bird diversity includes numerous endemic species of the Western Ghats – such as the Malabar Grey Hornbill, Nilgiri Woodpigeon, and White-bellied Treepie – making it vital for global biodiversity conservation. However, this natural treasure faces growing threats primarily from habitat destruction due to rapid urbanization, conversion of wetlands for development, and the expansion of agriculture, as well as the impact of invasive species. Therefore, urgent conservation efforts are necessary to protect key habitats like the bird sanctuaries and sacred groves, control environmental pollutants, mitigate climate change effects, and promote citizen-science initiatives like the Kerala Bird Atlas to safeguard our bird wealth for ecological balance and future generations.

Trills and Thrills: Birdwatching in India Activities – Let’s read and reflect:

The serenity of nature evokes tender feelings in most of us. The poem on p. 138 by W.B. Yeats is on his experience at Coole Park.
വായിച്ച് ആസ്വദിക്കാം
പ്രകൃതിയുടെ ശാന്തത നമ്മളിൽ മിക്കവരിലും ആർദ്രമായ വികാരങ്ങൾ ഉണർത്തുന്നു. W. B. സെയ്റ്റ്സിന്റെ 138-ാം പേജിലെ കവിത കൂൾ പാർക്കിലെ അദ്ദേഹത്തിന്റെ അനുഭവത്തെക്കുറിച്ചാണ്.

Trills and Thrills: Birdwatching in India Extra Questions and Answers

Question 1.
Is the field study of birds entirely without hazards in South India?
Answer:
No, it is not. Especially in South India there are hazards in the field study of birds especially from wild elephants.

Question 2.
In Parambikulam, Salim Ali saw a man in the midst of a large group of Kadar Adivasis. Who was he and how was he dressed?
Answer:
He was the Austrian anthropologist Baron Omar Rolf. He was barefooted, topless and wore crumpled khaki shorts.

Question 3.
Salim Ali recalls one particularly hair-raising incident when he was trying to have a good look at the yellow-naped Yuhina. Where and when did the incident happen?
Answer:
The incident happened along the Himalayan trail from Almora to the Lipu Lekh Pass when Salim Ali was on his way to Lake Manasarover and Mount Kailas. It was in 1945.

Question 4.
Was the yellow-naped Yuhina Ali saw a big bird?
Answer:
No, it wasn’t. It was a tiny bird.

Question 5.
How long has Salim Ali watched birds according to the story?
Answer:
According to the story, Salim Ali has watched birds for more than half a century.

Question 6.
Why does Salim Ali watch birds?
Answer:
Salim Ali watches birds chiefly for the pleasure and the elation they afford.

Students often refer to Kerala Syllabus 10th Standard English Textbook Solutions and Class 10 English Trills and Thrills: Birdwatching in India Summary in Malayalam & English Medium before discussing the text in class.

Class 10 English Trills and Thrills: Birdwatching in India Summary

Trills and Thrills: Birdwatching in India Summary in English

One of the questions they ask me is about the thrilling adventures in a lifetime of exploring for birds. My answer must seem disappointing. Ornithology as a hobby or profession is full of adventures, rewards and disappointments. It is one of the most peaceful of pursuits. It does not lack excitement and thrills.

Any birdwatcher can make exciting discoveries because of the rich variety of birds in India. It is refreshing to find that bird-watching as a hobby is growing.

I had grown up only on the traditional accounts of the nesting habits of birds. These were interesting for a keen photographer like me to record on film. While sitting hidden in a canvas hide, high up on stepladder, a few yards away from the nests, I noticed some uncommon happenings which showed that the birds had not read the textbooks.

The field of study of birds is a peaceful occupation. But it has its risks. In the elephant-ridden jungles in South India, I have often found myself in disturbing circumstances. A wild elephant suddenly coming near you is an unnerving experience. This is not uncommon in parts of Karnataka and Kerala. In fact there is hardly any danger from a wild elephant unless it happens to be a rogue or a female with a small calf.

My journey by the romantic Cochin Forest tramway was in February 1946. Soon after that it was dismantled to make the Parambikulam hydro-electric project. The huge dam and reservoir has submerged Parambikulam and all the beautiful places around it. We reached Parambikulam just as it was getting dark. In the veranda of the forest bungalow there was a large group of Kadar advasis. In their midst there was a barefooted ‘topless’ European male in khaki shorts.

He was an Austrian anthropologist – Baron Omar Rolf. He said ornithology is his second love. He wanted to come with me the next morning. Led by a forest guard, we were walking on a narrow animal path through dense tall grassland above 5 feet high. It was the right kind of habitat for the Broad-tailed Grass Warbler.

At a bend in the path, the forest guard suddenly stopped, excitedly pointing in front. I just saw an elephant coming down the same path from the opposite direction. I turned to flee as fast as I could, motioning to Omar who was a bit behind me to do the same.

I have not seen anyone act more quickly. He turned round and ran as fast as his long legs could carry him. He was far ahead of me now and it looked he would not stop running. When I got up with him, hot and panting, he asked me what it was. It was an amusing incident I always remembered whenever I saw a wild tusker looking at me. In this case, the elephant may not even have seen us. He moved into the grass before reaching the bend and was not seen again.

Ornithology sometimes entails risks of a different kind. I remember an incident along the Himalayan trail from Almora to the Lipu Lekh Pass on my way to Lake Manasarover and Mount Kailas in 1945. It was at a highly risky spot. The porters pitched the tent there, though I had walked ahead of them. Just at moment a tiny bird, a Yellow-naped Yuhina, got up to the top of a bush, some yards away on the hillside.

Just as I got it in the field of my glasses, it hopped a bit further up. So, to get a better view, I took a step back, entirely unmindful of where I was standing with my back to the abyss. I felt a small pebble slip from under my heel and heard a faint continuing clatter as it went rolling down the hill. Still unmindful of any danger, I just looked back over my shoulder. What I saw terrified me. In a moment I realised I was on the very edge – two inches more and I would gone the way the pebble had gone.

I have found it very interesting to chase birds in pleasant places. I have watched birds for more than half a century mainly for the joy they give. It may be a form of escapism. But it does not need any justification.

Trills and Thrills: Birdwatching in India Summary in Malayalam

അവർ എന്നോട് ചോദിക്കുന്ന ചോദ്യങ്ങളിലൊന്ന് പക്ഷികൾക്കായി പര്യവേക്ഷണം ചെയ്യുന്നതിന്റെ ആവേ ശകരമായ സാഹസികതകളെക്കുറിച്ചാണ് എന്റെ ഉത്തരം പലർക്കും നിരാശാജനകമായി തോന്നാം. ഒരു ഹോബി അല്ലെങ്കിൽ ഒരു ഇഷ്ടമുള്ള കാര്യം എന്ന നിലയിൽ പക്ഷിശാസ്ത്രം സാഹസികതകളും പ്രതി ഫലങ്ങളും നിരാശകളും നിറഞ്ഞതാണ്. ഇത് ഏറ്റവും സമാധാനപരമായ പ്രവർത്തനങ്ങളിൽ ഒന്നാണ്. എന്നാലും ഇത് ആവേശവും രസവും ഉള്ളതാണ്.

ഇന്ത്യയിലെ പക്ഷികളുടെ സമ്പന്നമായ വൈവിധ്യം കാരണം ഏതൊരു പക്ഷിനിരീക്ഷകനും ആവേശക രമായ കണ്ടെത്തലുകൾ നടത്താൻ കഴിയും. ഒരു ഹോബിയായി പക്ഷിനിരീക്ഷണം വളർന്നുകൊണ്ടിരി ക്കുകയാണെന്ന് കാണുന്നത് നവോന്മേഷദായകമാണ്.

പക്ഷികളുടെ കൂടുകെട്ടൽ ശീലങ്ങളെക്കുറിച്ചുള്ള പരമ്പരാഗത വിവരണങ്ങൾ മാത്രം കേട്ടാണ് ഞാൻ വളർന്നത്. എന്നെപ്പോലുള്ള ഒരു ഫോട്ടോഗ്രാഫർക്ക് സിനിമയിൽ റെക്കോർഡു ചെയ്യാൻ ഇവ രസകരമാ യിരുന്നു. കൂടുകളിൽ നിന്ന് ഏതാനും യാർഡുകൾ അകലെ. ഒരു കാൻവാസ് മറവിൽ, സ്റ്റെപ് ഏണിയിൽ ഉയരത്തിൽ ഇരിക്കുമ്പോൾ, പക്ഷികൾ പാഠപുസ്തകങ്ങൾ വായിച്ചിട്ടില്ലെന്ന് കാണിക്കുന്ന ചില അസാ ധാരണ സംഭവങ്ങൾ ഞാൻ ശ്രദ്ധിച്ചു.

പക്ഷികളെക്കുറിച്ചുള്ള പഠന മേഖല സമാധാനപരമായ ഒന്നാണ്. പക്ഷേ അതിന് അതിന്റേതായ അപക ടസാധ്യതകളുണ്ട്. ദക്ഷിണേന്ത്യയിലെ ആനകൾ നിറഞ്ഞ കാടുകളിൽ, ഞാൻ പലപ്പോഴും അസ്വസ്ഥ മായ സാഹചര്യങ്ങളിൽ അകപ്പെട്ടിട്ടുണ്ട്. ഒരു കാട്ടാന പെട്ടെന്ന് നിങ്ങളുടെ അടുത്തേക്ക് വരുന്നത് അസ്വ സ്ഥതയുണ്ടാക്കുന്ന ഒരു അനുഭവമാണ്. കർണാടക, ക, കേരള സംസ്ഥാനങ്ങളുടെ ചില ഭാഗങ്ങളിൽ ഇത് അസാധാരണമല്ല. വാസ്തവത്തിൽ, ഒരു തെമ്മാടിയോ ചെറിയ കുട്ടിയാനയോ ഉള്ള ഒരു പെൺആനയോ അല്ലെങ്കിൽ, ആനയിൽ നിന്ന് ഒരു അപകടവും ഉണ്ടാകില്ല.

കൊച്ചിൻ ഫോറസ്റ്റ് ട്രാംവേയിലൂടെയുള്ള എന്റെ യാത്ര 1946 ഫെബ്രുവരിയിലായിരുന്നു. അതിനുശേഷം താമസിയാതെ അത് പറമ്പിക്കുളം ജലവൈദ്യുത പദ്ധതിക്കായി പൊളിച്ചുമാറ്റി. ആ വലിയ അണക്കെട്ടും ജലസംഭരണിയും പറമ്പിക്കുളത്തെയും ചുറ്റുമുള്ള എല്ലാ മനോഹരമായ സ്ഥലങ്ങളെയും മുക്കിക്കളഞ്ഞു. ഇരുട്ട് വീഴാൻ തുടങ്ങിയപ്പോഴാണ് ഞങ്ങൾ പറമ്പിക്കുളത്ത് എത്തിയത്. വനബംഗ്ലാവിന്റെ വരാന്തയിൽ കാടർ ആദിവാസികളുടെ ഒരു വലിയ കൂട്ടം ഉണ്ടായിരുന്നു. അവരുടെ ഇടയിൽ കാക്കി ഷോർട്ട്സ് ധരിച്ച നഗ്നപാദനായ ഒരു യൂറോപ്യൻ പുരുഷൻ ഉണ്ടായിരുന്നു. അയാൾ ഷർട്ട് ധരിച്ചിരുന്നില്ല.

അദ്ദേഹം ഒരു ഓസ്ട്രിയൻ നരവംശശാസ്ത്രജ്ഞൻ -ബാരമൺ ഒമർ റോൾഫ് ആയിരുന്നു. പക്ഷിശാസ്ത്രം തന്റെ രണ്ടാമത്തെ പ്രണയമാണെന്ന് അദ്ദേഹം പറഞ്ഞു. പിറ്റേന്ന് രാവിലെ എന്നോടൊപ്പം വരാൻ അദ്ദേഹം ആഗ്രഹിച്ചു. ഒരു ഫോറസ്റ്റ് ഗാർഡിന്റെ നേതൃത്വത്തിൽ, 5 അടിയിലധികം ഉയരമുള്ള ഇടതൂർന്ന പുല്ലുക ളുള്ള പുൽമേടിലൂടെയുള്ള മൃഗപാതയിലൂടെ ഞങ്ങൾ നടക്കുകയായിരുന്നു. ബാഡ് ടെയിൽഡ് ഗ്രാസ് വാർബ്ലർ എന്ന പക്ഷിക്ക് അനുയോജ്യമായ ആവാസവ്യവസ്ഥയായിരുന്നു അത്.

പാതയിലെ ഒരു വളവിൽ, ഫോറസ്റ്റ് ഗാർഡ് പെട്ടെന്ന് നിന്നു. അയാൾ ആവേശത്തോടെ മുന്നിലേക്ക് കൈ ചൂണ്ടി. എതിർദിശയിൽ നിന്ന് ഒരു ആന അതേ പാതയിലൂടെ വരുന്നത് ഞാൻ കണ്ടു. കഴിയുന്നത്ര വേഗ ത്തിൽ ഓടിപ്പോകാൻ ഞാൻ തിരിഞ്ഞു. എന്റെ പിന്നിലായിരുന്ന ഒമറിനോട് അങ്ങനെ ചെയ്യാൻ ഞാൻ ആംഗ്യം കാണിച്ചു.

ഇത്രയും വേഗത്തിൽ ഓടുന്ന മറ്റാരെയും ഞാൻ കണ്ടിട്ടില്ല. അയാൾ തിരിഞ്ഞു നോക്കി. അയാളുടെ നീണ്ട കാലുകൾക്ക് താങ്ങാൻ കഴിയുന്നത്ര വേഗത്തിൽ ഓടി. അയാൾ ഇപ്പോൾ എന്നെക്കാൾ വളരെ മുന്നിലായിരുന്നു. അയാൾ ഓട്ടം നിർത്തില്ലെന്ന് തോന്നി. ഞാൻ അയാളുടെ അടുത്ത് എത്തുമ്പോൾ എനിക്ക് ചൂടും ശ്വാസമുട്ടലും അനുഭവപ്പെട്ടു. ഒരു കാട്ടാന എന്നെ നോക്കുന്നത് കാണുമ്പോഴെല്ലാം ഞാൻ എപ്പോഴും ഓർക്കുന്ന രസകരമായ ഒരു സംഭവമായിരുന്നു അത്. ഈ സാഹചര്യത്തിൽ ആന ഞങ്ങളെ കണ്ടിട്ടുണ്ടാകില്ല. വളവിൽ എത്തുന്നതിനു മുമ്പ് അവൻ പുല്ലിലേക്ക് നീങ്ങി. പിന്നെ കണ്ടില്ല.

പക്ഷിശാസ്ത്രം ചിലപ്പോൾ വ്യത്യസ്തമായ ഒരു തരത്തിലുള്ള അപകടസാധ്യതകൾ വരുത്തിവയ്ക്ക ന്നു. 1945-ൽ മാനസസരോവർ തടാകത്തിലേക്കും കൈലാസ് പർവതത്തിലേക്കുമുള്ള എന്റെ യാത്രാ മധ്യേ അൽമോറയിൽ നിന്ന് ലിപുലേഖ് ചുരത്തിലേക്കുള്ള ഹിമാലയൻ പാതയിലൂടെ നടക്കുമ്പോൾ ഉണ്ടായ ഒരു സംഭവം ഞാൻ ഓർക്കുന്നു. അത് വളരെ അപകടകരമായ ഒരു സ്ഥലത്തായിരുന്നു. ഞാൻ അവരുടെ മുന്നിൽ ഉണ്ടായിരുന്നെങ്കിലും, ചുമട്ടുതൊഴിലാളികൾ അവിടെ ഒരു കൂടാരം കെട്ടി. നിമിഷ ങ്ങൾക്കുള്ളിൽ ഒരു ചെറിയ പക്ഷി, മഞ്ഞനിറമുള്ള നൂഹിന, കുന്നിൻ ചെരുവിൽ കുറച്ച് യാർഡ് അകലെ യുള്ള ഒരു കുറ്റിക്കാട്ടിൽ കാണപ്പെട്ടു.

എന്റെ റ്റെലസ്കോപ്പിന്റെ പരിധിയിൽ അതിനെ കിട്ടിയപ്പോൾ, അത് കുറച്ചുകൂടി മുകളിലേക്ക് പോയി, അതിനാൽ, മികച്ച കാഴ്ചലഭിക്കാൻ, ഞാൻ ഒരു പടി പിന്നോട്ട് മാറി, അഗാധത്തിലേക്ക് പുറം തിരിഞ്ഞാണ് ഞാൻ നിൽക്കുന്നത്. എവിടെയാണെന്ന് പൂർണ്ണമായും ഞാൻ ശ്രദ്ധിച്ചില്ല. എന്റെ കുതികാൽക്കടിയിൽ നിന്ന് ഒരു ചെറിയ കല്ല് വഴുതി വീഴുന്നതായി എനിക്ക് തോന്നി, അത് കുന്നിൻ മുകളിൽ നിന്ന് താഴേക്കു ഉരുളുമ്പോൾ ഒരു നേരിയ തുടർച്ചയായ ശബ്ദം കേട്ടു. അപകടത്തെക്കുറിച്ച് ഇപ്പോഴും ഒന്നും ചിന്തിക്കാ തെ, ഞാൻ ഒന്ന് തിരിഞ്ഞുനോക്കി, ഞാൻ കണ്ടത് എന്നെ ഭയപ്പെടുത്തി. ഒരു നിമിഷം കൊണ്ട് ഞാൻ അപകടത്തിന്റെ വളരെ അരികിലാണെന്ന് എനിക്ക് മനസ്സിലായി. രണ്ട് ഇഞ്ച് കൂടി ഞാൻ പുറകോട്ടുപോ യെങ്കിൽ കല്ല് പോയ വഴിക്ക് ഞാനും അഗാധമായ ഗർത്തത്തിലേക്ക് പതിക്കുമായിരുന്നു.

സുഖകരമായ സ്ഥലങ്ങളിൽ പക്ഷികളെ നിരീക്ഷിക്കുന്നത് എനിക്ക് വളരെ രസകരമായി തോന്നി. അരനൂ റ്റാണ്ടിലേറെയായി ഞാൻ പക്ഷികളെ നിരീക്ഷിച്ചത് പ്രധാനമായും അവ നൽകുന്ന സന്തോഷത്തിനുവേ ണ്ടിയാണ്. അത് ഒരുതരം രക്ഷപ്പെടലായിരിക്കാം. പക്ഷേ അതിന് ഒരു ന്യായീകരണവും ആവശ്യമില്ല.

Class 10 English Trills and Thrills: Birdwatching in India by Salim Ali About the Author

Salim Ali (1896 – 1987) was an Indian ornithologist and naturalist. He is known as the “Birdman of India”. He was the first to conduct a systematic survey of birds in India. He has written many books on ornithology. He co- authored the famous book “Handbook of the Birds in India and Pakistan” with Sidney Dillon Ripley. Salim Ali was honoured with Padma Bhushan in 1958 and Padma Vibhushan in 1976. Several special kind of fauna like Salim Ali’s Fruit Bat and Salim Ali’s Dwarf Gecko are named after him. Two bird sanctuaries and institutions are named after him.

രചയിതാവിനെക്കുറിച്ച്

സലിം അലി (1896 – 1987) ഒരു ഇന്ത്യൻ പക്ഷിശാസ്ത്രജ്ഞനും പ്രകൃതി ശാസ്ത്രജ്ഞനുമായിരു ന്നു. “ഇന്ത്യയുടെ പക്ഷിമനുഷ്യൻ” എന്ന് അദ്ദേഹം അറിയപ്പെടുന്നു. ഇന്ത്യയിൽ പക്ഷികളെക്കു റിച്ച് ആദ്യമായി ഒരു വ്യവസ്ഥാപിത സർവേ നടത്തിയ വ്യക്തിയാണ് അദ്ദേഹം. പക്ഷിശാ സത്തെക്കുറിച്ച് അദ്ദേഹം നിരവധി പുസ്തകങ്ങൾ എഴുതിയിട്ടുണ്ട്. സിഡ്നി ഡില്ലൺ റിപ്ലിയു മായി സഹകരിച്ച് “ഹാൻഡ് ബുക്ക് ഓഫ് ദ് ബേർഡ്സ് ഇൻ ഇന്ത്യ ആൻഡ് പാകിസ്ഥാൻ” എന്ന പ്രശസ്ത പുസ്തകത്തിന്റെ രചയിതാവാണ് അദ്ദേഹം. 1958-ൽ പത്മഭൂഷണും 1976-ൽ പത്മവിഭൂ ഷണം നൽകി സലിം അലിയെ രാജ്യം ആദരിച്ചു. സലിം അലിയുടെ ഫ്രൂട്ട് ബാറ്റ്. സലിം അലി യുടെ കുള്ളൻ ഗെക്കോ തുടങ്ങിയ നിരവധി പ്രത്യേക ജന്തുജാലങ്ങൾക്ക് അദ്ദേഹത്തിന്റെ പേരാണ് നൽകിയിരിക്കുന്നത്. രണ്ട് പക്ഷി സങ്കേതങ്ങൾക്കും സ്ഥാപനങ്ങൾക്കും അദ്ദേഹത്തിന്റെ പേരാണ് നൽകിയിരിക്കുന്നത്.

Class 10 English Trills and Thrills: Birdwatching in India Vocabulary

glimpse – look, നോട്ടം
exotic – non-native, അസാധാരണമായ
species – a group of living organisms consisting of similar individuals, ഇനം, വർഗം
habitat – natural home, ആവാസവ്യസ്ഥ
trills – quavering or vibratory sounds, വിറയലുളള ശബ്ദങ്ങൾ
thrills – excitements, ആവേശം
ornithology – scientific study of birds, പക്ഷിശാസ്ത്രം
pursuits – activities of specified kinds for recreation, ഉദ്യമം
rapidly – quickly, വേഗത്തിൽ
accounts – descriptions, വിവരണങ്ങൾ
concealed – hidden, ഒളിപ്പിച്ചു വച്ച
hide – skin, തൊലി
perched – sat, ഇരുന്നു
hazards – risks, അപകടങ്ങൾ
for instance – for example, ഉദാഹരണത്തിന്
frequently – often, ഇടക്കിടക്ക്
at close quarters – very near, വളരെ അടുത്ത്
unnerving – fearful, പേടിപ്പെടുത്തുന്ന
rogue – rascal, rowdy, തെമ്മാടി
romantic – picturesque, beautiful, നല്ല ഭംഗിയുള്ള
tramway – a set of rails which forms the route for a tram, ട്രാം മാത
tram – a passenger vehicle, like a small train, running on rails, ട്രാം, ചെറിയ തീവണ്ടി പോലെയുള്ള റെയിലിൽ ഓടുന്ന വണ്ടി
dismantled – broken up and removed, പൊളിച്ചു മാറ്റി
gigantic – huge, വലിയ
reservoir – where water is collected, വെള്ള സംഭരണി
submerged – covered by water, വെള്ളത്തിൽ മുങ്ങിപ്പോകുക
Kadar adivasis – a group of adivasis known as “Kadar”, കാടർ ആദിവാസികൾ
topless – wearing nothing on the upper part of the body, ഷർട്ടില്ലാതെ
crumpled – crushed to form wrinkles and creases, ഇസ്തിരിയിടാത്ത
anthropologist – a scientist who studies humankind and human culture, നരവംശശാസ്ത്രജ്ഞൻ
keenness – interest, താൽപ്പര്യം
stalking – walking stealthily, ഒളിച്ചു നടക്കുക
trail – narrow path, ഇടുങ്ങിയ വഴി
Grass Warbler – a kind of small bird ഒരു തരം പക്ഷി, പുൽക്കുരുവി
ducked – turned quickly, കുനിഞ്ഞു
glimpsed – saw, കണ്ടു
tusker – a male elephant, കൊമ്പനാന
striding – walking majestically രാജാവിനെപ്പോലെ നടക്കുക
agility – speed, വേഗത
sprinted – ran, ഓടി
panting – breathing with short, quick breaths; out of breath, വീർപ്പുമുട്ടൽ
amusing – interesting, താൽപര്യം ജനിപ്പിക്കുന്ന
veered – changed direction suddenly, പെട്ടെന്ന് ദിശ മാറ്റുക.
entail – become necessary, അനിവാര്യമാക്കുക
scarp – cliff, steep slope കുത്തനെയുള്ള ചരിവ്
pitched – put, placed, കെട്ടി, വച്ചു
Yellow-naped Yuhina – a kind of bird, മഞ്ഞപ്പിടലി മരംകൊത്തി
flanking – to be on the side, ഒരു സൈഡിലായിരിക്കുക
hopped – leaped, ചാടി
abyss – a deep, immeasurable space, gulf, or cavity, ഗർത്തം
clatter – noise of things when moving ശബ്ദം
untoward – something bad, അനിഷ്ടകരമായ
made my hair stand on head – terrified, shocked, horrified പേടിക്കുമ്പോൾ ഉണ്ടാകുന്ന പ്രതിഭാസം
rollicking – exuberantly lively and amusing, രസാവഹമായ
elation – happiness, സന്തോഷം
afforded – gave, തന്നു
escapism – running away from reality സത്യത്തിൽ നിന്നും ഒളിച്ചോടൽ
justification – give a reason or explanation for doing something ന്യായീകരണം

Parents often use SCERT Class 10 English Solutions Unit 4 Chapter 1 Shakuntalam Questions and Answers Activities Notes Pdf to assist their kids with homework.

Class 10 English Shakuntalam Question Answer Notes Kerala Syllabus

Std 10 English Shakuntalam Question Answer

Class 10 English Unit 4 Question Answer

INTRODUCTION (ആമുഖം)

This Unit has a play, an autobiography and a poem. At the beginning of the Unit there are four lines taken from William Wordsworth’s poem “The Tables Turned”. The lines mean that a single moment or experience in a fresh, springtime forest can teach us more about humanity, morality, and life’s profound truths than all the wisdom found in the books and all the teachings scholars can give us. William Wordsworth was a great lover of Nature and he is called “The High Priest of Nature”.

ഈ യൂണിറ്റിൽ ഒരു നാടകം, ഒരു ആത്മകഥ, ഒരു കവിത എന്നിവയുണ്ട്. യൂണിറ്റിന്റെ തുടക്കത്തിൽ വില്യം വേഡ്സ്വർത്തിന്റെ “ദ ടേബിൾസ് ടേൺഡ്” എന്ന കവിതയിൽ നിന്ന് എടുത്ത നാല് വരികളുണ്ട്. പുസ്തകങ്ങളിൽ കാണുന്ന എല്ലാ ജ്ഞാനത്തേക്കാളും പണ്ഡിതന്മാർ നൽകുന്ന എല്ലാ പഠിപ്പിക്കലുക ളേക്കാളും മനുഷ്യത്വം, ധാർമ്മികത, ജീവിതത്തിന്റെ ആഴമേറിയ സത്യങ്ങൾ എന്നിവയെക്കുറിച്ച് ഒരു പുതിയ വസന്തകാല വനത്തിലെ ഒരു നിമിഷം അല്ലെങ്കിൽ അനുഭവം നമ്മെ പഠിപ്പിക്കും എന്നാണ് വരി കൾ അർത്ഥമാക്കുന്നത്. വില്യം വേർഡ്സ്വർത്ത് പ്രകൃതിയുടെ വലിയ സ്നേഹിയായിരുന്നു. അദ്ദേഹത്തെ “പ്രകൃതിയുടെ മഹാപുരോഹിതൻ” എന്ന് വിളിക്കുന്നു.

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Read the 4 lines from the poem “On the Grasshopper and the Cricket” by John Keats, given on p. 120 of your text.
ജോൺ കീറ്റ്സിന്റെ “ഓൺ ദ് ഗ്രാസ് ഹോപ്പർ ആൻഡ് ദ് ക്രിക്കറ്റ്” എന്ന കവിതയിലെ 4 വരികൾ വായിക്കുക.

Answers to the questions on p. 120:
നിങ്ങളുടെ പുസ്തകത്തിലെ 120-ാം പേജിലെ ചോദ്യങ്ങൾക്കുള്ള ഉത്തരങ്ങൾ

Question 1.
What in your opinion is the “poetry of earth”?
നിങ്ങളുടെ അഭിപ്രായത്തിൽ “ഭൂമിയുടെ കവിത” എന്താണ്?
Answer:
The “poetry of earth” is the singing of the grasshopper, hopping from hedge to hedge in the newly mown meadows. His song represents the vibrant, joyful music of nature, which flourishes even when birds are silenced by the climate. His song proves that the “poetry of earth is never dead.”
“ഭൂമിയുടെ കവിത” എന്നത് പുൽമേടുകളിൽ ചെടികളിൽ നിന്ന് ചെടികളിലേക്കു ചാടി നടക്കുന്ന വെട്ടുക്കിളിയുടെ സംഗീതാലാപനമാണ്. അവയുടെ സംഗീതം പ്രകൃതിയുടെ ഊർജ്ജസ്വലമായ, ആഹ്ലാദകരമായ സംഗീതത്തെ പ്രതിനിധീകരിക്കുന്നു. അത് കാലാവസ്ഥയാൽ പക്ഷികൾ നിശബ്ദ മാകുമ്പോഴും തഴച്ചുവളരുന്നു. “ഭൂമിയുടെ കവിത ഒരിക്കലും മരിക്കുന്നില്ല” എന്ന് അവയുടെ ഗാനം തെളിയിക്കുന്നു.

Question 2.
List a few sounds of nature that you love to listen to:
നിങ്ങൾ കേൾക്കാൻ ഇഷ്ടപ്പെടുന്ന പ്രകൃതിയുടെ കുറച്ച് ശബ്ദങ്ങൾ പട്ടികപ്പെടുത്തുക.
a) The singing of birds
പക്ഷികളുടെ ആലാപനം

b) The rustling of leaves in the wind
കാറ്റിൽ ഇലകൾ കൂട്ടിയുരയുന്ന ശബ്ദം

c) The rain drops drumming on the roof
മഴത്തുള്ളികൾ മേൽക്കൂരയിൽ താളത്തിൽ വീഴുന്നത്.

d) The sound of waves crashing on the shore
തീരത്ത് തിരമാലകൾ ആഞ്ഞടിക്കുന്ന ശബ്ദം

Question 3.
Isn’t nature a great influence on our lives? Have you ever felt a deep connection with nature? You may describe your experience.
പ്രകൃതി നമ്മുടെ ജീവിതത്തിൽ വലിയ സ്വാധീനം ചെലുത്തുന്നില്ലേ? പ്രകൃതിയുമായി ആഴ ത്തിലുള്ള ബന്ധം നിങ്ങൾക്ക് എപ്പോഴെങ്കിലും തോന്നിയിട്ടുണ്ടോ? നിങ്ങളുടെ അനുഭവം വിവരി ക്കുക.
Answer:
Nature certainly has a great influence on our lives. It is nature that gives us air to breathe, water to drink and the sunlight during the day. I have always felt a deep connection with nature. I enjoy the sunrise and the sunset, the full moon and the star-studded sky, the twittering butterflies, the blooming flowers, the chirping birds and the wonderful animals around. To me nature is like a mother who nurtures us and makes us happy with its myriad sights and sounds. If we pollute nature it will end in calamity.
പ്രകൃതിക്ക് തീർച്ചയായും നമ്മുടെ ജീവിതത്തിൽ വലിയ സ്വാധീനമുണ്ട്. ശ്വസിക്കാൻ വായുവും കുടിക്കാൻ വെള്ളവും പകൽ സമയത്ത് സൂര്യപ്രകാശവും നൽകുന്നത് പ്രകൃതിയാണ്. പ്രകൃതിയു മായി ആഴത്തിലുള്ള ബന്ധം എനിക്ക് എപ്പോഴും തോന്നിയിട്ടുണ്ട്. സൂര്യോദയവും സൂര്യാസ്തമ യവും, പൂർണ്ണ ചന്ദ്രനും നക്ഷത്രനിബിഡമായ ആകാശവും പാറിപ്പറക്കുന്ന ചിത്രശലഭങ്ങളും, വിരി യുന്ന പൂക്കളും, ചിലമ്പുന്ന പക്ഷികളും ചുറ്റുമുള്ള അത്ഭുതകരമായ മൃഗങ്ങളുടേയും കാഴ്ച ഞാൻ ആസ്വദിക്കുന്നു. എന്നെ സംബന്ധിച്ചിടത്തോളം പ്രകൃതി നമ്മെ പോറ്റിവളർത്തുന്ന ഒരു അമ്മയെ പ്പോലെയാണ്, അതിന്റെ അസംഖ്യം കാഴ്ചകളും ശബ്ദങ്ങളും നമ്മെ സന്തോഷിപ്പിക്കുന്നു. നമ്മൾ പ്രകൃതിയെ മലിനമാക്കിയാൽ അത് ദുരന്തത്തിൽ അവസാനിക്കും.

Let’s read and Enjoy

Introduction:
The clear blue sky, sparkling rivers and fragrant flowers speak to us of the joy and peace in nature. We should live in harmony with nature. If we break our bond with nature it will end in disaster. Our ancestors loved and respected nature and co-existed with it.

തെളിഞ്ഞ നീലാകാശവും തിളങ്ങുന്ന നദികളും സുഗന്ധമുള്ള പൂക്കളും പ്രകൃതിയിലെ സന്തോഷത്തെയും സമാധാനത്തെയും കുറിച്ച് നമ്മോട് സംസാരിക്കുന്നു. പ്രകൃതിയോട് ഇണങ്ങി ജീവിക്കണം. പ്രകൃതിയു മായുള്ള നമ്മുടെ ബന്ധം വിച്ഛേദിച്ചാൽ അത് ദുരന്തത്തിൽ അവസാനിക്കും. നമ്മുടെ പൂർവ്വികർ പ്രക തിയെ സ്നേഹിക്കുകയും ബഹുമാനിക്കുകയും ചെയ്തു.

10th Class English Shakuntalam Question Answer – Textual Questions And Answers

a) “Fear shrinks to half the body small…” explain.
“ഭയം ശരീരത്തെ പകുതിയായി ചുരുക്കുന്നു…” വിശദീകരിക്കുക.
Answer:
It means that when fear grips something or someone, its/his body becomes smaller in
size.
എന്തിനെയെങ്കിലും അല്ലെങ്കിൽ ആരെയെ ങ്കിലും ഭയം പിടികൂടുമ്പോൾ, അതിന്റെ അവന്റെ ശരീരം ചെറുതായിത്തീരുന്നു എന്നാണ് ഇതിനർത്ഥം.

b) “I can hardly keep him in sight.” Why does the king say so?
“എനിക്ക് അവനെ കാണാൻ കഴിയില്ല.” എന്തുകൊണ്ടാണ് രാജാവ് അങ്ങനെ പറയുന്നത്?
Answer:
The king says so because the deer jumps so often and so high that it looks as if it is flying and not running. So he can’t keep the deer in sight.
മാൻ പലപ്പോഴും ചാടുകയും ഉയരത്തിൽ പായു കയും ചെയ്യുന്നതിനാലാണ് രാജാവ് അങ്ങനെ പറയുന്നത്. അതുകൊണ്ട് മാനിനെ ശരിക്കു കാണാൻ അദ്ദേഹത്തിന് കഴിയുന്നില്ല.

c) What is the tender form of the deer compared to?
മാനിന്റെ മൃദുവായ രൂപം എന്തുമായിട്ടാണ് താരതമ്യം ചെയ്യുന്നത്?
Answer:
The tender form of the deer is compared to a blossom, a flower.
മാനിന്റെ മൃദുവായ രൂപത്തെ ഒരു പുഷ്പത്തോ ടാണ് താരതമ്യം ചെയ്യുന്നത്.

d) What is the hermit’s request to Dushyanta?
ദുഷ്യന്തനോട് സന്യാസിയുടെ അഭ്യർത്ഥന എന്താണ്?
Answer:
The hermit’s request to Dushyanta is not to hunt the deer and put back his arrow in the quiver.
മാനിനെ വേട്ടയാടാതെ തന്റെ അസ്ത്രം ആവ നാഴിയിൽ തിരികെ വെയ്ക്കണമെന്നാണ് ദുഷ്യ ന്തനോടുള്ള സന്യാസിയുടെ അപേക്ഷ.

e) The hermits refer to Dushyanta as ” …..a shining example of kings.” Why?
സന്യാസിമാർ ദുഷ്യന്തനെ “രാജാക്കന്മാ രുടെ ഉജ്ജ്വല ഉദാഹരണം.” എന്നാണ് വിശേ ഷിപ്പിക്കുന്നത് എന്തിന്?
Answer:
The hermits refer to Dushyanta as ” …..a shining example of kings” because he obeyed the hermit’s instruction and put the arrow back in its quiver.
സന്യാസിമാർ ദുഷ്യന്തനെ “രാജാക്കന്മാരുടെ ഉജ്ജ്വല ഉദാഹരണം.” എന്ന് വിളിക്കുന്നു. കാരണം അവൻ സന്യാസിയുടെ നിർദ്ദേശം അനുസരിക്കുകയും അമ്പ് അതിന്റെ ആവനാ ഴിയിൽ തിരികെ വയ്ക്കുകയും ചെയ്തു.

f) What aspect of the king’s character is revealed when he says, “We must not disturb the hermitage”.
“നമ്മൾ ആശ്രമത്തെ ശല്യപ്പെടുത്തരുത് എന്ന് പറയുമ്പോൾ രാജാവിന്റെ സ്വഭാവ ത്തിന്റെ ഏത് വശമാണ് വെളിപ്പെടുന്നത്?
Answer:
When he says, “We must not disturb the hermitage”, it reveals the gentle nature and humility of the king. Although he is the king he is willing to obey the instruction of the hermit.
നാം ആശ്രമത്തെ ശല്യപ്പെടുത്തരുത്” എന്ന് അദ്ദേഹം പറയുമ്പോൾ അത് രാജാവിന്റെ സൗമ്യ സ്വഭാവവും വിനയവും വെളിപ്പെടു ത്തുന്നു. അദ്ദേഹം രാജാവാണെങ്കിലും സന്യാ സിയുടെ നിർദ്ദേശം അനുസരിക്കാൻ അദ്ദേഹം തയ്യാറാണ്.

g) Whose voice does the king hear, to the right of the grove?
തോട്ടത്തിന്റെ വലതു വശത്ത് രാജാവ് ആരുടെ ശബ്ദമാണ് കേൾക്കുന്നത്?
Answer:
To the right of the grove, the king hears the voice of Shakuntala.
തോപ്പിന്റെ വലതുവശത്ത് രാജാവ് ശകുന്തള യുടെ ശബ്ദമാണ് കേൾക്കുന്നത്.

h) ‘…we shall not be working for a reward,’ says Priyamvada. What does she mean by this?
…..ഞങ്ങൾ പ്രതിഫലത്തിനായി പ്രവർത്തി ക്കില്ല; പ്രിയംവദ പറയുന്നു. അവൾ എന്താണ് ഇത് കൊണ്ട് ഉദ്ദേശിക്കുന്നത്?
Answer:
By this she means by watering those trees whose flowering time is past, they are not working for a reward. It is a better deed because it is done out of love and not for any reward.
ഇതിലൂടെ അവൾ അർത്ഥമാക്കുന്നത് പൂവി ടുന്ന സമയം കഴിഞ്ഞുപോയ വൃക്ഷങ്ങൾക്ക് വെള്ളം നനയ്ക്കുക എന്നതാണ്. അവർ അപ്പോൾ പ്രതിഫലത്തിനായിട്ടല്ല പ്രവർത്തിക്കു ന്നത്. ഒരു പ്രതിഫലത്തിനും വേണ്ടിയല്ല. സ്നേഹം കൊണ്ടാണ് ഇത് ചെയ്യുന്നതെന്നതി നാൽ ഇത് ഒരു മികച്ച കർമ്മമാണ്.

i) Shakuntala feels that the mango tree is trying to tell her something. Do you think nature communicates with us? How?
മാവ് തന്നോട് എന്തോ പറയാൻ ശ്രമിക്കുന്ന തായി ശകുന്തളയ്ക്ക് തോന്നുന്നു. പ്രകൃതി നമ്മോട് ആശയവിനിമയം നടത്തുമെന്ന് നിങ്ങൾ കരുതുന്നുണ്ടോ? എങ്ങനെ?
Answer:
I do think that nature communicates with us. She does that through signs. When we see dark clouds in the sky we know it will rain.
പ്രകൃതി നമ്മോട് ആശയവിനിമയം നടത്തു മെന്ന് ഞാൻ കരുതുന്നു. അവൾ അത് അടയാ ളങ്ങളിലൂടെ ചെയ്യുന്നു. ആകാശത്ത് ഇരുണ്ട മേഘങ്ങൾ കാണുമ്പോൾ മഴ പെയ്യുമെന്ന് നമു
ക്കറിയാം.

j) The jasmine vine, the mango tree and Shakuntala blend in a unified image. Have you come across works of art in which humans are depicted in harmony with nature? List a few that you remember.
മുല്ലപ്പൂവ്, മാമ്പഴം, ശകുന്തള എന്നിവ ഒരു ഏകീകൃത ചിത്രത്തിൽ കൂടിച്ചേരുന്നു. പ്രക തിയുമായി ഇണങ്ങി മനുഷ്യരെ ചിത്രീകരി ക്കുന്ന കലാസൃഷ്ടികൾ നിങ്ങൾ കണ്ടി ട്ടുണ്ടോ? നിങ്ങൾ ഓർക്കുന്ന ചിലത് പട്ടിക പ്പെടുത്തുക.
Answer:
A few works of art in which humans are depicted in harmony with nature that I remember are:
മനുഷ്യനെ പ്രകൃതിയുമായി ഇണങ്ങി ചിത്രീ കരിച്ചിരിക്കുന്ന ചില കലാസൃഷ്ടികൾ ഞാൻ ഓർക്കുന്നു.

a) Kerala Mural Paintings: They are found mainly in temples. Although they may feature gods and goddesses, a key characteristic is the integration of human and divine figures with lush natural elements.
കേരള മ്യൂറൽ പെയിന്റിംഗുകൾ അവ പ്രധാ നമായും ക്ഷേത്രങ്ങളിലാണ് കാണപ്പെടുന്നത്. അവയിൽ ദേവന്മാരെയും ദേവതകളെയും ആണ് പ്രധാനമായി അവതരിപ്പിക്കുന്നതെങ്കി ലും, സമൃദ്ധമായ പ്രകൃതി ഘടകങ്ങളുമായി മനുഷ്യരുടെയും ദൈവീക രൂപങ്ങളുടെയും സമന്വയമാണ് അവയുടെ സ്വഭാവം.

b) “Harmonious life of nature and man,” by Aiswaria A.K. beautifully depicts farmers working in a paddy field. The artist’s statement highlights how agriculture is an “unavoidable part of our culture and traditions” and that farmers are a “true sign of harmony between man and nature.”
“പ്രകൃതിയുടെയും മനുഷ്യന്റെയും യോജിപ്പുള്ള ജീവിതം” എന്ന കലാസൃഷ്ടി യിൽ ഐശ്വര്യ എ. കെ. നെൽവയലിൽ ജോലി ചെയ്യുന്ന കർഷകരെ മനോഹരമായി ചിത്രീ കരിക്കുന്നു. കൃഷി എങ്ങനെ “നമ്മുടെ സംസ്കാരത്തിന്റെയും പാരമ്പര്യത്തിന്റെയും ഒഴിവാക്കാനാകാത്ത ഭാഗമാണ്” എന്നും കർഷ കർ “മനുഷ്യനും പ്രകൃതിയും തമ്മിലുള്ള യോജിപ്പിന്റെ യഥാർത്ഥ അടയാളം” ആണെന്നും കലാകാരിയുടെ പ്രസ്താവന എടുത്തുകാണിക്കുന്നു.

c) Tribal Art from the Western Ghats: The art of indigenous communities in the Western Ghats is a good example of humans in harmony with nature. Their art forms are often rooted in their daily life and their intimate knowledge of the forest.
പശ്ചിമഘട്ടത്തിൽ നിന്നുള്ള ഗോത്രകലകൾ ; പഞ്ചിമഘട്ടത്തിലെ തദ്ദേശീയ സമൂഹങ്ങളുടെ കല പ്രകൃതിയുമായി ഇണങ്ങി നിൽക്കുന്ന മനു ഷ്യരുടെ ഉത്തമ ഉദാഹരണമാണ്. അവരുടെ കലാരൂപങ്ങൾ പലപ്പോഴും അവരുടെ ദൈനം ദിന ജീവിതത്തിലും കാടിനെക്കുറിച്ചുള്ള അവ രുടെ അടുത്ത അറിവിലും വേരൂന്നിയതാണ്.

d) What unseasonal quality of the spring creeper does Shakuntala notice?
സിംഗ് കീപ്പറിന്റെ ഏത് കാലാനുസൃതമ ല്ലാത്ത ഗുണമാണ് ശകുന്തള ശ്രദ്ധിക്കുന്നത്?
Answer:
The unseasonal quality of the spring creeper that Shakuntala notices is the spring creeper covered with buds down to the very root. ശകുന്തള ശ്രദ്ധിക്കുന്ന സിംഗ് കീപ്പറിന്റെ കാലാനുസൃതമല്ലാത്ത ഗുണം വേരുവരെ മുകു ളങ്ങളാൽ അത് പൊതിഞ്ഞു നിൽക്കുന്നു എന്നതാണ്.

10th Std English Shakuntalam Question Answer Notes – Let’s Rewind

Question 1.
Describe the king’s change in attitude after meeting the hermits.
Answer:
King was planning to kill the deer and he was pursuing it with his arrow ready. But after meeting the hermits he changed his mind. He put his arrow back in the quiver. He even thanked the hermit for his blessing.

Question 2.
What would have prompted Sage Kanva to leave the hermitage under Shakuntala’s supervision?
Answer:
Sage Kanva’s desire to go to Somathirtha prompted him to leave the hermitage under Shakuntala’s supervision. Somathirtha is a sacred location where bathing is believed to bestow spiritual blessings and purification. Another reason could be that he wanted Dushyanta to meet Shakuntala in his absence so that they can mingle freely without his interference.

Question 3.
How does the king guess that there is a hermitage in the vicinity?
Answer:
The king guesses that there is a hermitage in the vicinity because he sees rice grains dropped from the bills of parrot chicks beneath the trees. There are also pounding stones on which a little almond oil sticks. Again there are trustful deer that do not run away even when the king approaches.

Question 4.
“Father Kanva cares more for the hermitage trees than he does for you,” says one of Shakuntala’s friends. How are the plants and trees treated by the inmates of the hermitage? Do we try to maintain this attitude to nature? Explain.
Answer:
The statement of Shankuntala’s friend “Father Kanva cares more for the hermitage trees than he does for you” shows how much he loves nature and its flora and fauna. Other inmates also take good care of the plants and trees there. We see them watering the plants and trees, even those whose flowering time is past. We do not have this kind of attitude to nature. Many of us pollute it instead of taking care of it.

Question 5.
Pick out instances from the play that show the harmonious coexistence of all beings.
Answer:
a) The care given by Father Kanva and other inmates to the plants and animals.
b) The jasmine creeping over the mango tree.
c) The deer that are not afraid of the human presence.
d) The parrot chicks dropping grains of rice from their bills.
e) The flowering and fruitioning of the plants and trees which help people and other creatures.

Shakuntalam Activities

10th English Shakuntalam Question Answer – Let’s recall and recreate:

Activity 1

Question 1.
Revisit the extract and identify the words of the character which express their love of nature, and complete the table given.


Answer:

Character	Dialogue	Context
Dushyanta	• His neck in beauty bends As backward looks he sends • Rice grains dropped from the bills of parrot chicks beneath the trees. Trustful deer that do not run away as we draw near.	• the attractive appearance of the deer • the surroundings of the hermitage
Hermits	• O King, this deer belongs to the hermitage and must not be killed. • Restore your arrow to the quiver.	• The king trying to shoot the deer with his arrow. • The arrow is in the hand of the king.
Shakuntala	• I feel like a real sister to them. • Oh, girls, that mango tree is trying to tell me something with his branches that move in the wind like fingers. • What a pretty pair they make. The jasmine shows her youth in her fresh flowers, and the mango tree shows his strength in ripening fruit.	• Her sisterly affection for the trees. • The branches of the mango tree moving in the wind. • The image of the young jasmine vine and the strong mango tree.
Anasuya	• It seems to me, dear, that Father Kanva cares more for the hermitage trees than he does for you. You are delicate as a jasmine blossom, yet he tells you to fill the trenches about the trees. • Oh, Shakuntala! Here is the jasmine vine that you named Light of the Grove. She has chosen the mango tree as her husband!	• Shakuntala watering trees. • The jasmine vine clinging to the mango tree.
Priyamvada	• Now let’s sprinkle those whose flowering time is past. That will be a better deed, because we shall not be working for a reward. • And I have something pleasant to tell you. You are to be married soon.	• Shakuntala waters trees. • Shakuntala looking like a jasmine vine.

Question 2.
King Dushyanta is a fearless warrior. He has many other qualities also. List instances from the play that reveal his traits of character.

Answer:

Instances from the play	Character traits
(a) And even now His neck in beauty bends ………………….	appreciates beauty
(b) The path he takes is strewed With blades of grass half chewed.	willingness to overcome difficulties
(c) Bows to the hermit and puts the arrow in humility its quiver and thanks him for his blessing.	humility
(d) Do you not see? Why, here are rice grains, dropped from the bills of parrot chicks Beneath the trees; and pounding stones where sticks A little almond oil; and trustful deer That do not run away as we draw near.	power of observation and logical thinking.
(e) We must not disturb the hermitage, Stop respect for others; good behaviour here while I dismount.	respect for others; good behaviour
(f) But the flattery is true, her arms are tender shoots; her lips are blossoms red and warm; bewitching youth begins to flower, in beauty on her form.	poetic quality; an enthusiastic lover

Now, develop a character sketch of the king.
Answer:
In the play we see King Dushyanta in a chariot chasing a deer with an arrow ready to shoot. Even then he appreciates the beauty of the deer. He loves hunting. But when a hermit tells him the deer he is chasing belongs to the hermitage and he should not kill the dear, the king readily puts the arrow back in its quiver. This shows his humility. He is a man of logic. When he sees the grains dropped from the bills of parrot chicks, the pounding stones having almond oil on them and the deer not running away when they approach, he guesses there is a hermitage nearby. He respects others and shows good behaviour. When he comes near the hermit – age, he dismounts from his chariot and walks toward it so that he does not disturb the calm and quiet of the place and the people. He has poetic qualities and we see this in his description of Shakuntala. He is a young man ready to fall in love with Shakuntala! On the whole he is a lovable person.

Question 3.
As part of a student exchange programme, you engage in a session on theatre with a group of class 10 learners from your neighbouring state. Prepare an introduc- tion to Shakuntalam for their benefit. You may highlight the theatrical techniques, the setting, characters and sequence of events.
Answer:
In this session, we are going to tell you something about the play “Shakuntalam” by Kalidasa. Kalidasa was a famous playwright who lived between the 4^th and 5^th century AD.

It is the story of King Dushyanta meeting Shakuntala and falling in love with her. Shakuntala is the daughter of Sage Kanva. She lives in the hermitage of Kanva. In the hermitage we find the beauty of nature at its best. We see trees and plants growing luxuriously, taken care of by Sage Kanva, Shakuntala and her friends – Anasuya and Priyamvada.

We see deer, parrots, bees and other things moving about freely in the garden. We see the beautiful jasmine vine creeping on the mango tree. They look like husband and wife. The beautiful jasmine vine is the wife and the strong mango tree is the husband.

In the beginning of the play we see Dushyanta chasing a deer. He is ready to shoot it with his arrow when a hermit tells him not to shoot it, as it belongs to the hermitage. The king obeys the hermit and puts his arrow back in is quiver. Dushyanta does not want to disturb the calm and quiet of the hermitage. He dismounts from his chariot and walks towards the hermitage. He sees Shakuntala and her friends Anasuya and Priyamvada watering the trees and plans, and chitchatting in a pleasant way. Dushyanta is enamoured by the beauty of Shakuntala and falls in love with her. It is a story of love at first sight.

Question 4.
Inspired by the blissful coexistence with nature displayed by the characters in the play Shakuntalam, a group of nature lovers set up an ecotourism project. )
a) Many tourists visit the ecotourism project. Certain guidelines are to be displayed at the entrance to the site. Discuss in groups and frame the guidelines. One has been done for you.
• Don’t litter your surroundings.
• ………………………………..
• ………………………………..
• ………………………………..
• ………………………………..
Answer:
• Stay on Marked Trails
• Observe Wildlife from a Distance.
• Minimize Noise and Group Size
• Take only pictures, nothing else!

(b) Entries are invited for the content of a blog on your ecotourism project, Develop the content for the blog.

A blog is a website or web page maintained by an individual or group. The content is usually informal and updated regularly.

What are the features of a blog?
• an attracgtive template
• cogent presentation of the content
• easy navigation
• space for comments
• ……………………………..
• ……………………………
Answer:
Features of a blog:
• An attractive template, a pre-designed framework that dictates the visual layout and the structural organization of the content.
• Clear, logical and convincing presentation.
• Easy navigation.
• Space for comments.
• Simple, straightforward language.
• Purposeful – giving information, giving a lesson, telling something new.

(c) Now, discuss in groups and develop the content of the blog on the ecotourism project.

Answer:
WOODLAND WONDERS
I want to tell you about my Ecotourism Project! Before embarking upon this project, I thought “tourism” just meant going on a fun vacation. Now I know better! Ecotourism is responsible travel to natural areas that conserves the environment and improves the well-being of local people. It’s about minimizing our impact on the environment, educating the traveller and Learning about the ecosystem and the culture of the people there.

My Project is “The Silent Valley Eco-Hut”. For it, I designed an imaginary ecotourism destination near a local forest reserve. The key features of my project are:
a) Sustainable Accommodation: The huts are built using locally sourced, renewable materials like bamboo and recycled wood. We’d use solar panels for electricity and rainwater harvesting for water.

b) Guided Nature Walks: Instead of letting people wander, local guides would lead small groups. They would tell visitors about the local flora and fauna, like the rare plants, birds or animals found there.

c) No Plastic Zone: Tourists would be encouraged to use reusable bottles and cloth bags. Food would be served on biodegradable plates or banana or teak leaves.

d) Community Kitchen: All meals would use organic, locally grown produce. This directly supports the farmers in the nearby village and gives tourists a taste of authentic local foods. Conservation isn’t somebody else’s job. It is our responsibility.

Question 5.
On p. 130, there is the image of the mango tree embraced by the jasmine vine. Pick the lines from the play that describe the scene.

Lines from the extract describing the tree and the jasmine vine:
……………………………………………………………………………………………………
……………………………………………………………………………………………………
……………………………………………………………………………………………………
Answer:
Anasuya: Oh Shakuntala! Here is the jasmine vine that you named Light of the Grove. She has chosen the mango tree as her husband. Shakuntala (approaches and looks at it, joyfully): What a pretty pair they make! The jasmine shows her youth in her fresh flowers, and the mango tree shows his strength in his ripening fruit. (She stands gazing at them.)

Now prepare a description on the scene in your words.
……………………………………………………………………………………………………
……………………………………………………………………………………………………
……………………………………………………………………………………………………
Answer:
In the scene we see Shakuntala and her friends Anasuya and Priyamvada watering the plants and trees in the hermitage. They are chitchatting and enjoying themselves. Suddenly Shakuntala sees the branches of the mango tree moving in the wind. She interprets as the mango tree calling her to tell her something. On the mango tree a jasmine vine has crept. Anasuya says that the jasmine embracing the mango tree was named Light of the Grove by Shakuntala. She adds that the jasmine vine has chosen the mango tree as her husband. The king describes Shakuntala. Her arms are tender shoots; her lips are red and warm flowers. She is a bewitching youth beginning to flower in the beauty of her own.

Shakuntala goes to the mango tree. She says the mango tree and the jasmine vine make a pretty pair. The jasmine shows her youth in her fresh flowers, and the mango tree shows his strength in his ripening fruit. She stands looking at them in wonder.

Question 6.
You plan to make a short video on scenes from the play Shakuntalam. Prepare a story board for the video.
Read the notes and explanations given on p. 130.
A storyboard is an outline of the shots and frames of a video. It includes a series of images (frames) that represent each shot of the story.

Here is a sample storyboard based on the segment The Bee Attacks from the play ‘Shãkunta1am’.

Now list the features of a story board that you have identified.
• Sketches: Simple drawings portraying what should appear on the screen.
• …………. : …………………………………………………………
• …………. : …………………………………………………………
• …………. : …………………………………………………………
• …………. : …………………………………………………………
A storyboard helps a director to:
• visualise an idea
• communicate the idea clearly
• identify the limitations of the script /other issues
• plan shots and engage in experimentation

Revisit the play Shakuntalam and identify the scenes to be made into short videos, each of two minutes’ duration.
• King Dushyanta’s hunt being stopped by the hermits
• The conversation with the charioteer
• …………………………………………………….
• …………………………………………………….
• …………………………………………………….
Now, form groups and develop a storyboard for the selected scene. Your teacher will help you. Present the storyboard in class, create short videos and upload them to your school blog, or to a social media platform.

Answer:
Now list the features of a story board that you have identified.
a) Sketches – Simple drawings portraying what should appear on the screen.

b) Captions – Brief descriptions that provides additional context. This text explains what is happening in the scene, the characters’ movements, and any important details that aren’t clear from the drawing alone.

c) Camera angles and shots: Information about the camera’s position (e.g., high-angle, low-angle) and the type of shot (e.g., close-up, wide shot).

d) Dialogue: The script of dialogue that will be spoken in the scene are usually written below or next to the panel.

e) Sound and music cues: Notes about any sound effects or background music that should be included.

Revisit the play Shakuntalam and identify the scenes to be made into short Doble videos.

• King Dushyanta’s hunt being stopped by the by hermits.
• The conversation with the charioteer.
• Shakuntala and her friends watering and chitchatting.
• King Dushyanta gazing at “Shakuntala” and his aside.
• The jasmine vine embracing the mango tree.
• The bee attack.

Shakuntalam Activities – Let’s read and reflect:

(Some of us spend hours in the forest to see an exotic bird or animal. Do you engage in bird- watching or wildlife photography? Have you identified the species of animals or birds that have chosen your locality for their habitat? The following is an excerpt from “The Fall of a Sparrow” by Salim Ali who is known as the ‘birdman of India’.
(നമ്മളിൽ ചിലർ മണിക്കൂറുകളോളം കാട്ടിൽ ചിലവഴിക്കുന്നത് അസാധാരണമായ പക്ഷികളെയോ മൃഗ ങ്ങളെയോ കാണാൻ വേണ്ടിയാണ്. നിങ്ങൾ പക്ഷിനിരീക്ഷണത്തിലോ വന്യജീവി ഫോട്ടോഗ്രാഫിയിലോ ഏർപ്പെടാറുണ്ടോ? അവയുടെ ആവാസവ്യവസ്ഥയ്ക്കായി തിരഞ്ഞെടുത്ത നിങ്ങളുടെ പ്രദേശത്തുള്ള മൃഗ ങ്ങളുടെയോ പക്ഷികളുടെയോ ഇനങ്ങൾ നിങ്ങൾ തിരിച്ചറിഞ്ഞിട്ടുണ്ടോ? ഇന്ത്യയുടെ പക്ഷി മനുഷ്യൻ എന്നറിയപ്പെടുന്ന സലിം അലിയുടെ “ദി ഫോൾ ഓഫ് എ സ്പാരോ എന്ന പുസ്തകത്തിൽ നിന്നുള്ള ഒരു ഉദ്ധരണിയാണ് താഴെ കൊടുത്തിരിക്കുന്നത്.

Shakuntalam Extra Questions and Answers

Question 1.
While pursuing the deer in his chariot, what did King Dushyanta have in his hand?
Answer:
While pursuing the deer in his chariot, King Dushyanta had a bow and arrow in his hand.

Question 2.
According to the charioteer, how did the deer get a lead?
Answer:
According to the charioteer, the deer got a lead because as the ground was rough he was holding the horses back.

Question 3.
Who are Shakuntala’s friends?
Answer:
Shakuntala’s friends are Anasuya and Priyamvada.

Question 4.
Whom does Shakuntala call a flatterer?
Answer:
Shakuntala calls Anasuya a flatterer.

Question 5.
How does the jasmine show her youth and the mango tree its strength?
Answer:
The jasmine shows her youth in her fresh flowers and the mango tree shows his strength in his ripening fruit.

Question 6.
Priyamvada tells Shakuntala that she has something pleasing to tell her. What was the pleasing thing?
Answer:
The pleasing thing was that Shakuntala would be married soon.

Students often refer to Kerala Syllabus 10th Standard English Textbook Solutions and Class 10 English Shakuntalam Summary in Malayalam & English Medium before discussing the text in class.

Class 10 English Shakuntalam Summary

Shakuntalam Summary in English

Characters:
King Dushyanta
Charioteer
Shakuntala, foster child of Sage Kanva
Anasuya and Priyamvada (friends of Shakuntala)

Dushyanta and the Deer
King Dushyanta is on a hunting expedition. He is pursuing a deer. He is in a chariot, driven by his charioteer. The king has a bow and arrow in his hand.

The charioteer tells the king that he is chasing a spotted deer. The king wants the deer to stop running. The king tells him that the deer has been making the chase a long one. The deer often looks behind and the king sees that he, the deer, has a beautiful neck. The deer is afraid and his body has shrunk. He is afraid of being hit by the arrow.

The king continues to speak. The path the deer takes is strewn with blades of grass that is half chewed. He jumps so often so sigh, it looks as if he is not running but flying. Even though the king is trying his best, he can’t keep the deer in sight.

The charioteer tells the king that he had been making the horses run slow as the ground was rough. That gave the deer an advantage and he could get a lead. Now they are on level ground and they can easily overtake the deer.

The Warning
A voice behind the scenes announces to the king that the deer belongs to the hermitage and it must not be killed. The charioteer tells the king that two hermits have come to save the deer from the arrow of the king. The king asks the charioteer to stop the chariot. A hermit comes with his pupil and tells the king that the deer belongs to the hermitage.

Why should the deer die like flowers in the fire? How can the gentle life of the deer endure the sharp arrow? Put your arrow back in its quiver. You are given the weapons to deliver the broken hearted and not to kill the innocent.

The king being a gentle person puts the arrow back in its quiver. The hermit is happy and says he has behaved like a king. The king thanks him for his blessing. Now the two hermits speak together: “King, we are on our way to gather firewood. Here on the bank of river Malini, you may see the hermitage of Kanva. Shakuntala is the guardian there. Please enter the hermitage and receive a welcome.” The king wants to know if Kanva is there. The two hermits tell him that Kanva has gone to Somathirtha, asking her daughter to welcome the guests. The king says he will see her and she will see his devotion and report it to her father. He asks the charioteer to drive to the hermitage as its sight will purify them. The king looks out and says anybody can realise that this is the precinct of a pious grove. There are rice grains, beneath the trees dropped from the bills of parrot chicks. There are also pounding stones on which there is almond oil. The trustful deer doesn’t run away as they approach it.

The charioteer fully agrees with the king. The king asks the charioteer to stop the chariot as he does not want to disturb the place. He gets down from the chariot. He looks at himself and says one should wear modest clothes when entering a hermitage. He asks the charioteer to keep the jewels and the bow.
The king walks further and sees the hermitage. He decides to enter.

Shakuntala and the jasmine Vine
A voice behind the scene asks the girls to come that way. King hears the voice. He thinks it is coming from the right of the grove. He walks and looks out. He sees hermit girls with watering pots.

They are coming in this direction to water the young trees there. He draws back into the shade and waits for them. He gazes at them. Shakuntala comes with two friends – Anasuya and Priyamvada. Anasuya tells Shakuntala that her father Kanva cares more for the hermitage trees than her. She adds that Shakuntala is as delicate as a jasmine flower yet he tells her to fill the trenches around the trees.

Shakuntala says she fills the trenches not because her father told her to do it but because she feels the trees are her sisters. Priyamvada says they had watered the plants that flower in the summer. Now they should sprinkle those plants whose flowering time is past. That will be better because then they will not be working for a reward. Shakuntala appreciates the idea. The king says to himself that Shakuntala must be the daughter of Kanva.

Shakuntala looks ahead. She sees the branches of a mango tree moving in the wind. She says to the girls that the mango tree is calling her as if it wants to tell her something. She must go to the tree. She then goes ahead to the tree.

Priyamvada asks her to stop. When Shakuntala asks why, she says there she looks like a vine clinging to the mango tree. Anasuya tells Shakuntala that it is the same jasmine vine that she had named “Light of the Grove”. The vine has chosen the mango tree as her husband! Shakuntala tells Anasuya now she knows why others call her a flatterer.

The king says the flattery is true. Her arms are tender shoots; lips blossoms red and warm. The beautiful youth is beginning to flower in beauty and form. Shakuntala comes near the mango tree on which the jasmine vine is creeping. She says they make a pretty pair. The jasmine shows her youth in her fresh flowers and the mango tree shows his strength in his ripening fruit. She stands looking at them.

Anasuya tells Shakuntala that she has forgotten the spring creeper her father himself tended as he tended her. Shakuntala goes to that creeper and says it is wonderful. She tells Priyamvada that she has something to say to her. Priyamvada asks what it is. Shakuntala tells her that it is out of season, but the spring creeper is covered with buds down to the very root. Both the friends come running to see it.

Looking at it joyfully, Priyamvada says that she has something pleasant to tell Shakuntala. She is to be married soon. Anasuya says that is why Shakuntala waters the spring creeper so lovingly. Shakuntala says she is her sister and that is why she waters her and she tips her watering pot.

The Bee Attacks
Shakuntala says that a bee has left the jasmine vine and she is flying into her face. She appears annoyed at the bee. She asks the girls to save her from the dangerous bee. The two girls tell her they are nobody to save her. She should call King Dushyanta as he is the protector of the grove. (Translated by Arthur W. Ryder)

Shakuntalam Summary in Malayalam

കഥാപാത്രങ്ങൾ:
ദുഷ്യന്ത് രാജാവ്
സാരഥി
ശകുന്തള, കണ്വമുനിയുടെ വളർത്തുപുത്രി
അനസൂയയും പ്രിയംവദയും (ശകുന്തളയുടെ സുഹൃത്തുക്കൾ)

ദുഷ്യന്തനും മാനും
ദുഷ്യന്ത രാജാവ് വേട്ടയാടുകയാണ്. അദ്ദേഹം ഒരു മാനിനെ പിന്തുടരുകയാണ്. അദ്ദേഹം ഒരു രഥത്തിലാ ണ്, അദ്ദേഹത്തിന്റെ സാരഥി അത് ഓടിക്കുന്നു. രാജാവിന്റെ കയ്യിൽ വില്ലും അമ്പും ഉണ്ട്.

അദ്ദേഹം ഒരു പുള്ളിമാനിനെ പിന്തുടരുകയാണെന്ന് സാരഥി രാജാവിനോട് പറയുന്നു. മാൻ ഓട്ടം നിർത്ത ണമെന്ന് രാജാവ് ആഗ്രഹിക്കുന്നു. മാൻ വളരെ സ്പീഡിലാണ് ഓടുന്നത് എന്ന് രാജാവ് അവനോട് പറ യുന്നു. മാൻ പലപ്പോഴും പുറകിലേക്ക് നോക്കുന്നു, മാനിന് മനോഹരമായ കഴുത്തുള്ളതായി രാജാവ് കാണുന്നു. മാൻ ഭയന്ന് അതിന്റെ ശരീരം ചുരുങ്ങി. അമ്പ് ഏൽക്കുമെന്ന് അത് ഭയപ്പെടുന്നു.

രാജാവ് തുടർന്നു സംസാരിച്ചു. മാൻ കടന്നുപോകുന്ന പാതയിൽ പാതി കടിച്ച് പുല്ലുകൾ ചിതറിക്കിടക്കു ന്നു. മാൻ പലപ്പോഴും ഉയർന്നു ചാടുന്നു. അവൻ ഓടുകയല്ല പറക്കുകയാണെന്ന് തോന്നുന്നു. രാജാവ് പരമാവധി ശ്രമിച്ചിട്ടും മാനിനെ ശരിയായി കാണാൻ കഴിയുന്നില്ല.

നിലം പരുക്കനായതിനാൽ കുതിരകളെ പതുക്കെ ഓടിക്കുകയാണെന്ന് സാരഥി രാജാവിനോട് പറയു ന്നു. അത് മാനിന് മുൻതൂക്കം നൽകുകയും അത് ലീഡ് നേടുകയും ചെയ്തു. ഇപ്പോൾ അവർ നിരപ്പായ സ്ഥലത്താണ്, അവർക്ക് മാനുകളെ എളുപ്പത്തിൽ മറികടക്കാൻ കഴിയും.

മുന്നറിയിപ്പ്
തിരശ്ശീലയ്ക്ക് പിന്നിലെ ഒരു ശബ്ദം രാജാവിനോട് മാൻ ആശ്രമത്തിൽ പെട്ടതാണെന്നും അതിനെ കൊല്ലാൻ പാടില്ലെന്നും അറിയിക്കുന്നു. രാജാവിന്റെ അസ്ത്രത്തിൽ നിന്ന് മാനിനെ രക്ഷിക്കാൻ രണ്ട് സന്യാസിമാർ വന്നിട്ടുണ്ടെന്ന് സാരഥി രാജാവിനോട് പറയുന്നു. രഥം നിർത്താൻ രാജാവ് സാരഥിയോട് ആവശ്യപ്പെടു ന്നു. ഒരു സന്യാസി തന്റെ ശിഷ്യനുമായി വന്ന് മാൻ ആശ്രമത്തിന്റേതാണെന്ന് രാജാവിനോട് പറയുന്നു.

“തീയിൽ പൂക്കൾ കരിയുന്നപോലെ മാൻ എന്തിന് മരിക്കണം? മൂർച്ചയേറിയ അസ്ത്രത്തെ മാനിന്റെ സൗമ്യത എങ്ങനെ സഹിക്കും? നിങ്ങളുടെ അമ്പ് അതിന്റെ ആവനാഴിയിൽ തിരികെ വയ്ക്കുക. ഹൃദയം തകർന്നവരെ വിടുവിക്കാനാണ് നിങ്ങൾക്ക് ആയുധങ്ങൾ നൽകിയിരിക്കുന്നത്. നിരപരാധികളെ കൊല്ലാനല്ല.”

രാജാവ് സൗമ്യനായതിനാൽ അമ്പ് അതിന്റെ ആവനാഴിയിൽ തിരികെ വെക്കുന്നു. സന്യാസി സന്തോഷ വാനാണ്. അദ്ദേഹം ഒരു രാജാവിനെപ്പോലെയാണ് പെരുമാറിയതെന്ന് പറയുന്നു. സന്യാസിയുടെ അനു ഗ്രഹത്തിന് രാജാവ് നന്ദി പറയുന്നു. ഇപ്പോൾ രണ്ട് സന്യാസിമാരും ഒരുമിച്ച് സംസാരിക്കുന്നു. “രാജാ ഞങ്ങൾ ഒരുമിച്ചാണ് വിറക് ശേഖരിക്കാൻ പോകുന്നത്. ഇവിടെ മാലിനി നദീതീരത്ത് നിങ്ങൾക്ക് കണ്വമഹർഷിയുടെ ആശ്രമം കാണാം. ശകുന്തളയാണ് അവിടെ കാവൽക്കാരി. ദയവായി ആശ്രമത്തിൽ പ്രവേശിച്ച് സ്വാഗതം സ്വീകരിക്കുക. കണ്വമഹർഷി അവിടെ ഉണ്ടോ എന്ന് രാജാവിന് അറിയണം. അതി ഥികളെ സ്വാഗതം ചെയ്യാൻ മകളോട് ആവശ്യപ്പെട്ട് കണ്വമഹർഷി സോമതീർത്ഥത്തിലേക്ക് പോയതായി രണ്ട് സന്യാസിമാർ പറഞ്ഞു. രാജാവ് അവളെ കാണുമെന്നും അവൾ അവന്റെ ഭക്തി കണ്ട് അത് അവ ളുടെ പിതാവിനെ അറിയിക്കുമെന്നും പറയുന്നു. ആശ്രമത്തിലെ കാഴ്ചകൾ, അവരെ ശുദ്ധീകരിക്കുമെന്ന തിനാൽ ആശ്രമത്തിലേക്ക് ഓടിക്കാൻ അദ്ദേഹം സാരഥിയോട് ആവശ്യപ്പെടുന്നു. രാജാവ് പുറത്തേക്ക് നോക്കി പറഞ്ഞു. ഇത് ഒരു പുണ്യതോട്ടത്തിന്റെ പരിസരമാണെന്ന് ആർക്കും മനസ്സിലാകും. തത്തക്കു ഞ്ഞുങ്ങളുടെ കൊക്കുകളിൽ നിന്ന് കൊഴിഞ്ഞ നെൽമണികൾ മരങ്ങൾക്കു താഴെയുണ്ട്. ബദാം ഓയിൽ പുരണ്ട് കല്ലുകളും അവിടെ ഉണ്ട്. വിശ്വസ്തനായ മാൻ അതിനെ സമീപിക്കുമ്പോൾ പേടിച്ച് ഓടിപ്പോകു ന്നില്ല.

സാരഥി രാജാവിനോട് പൂർണ്ണമായും യോജിക്കുന്നു. ഈ സ്ഥലത്തെ ശല്യപ്പെടുത്താൻ ആഗ്രഹിക്കാത്തതി നാൽ രഥം നിർത്താൻ രാജാവ് സാരഥിയോട് ആവശ്യപ്പെടുന്നു. അദ്ദേഹം രഥത്തിൽ നിന്ന് ഇറങ്ങുന്നു. ആശ്രമത്തിൽ പ്രവേശിക്കുമ്പോൾ മാന്യമായ വസ്ത്രം ധരിക്കണമെന്ന് അദ്ദേഹം തന്നെത്തന്നെ നോക്കി പറയുന്നു. തന്റെ ആഭരണങ്ങളും വില്ലും സൂക്ഷിക്കാൻ അദ്ദേഹം സാരഥിയോട് ആവശ്യപ്പെടുന്നു. രാജാവ് മുന്നോട്ട് നടന്ന് ആശ്രമം കാണുന്നു. അദ്ദേഹം അതിൽ പ്രവേശിക്കാൻ തീരുമാനിക്കുന്നു.

ശകുന്തളയും മുല്ലച്ചെടിയും
പിന്നിൽ നിന്ന് ഒരു ശബ്ദം പെൺകുട്ടികളോട് ആ വഴിക്ക് വരാൻ ആവശ്യപ്പെടുന്നു. രാജാവ് ശബ്ദം കേൾക്കുന്നു. അത് തോപ്പിന്റെ വലതുഭാഗത്ത് നിന്നാണ് വരുന്നതെന്ന് അദ്ദേഹം കരുതുന്നു. അദ്ദേഹം പുറത്തേക്ക് നോക്കി നടക്കുന്നു. വെള്ളമൊഴിക്കുന്ന പാത്രങ്ങളുമായി സന്യാസി പെൺകുട്ടികളെ അദ്ദേഹം കാണുന്നു. അവിടെയുള്ള ഇളം മരങ്ങൾ നനയ്ക്കാൻ അവർ ഈ ദിശയിൽ വരുന്നു. അദ്ദേഹം തണലി ലേക്ക് തിരികെ വരുകയും അവർക്കായി കാത്തിരിക്കുകയും ചെയ്യുന്നു. അദ്ദേഹം അവരെ നോക്കുന്നു. അനസൂയ, പ്രിയംവദ എന്നീ രണ്ട് സുഹൃത്തുക്കളോടൊപ്പമാണ് ശകുന്തള വരുന്നത്.

തന്റെ പിതാവ് കണ്വമഹർഷിക്ക് അവളേക്കാൾ കൂടുതൽ കരുതൽ മരങ്ങളോടാണെന്ന് ശകുന്തളയോട് അനസൂയ പറയുന്നു. ശകുന്തള മുല്ലപ്പൂപോലെ മൃദുലയാണെങ്കിലും മരങ്ങൾക്ക് ചുറ്റുമുള്ള കിടങ്ങുകൾ നിറയ്ക്കാൻ അവളുടെ പിതാവ് അവളോട് പറയുന്നുവെന്നും അവൾ കൂട്ടിച്ചേർത്തു.

അച്ഛൻ തന്നോട് അത് ചെയ്യാൻ പറഞ്ഞതുകൊണ്ടല്ല. മരങ്ങൾ തന്റെ സഹോദരിമാരാണെന്ന് തോന്നിയ തുകൊണ്ടാണ് താൻ കിടങ്ങുകൾ നിറയ്ക്കുന്നതെന്ന് ശകുന്തള പറയുന്നു. വേനലിൽ പൂക്കുന്ന ചെടി കൾക്ക് വെള്ളം നൽകിയിരുന്നതായി പ്രിയംവദ പറയുന്നു. ഇപ്പോൾ അവർ പൂവിടുന്ന സമയം കഴിഞ്ഞ ചെടികൾ തളിക്കണം. അത് മികച്ച ഒരു പ്രവർത്തിയായിരിക്കും, കാരണം അവർ പ്രതിഫലത്തിനായിട്ടല്ല പ്രവർത്തിക്കുന്നത്. ശകുന്തള ഈ ആശയത്തെ അഭിനന്ദിക്കുന്നു. ശകുന്തള കണ്വമഹർഷിയുടെ മകളാ യിരിക്കും എന്ന് രാജാവ് സ്വയം പറയുന്നു.

ശകുന്തള മുന്നിലേക്ക് നോക്കി. ഒരു മാവിന്റെ ശിഖരങ്ങൾ കാറ്റിൽ ചലിക്കുന്നത് അവൾ കാണുന്നു. മാവ് തന്നോട് എന്തോ പറയണമെന്ന മട്ടിലാണ് അവളെ വിളിക്കുന്നതെന്ന് അവൾ പെൺകുട്ടികളോട് പറയു ന്നു. അവൾ മരത്തിന്റെ അടുത്തേക്ക് പോകണം. എന്നിട്ട് അവൾ മരത്തിന്റെ അടുത്തേക്ക് പോകുന്നു.

പ്രിയംവദ അവളോട് നിൽക്കാൻ ആവശ്യപ്പെട്ടു. ശകുന്തള എന്തിനാണെന്ന് ചോദിച്ചപ്പോൾ, അവൾ പറ ഞ്ഞു, അവൾ മാവിൽ പറ്റിപ്പിടിച്ചിരിക്കുന്ന ഒരു വളിയെപ്പോലെയാണ്. അവൾ “ലൈറ്റ് ഓഫ് ദ് ഗ്രോവ്” എന്ന് പേരിട്ട അതേ മുല്ലയാണ് ഇതെന്നും അനസൂയ ശകുന്തളയോട് പറയുന്നു. മുല്ല തന്റെ ഭർത്താവായി തിരഞ്ഞെടുത്തത് ആ മാവിനെയാണ്. മറ്റുള്ളവർ അനസൂയയെ മുഖസ്തുതിക്കാരി എന്ന് വിളിക്കുന്നത് എന്തുകൊണ്ടാണെന്ന് ഇപ്പോൾ തനിക്കറിയാമെന്ന് ശകുന്തള അനസൂയയോട് പറയുന്നു.

അവളുടെ മുഖസ്തുതി സത്യമാണെന്ന് രാജാവ് പറയുന്നു. ശകുന്തളയുടെ കൈകൾ ഇളം തണ്ടുകളാണ്. ചുണ്ടുകൾ പൂക്കൾപോലെ ചുവന്നും ചൂടുള്ളതും ആണ്. സുന്ദരിയും യൗവനവുമുള്ള അവൾ സൗന്ദര്യ ത്തിലും രൂപത്തിലും പൂവിടാൻ തുടങ്ങിയിരിക്കുന്നു. മുല്ലപ്പൂ വള്ളി കയറിയ മാവിന് സമീപം ശകുന്തള വരുന്നു. അവ മനോഹരമായ ഒരു ജോഡി ഉണ്ടാക്കുന്നുവെന്ന് അവൾ പറയുന്നു. മുല്ലപ്പൂ അവളുടെ പുത്തൻ പൂക്കളിൽ അവളുടെ യൗവനം കാണിക്കുന്നു. മാവ് അവന്റെ പഴുക്കുന്ന പഴങ്ങളിൽ അവന്റെ ശക്തി കാണിക്കുന്നു. അവൾ അവയെ നോക്കി നിൽക്കുന്നു.

അവളുടെ പിതാവ് സ്വന്തം കൈകൊണ്ടു വളർത്തിയിരുന്ന വസന്തകാല വള്ളിച്ചെടിയെ അവൾ മറന്നു പോയെന്ന് അനസൂയ ശകുന്തളയോട് പറയുന്നു. ശകുന്തള ആ വള്ളിച്ചെടിയുടെ അടുത്ത് ചെന്ന് അത് അത്ഭുതകരമാണെന്ന് പറയുന്നു. തനിക്കൊരു കാര്യം പറയാനുണ്ടെന്ന് പ്രിയംവദയോട് അവൾ പറഞ്ഞു. അതെന്താണെന്ന് പ്രിയംവദ ചോദിക്കുന്നു. അപ്പോൾ അവൾ പറയുന്നു, ഇത് സീസണല്ലെങ്കിലും സ്പ്രിംഗ് കീപ്പർ വേരുവരെ മുകുളങ്ങളാൽ മൂടപ്പെട്ടിരിക്കുന്നു. അവളുടെ സുഹൃത്തുക്കൾ രണ്ടുപേരും അത് കാണാൻ ഓടി വരുന്നു.

അത് സന്തോഷത്തോടെ നോക്കിയിട്ട് പ്രിയംവദ പറഞ്ഞു, അവൾക്ക്, ശകുന്തളയോട് എന്തോ പറയാനു ണ്ടെന്ന്, അവൾ ഉടൻ വിവാഹിതയാകും. അതുകൊണ്ടാണ് ശകുന്തള വള്ളിച്ചെടിയെ സ്നേഹത്തോടെ നനയ്ക്കുന്നതെന്ന് അനസൂയ പറയുന്നു. ശകുന്തള പറയുന്നു. ആ ചെടി അവളുടെ സഹോദരിയാണെന്നും അതുകൊണ്ടാണ് അതിന് അവൾ വെള്ളം നനയ്ക്കുന്നത് എന്നും. അവൾ നനയ്ക്കുന്ന പാത്രത്തിൽ നിന്നും വെള്ളം ചെടിക്കൊഴിക്കുന്നു.

തേനീച്ച ആക്രമിക്കുന്നു
മുല്ലപ്പൂവള്ളി വിട്ട് ഒരു തേനീച്ച തന്റെ മുഖത്തേക്ക് പറക്കുന്നുണ്ടെന്ന് ശകുന്തള പറയുന്നു. അവൾക്കും തേനീച്ചയോട് ദേഷ്യം തോന്നുന്നു. അപകടകരമായ തേനീച്ചയിൽ നിന്ന് തന്നെ രക്ഷിക്കാൻ അവൾ പെൺകുട്ടികളോട് ആവശ്യപ്പെടുന്നു. അവളെ രക്ഷിക്കാൻ തങ്ങൾ ആരുമല്ലെന്ന് പെൺകുട്ടികൾ പറയു ന്നു. തോട്ടത്തിന്റെ സംരക്ഷകനായതിനാൽ അവൾ ദുഷ്യന്തൻ രാജാവിനെ വിളിക്കണം.
(വിവർത്തനം ചെയ്തത് ആർതർ ഡബ്ല്യൂ. റൈഡർ)

Class 10 English Shakuntalam by Arthur William Ryder About the Author

We don’t know exactly when Kalidas was born or died. It is believed that he lived between the 4^th and 5^th century AD during the Gupta Empire. He wrote in classical Sanskrit. He is ancient India’s greatest poet and playwright. His well known works are: “Kumamarasambhava”, “Raghuvamsa” “Abhijnana Shakuntalam” and “Meghadoota”.

രചയിതാവിനെക്കുറിച്ച്
കാളിദാസൻ എപ്പോഴാണ് ജനിച്ചതെന്നോ മരിച്ചതെന്നോ കൃത്യമായി അറി യില്ല. എ.ഡി. നാലു, അഞ്ചു നൂറ്റാണ്ടിനും ഇടയിൽ ഗുപ്ത സാമ്രാജ്യത്തിന്റെ കാലത്ത് അദ്ദേഹം ജീവിച്ചിരുന്നതായി കരുതപ്പെടുന്നു. ക്ലാസിക്കൽ സംസ്ക തത്തിലാണ് അദ്ദേഹം എഴുതിയത്. പുരാതന ഇന്ത്യയിലെ ഏറ്റവും വലിയ കവിയും നാടകകൃത്തുമാണ് അദ്ദേഹം. അദ്ദേഹത്തിന്റെ അറിയപ്പെടുന്ന കൃതി കൾ ഇവയാണ്. “കുമാരസംഭവം”, “രഘുവംശം”, “അഭിജ്ഞാന ശാകുന്തളം, “മേഘദൂത”

Arthur William Ryder (1877 – 1938) was a professor of Sanskrit in the University of California, Berkeley. He is best known for translating several Sanskrit books into English, notably “Abhijnana Shakuntalam”, “Meghadoota”, “The Bhagwat Gita” and “The Panchatantra”.
ആർതർ വില്യം റൈഡർ (1877-1938) ബെർക്കിയിലെ കാലിഫോർണിയ സർവകലാശാലയിൽ സംസ്കൃതം പ്രൊഫസറായിരുന്നു. “അഭിജ്ഞാന ശാകുന്തളം”, “മേഘദൂത്” “ദ് ഭഗവത് ഗീത”, “പഞ്ച തന്ത്രം” എന്നിങ്ങനെ നിരവധി സംസ്കൃത പുസ്തകങ്ങൾ ഇംഗ്ലീഷിലേക്ക് വിവർത്തനം ചെയ്തതി നാണ് അദ്ദേഹം കൂടുതൽ അറിയപ്പെടുന്നത്.

Class 10 English Shakuntalam Vocabulary

sage – sanyasi, teacher, സന്യാസി
pursuing – following, പിൻതുടരുക
shafts – arrows, അമ്പുകൾ
shrinks – reduces in size, ചുരുങ്ങുക
strewed – scatter, spread, ചിതറിക്കിടക്കുക
leaps – jumps, ചാടുക
hermitage – a place where a hermit lives, ashram, ആശ്രമം
tender – smooth, soft, മൃദുവായ
expire – die, മരിക്കുക, കാലഹരണപ്പെടുക
blossoms – flowers, പൂക്കൾ
endure – suffer, സഹിക്കുക
quiver – a case for arrows, ആവനാഴി
Malini – a river, ഒരു നദി
precinct – the area within the walls of a place, ചുറ്റുപാട്
grove – a fruit garden, പഴത്തോട്ടം
pounding stone – a stone for powdering things, ഇടിക്കാനുള്ള കല്ലുകൾ
dismount – alight, come down, ഇറങ്ങുക
garments – dresses, clothes, വസ്ത്രങ്ങൾ
gazing – looking, നോക്കുക
delicate – fine, beautiful, ഭംഗിയുള്ള
trenches – long, narrow ditches, കിടങ്ങുകൾ
bidding – asking, ചെയ്യാൻ പറയുക
vine – creeping plant, പടർന്നു കയറുന്ന വള്ളി
clinging – holding, മുറുകെ പിടിക്കുക
flatterer – one who praises others without proper reason, മുഖസ്തുതിക്കാരി
bewitching – attracting, ആകർഷിക്കുന്ന
approaches – comes near, അടുത്തു വരുക
tended – took care of, പരിചരിക്കുക
dreadful – fearful, പേടിപ്പിക്കുന്ന

Students often refer to SSLC Maths Textbook Solutions and Class 10 Maths Chapter 10 Circles and Lines Important Extra Questions and Answers Kerala State Syllabus to clear their doubts.

SSLC Maths Chapter 10 Circles and Lines Important Questions and Answers

Circles and Lines Class 10 Extra Questions Kerala Syllabus

Circles and Lines Class 10 Kerala Syllabus Extra Questions

Question 1.
AB and CD intersect at P. O is the centre of the circle. CD is the diameter of the circle. OP = 2, PA = 9, PB = 5, then what is its radius?

(a) 7
(b) 8
(c) 4
(d) 3
Answer:
(a) 7
(r + 2)(r – 2) = 9 × 5
⇒ r² – 2² = 45
⇒ r² = 49
⇒ r = 7

Question 2.
AB is the diameter of the circle. P divides the diameter of the circle into a and b as shown in the diagram. PC is perpendicular to AB. The point P divides AB in the ratio a : b. The area of the square with side PC is:

(a) a × b
(b) a + b
(c) a / b
(d) ab
Answer:
(a) a × b
a × b = PC²

Question 3.
In the diagram, AB is the diameter of the semicircle, PQ = √14, and RS = √18 are perpendicular to AB. What is the length of the line AB?

(a) 10
(b) 9
(c) 12
(d) 15
Answer:
(b) 9
\(\sqrt{18}=\sqrt{6 \times 3}\)
\(\sqrt{14}=\sqrt{7 \times 2}\)
AB = 9

Question 4.
Read the two statements given below:
In a circle, the length of the chord AB is 16 cm. The length of the perpendicular drawn from the center of the circle to AB is 6 cm.
Statement (1): The radius of the circle is 10 cm.
Statement (2): The perpendicular drawn from the center to the circle to the chord bisects the chord.
Choose the correct answer from those given below.
(a) Statement (1) is true, Statement 2 is false
(b) Statement (2) is true, Statement 1 is false
(c) Both are true. Statement 1 is the reason for Statement 2
(d) Both are true. Statement 1 is not the reason for Statement 2
Answer:
(c) Both are true. Statement 1 is the reason for Statement 2.

Question 5.
Read the two statements given below:
In a circle, two chords AB and CD meet at a point P inside the circle. AP = 4 cm, PB = 6 cm, CP = 3 cm, PD = ?
Statement (1): PD = 8 cm
Statement (2): If two chords intersect inside a circle, then AP × PB = CP × PD
Choose the correct answer from those given below.
(a) Statement 1 is correct, Statement 2 is incorrect
(b) Statement 2 is correct, Statement 1 is incorrect
(c) Both are correct. Statement A is the reason for Statement 2
(d) Both are correct. Statement 1 is not the reason for Statement 2
Answer:
(c) Both are correct. Statement 1 is the reason for Statement 2.

Question 6.
In the diagram, a line passing through the center of a circle divides a chord into two. What is the radius of the circle?

Answer:
If we extend the line OP that intersects the two ends of the circle, we get another chord CD.

AB and CD are two chords that meet at the point P.
Therefore, PA × PB = PC × PD.
That is, 4 × 6 = (r + 5) × (r – 5)
⇒ 24 = r² – 5²
⇒ 24 = r² – 25
⇒ 24 + 25 = r²
⇒ 49 = r²
⇒ r = ±7
r is the radius of the circle; it will always be a positive number.
⇒ r = 7 cm

Question 7.
In the diagram, a line drawn from the center of a circle intersects a chord of the circle. What are the lengths of the two parts of the chord?

Answer:

If we extend the line OP that intersects the two ends of the circle, we get a chord CD.
AB and CD are two chords that meet at the point P.
So, AP × PB = CP × PD
AP × PB = 4 × 10
AP × PB = 40
Also AP + PB = 13 cm
The product of two numbers is 40, and their sum is 13 is 8 and 5.
Therefore AP = 5 cm, PB = 8 cm

Question 8.
The chords AB and CD intersect at point P. AB = 17 cm, PA = 9 cm, PD = 12 cm.

(a) What is the length of PB?
(b) Calculate the length of PC?
Answer:
(a) AB = PA + PB
⇒ 17 = 9 + PB
⇒ PB = 17 – 9 = 8 cm

(b) PA × PB = PC × PD
⇒ 9 × 8 = PC × 12
⇒ PC = 6 cm

Question 9.

(a) In the diagram, AB is the diameter of the semicircle. PC is perpendicular to AB. If AP = 5 cm, PB = 3 cm, calculate the length of PC.
(b) Draw a square with an area of 15 square centimeters.
Answer:
(a) AP × PB = PC × PC
⇒ 5 × 3 = PC × PC
⇒ 15 = PC²
⇒ PC = √15
(b)

Question 10.
Chords AB and CD intersect at point P. AB = 13 cm, CD = 15 cm, PD = 3 cm.

(a) What is the length of PC?
(b) If PB = x, find PA?
(c) What is the length of PB?
Answer:
CD = PC + PD
⇒ 15 = PC + 3
⇒ PC = 15 – 3 = 12
(a) AB = PA + PB
⇒ 13 = PA + x
⇒ PA = 13 – x

(b) PC × PD = PA × PB
⇒ 12 × 3 = (13 – x) × x
⇒ 36 = 13x – x²
⇒ x² – 13x + 36 = 0
⇒ x = 4 or x = 9
Therefore PB = 4 cm

Question 11.
Draw a rectangle with an area of 24 square centimeters. Draw a square of area equal to the area of this rectangle.
Answer:
Draw a rectangle with length 6 cm and height 4 cm.
Add the height of the rectangle to the length of the base.
The new length is 6 + 4 = 10 cm.
Now draw a semicircle with the bottom line as the diameter, extend the right side of the square, and meet it with the semicircle.
This line is the side of the square.
BSTF is the required square.

Question 12.
In the diagram, AB is the diameter of the semicircle, and PC is the perpendicular to AB. AC = 5√29 cm, PA = 25 cm.

(a) What is the length of PC?
(b) What is the length of PB?
(c) What is the radius of the circle?
Answer:
(a) PC = \(\sqrt{(5 \sqrt{29})^2-25^2}\) = 10 cm

(b) PA × PB = PC²
⇒ 25 × PB = 10²
⇒ PB = 4 cm

(c) AB = 25 + 4 = 29
Radius = 14.5 cm

Question 13.
If the chords AB and CD are extended, they intersect at P outside the circle. If PA = PC, then prove AB = CD.

Answer:
PA × PB = PC × PD
PA and PC on both sides
⇒ PB = PD,
⇒ PA – AB = PC – CD
PA and PC on both sides
⇒ -AB = -CD
⇒ AB = CD

Question 14.
The chords AB and CD intersect at a point inside a circle at P as shown in the figure.

(a) Draw a diagram. Join AC and BD. Establish the similarity of triangles PAC and PBD and the relationship between PA, PB, PC, and PD.
(b) PA = 3 cm, PB = 5 cm. What is the area of the rectangle with sides PC and PD?
(c) Draw a rectangle with sides of length 5 cm and 3 cm. Draw another rectangle with one side 6 cm and an area equal to the area of the first rectangle.
Answer:
(a) Consider ∆PCA and ∆PBD
∠C = ∠B, ∠A = ∠D (Angles at the end of an arc are equal)
\(\frac{P A}{P D}=\frac{P C}{P B}\)
PA × PB = PC × PD

(b) 3 × 5 = PC × PD
PC × PD = 15
The area of the rectangle with sides PC, PD is 15 sq.cm

(c) Consider a rectangle with length 5 cm and width 3 cm.
Consider another length of 6 cm.
In the square, we can draw a line extending 3 cm from the bottom edge to the left side and 6 cm from the left edge to the bottom:
Now draw a circle through the left, right, and bottom corners.
Draw a line through the rectangle where the left side intersects the circle.
Now, let’s draw the required square by marking the length of the rectangle we have obtained.

Here, ABCD is the required rectangle.

Question 15.
See the diagram below. AB is the diameter ofthe  semicircle, and PC is perpendicular to AB. You can see the triangles ACB, PCB.

(a) ∆ACB and ∆PCB are similar triangles, and prove PA × PB = PC².
(b) Construct a diagram such that PC = √12 cm, and a square of area 12 sq.cm
(c) Draw a chord of length √48 cm in a circle in your construction.
Answer:
Draw AC and BC.
Consider ∆ACB and ∆ACP
In triangle ACB, ∠C = 90°
(a) If ∠PAC = x, then ∠ACP = 90 – x, ∠BCP = x, ∠PBC = 90 – x
The two angles of the triangle ABC are equal to the two angles of the triangle ACP.
The triangle ABC and the triangle ACP are similar triangles.
Sides opposite to equal angles are proportional.
\(\frac{P A}{P C}=\frac{P C}{P B}\)
⇒ PA × PB = PC²

(b) \(\sqrt{12}=\sqrt{6 \times 2}\)
Draw a semicircle with diameter AB = 6 + 2 = 8 cm
Mark the point P.
Length of PC = √12

Complete the square with side PC.
The area of the square will be (√12)² = 12

(c) Draw the circle.
Extend the line CP to D.
Length of the chord CD is √48.

CD = 2 × √12 = √48.

Students often refer to SSLC Maths Textbook Solutions and Class 10 Maths Chapter 9 Polynomials and Equations Important Extra Questions and Answers Kerala State Syllabus to clear their doubts.

SSLC Maths Chapter 9 Polynomials and Equations Important Questions and Answers

Polynomials and Equations Class 10 Extra Questions Kerala Syllabus

Polynomials and Equations Class 10 Kerala Syllabus Extra Questions

Question 1.
At what points does the graph of the polynomial 4x² – 8x + 3 cross the x-axis?
(a) \(\left(\frac{3}{4}, 0\right),\left(\frac{1}{2}, 0\right)\)
(b) \(\left(\frac{3}{2}, 0\right),\left(\frac{1}{2}, 0\right)\)
(c) \(\left(\frac{3}{4}, 0\right),\left(\frac{1}{6}, 0\right)\)
(d) \(\left(\frac{7}{2}, 0\right),\left(\frac{3}{2}, 0\right)\)
Answer:
(b) \(\left(\frac{3}{2}, 0\right),\left(\frac{1}{2}, 0\right)\)
4x² – 8x + 3 = 0
a = 4, b = -8, c = 3

So, the points at which the graph of the polynomial 4x² – 8x + 3 crosses the x-axis are \(\left(\frac{3}{2}, 0\right),\left(\frac{1}{2}, 0\right)\)

Question 2.
The sum and product of the solutions of a second-degree polynomial are -10 and \(\frac {5}{2}\) respectively. Then the polynomial is:
(a) 2x² – 20x + 10
(b) 2x² – x + 5
(c) 2x² – 20x + 5
(d) x² – 20x + 5
Answer:
(c) 2x² – 20x + 5
Sum of the solutions is -10, and its product is \(\frac {5}{2}\)
a + b = -10, ab = \(\frac {5}{2}\)
Polynomial = x² + (a + b)x + ab
= x² – 10x + \(\frac {5}{2}\)
= 2x² – 20x + 5

Question 3.
Consider the following statements.
Statement (A): x² + 20x + 96 is the polynomial with the sum of solutions 8 and their product 12.
Statement (B): If a, b are the solutions of a polynomial, then the polynomial is x² + (a + b)x + ab.
(a) Statement A is true, Statement B is false.
(b) Statement B is true, Statement A is false.
(c) Both statements are true. Statement B is the reason for Statement A.
(d) Both statements are true. Statement B is not the reason for Statement A.
Answer:
(b) Statement B is true, Statement A is false.
Statement B is true.
The sum of solutions 8 and their product 12.
⇒ a + b = 8, ab = 12
Polynomial = x² + (a + b)x + ab = x² + 8x + 12
∴ Statement A is false.

Question 4.
Consider the following statements.
Statement (A): The area of a rectangle with sides (x + 5) and (x – 3) is a second-degree polynomial.
Statement (B): A second-degree polynomial can be written as the product of two first-degree polynomials.
(a) Statement A is true, Statement B is false.
(b) Statement B is true, Statement A is false.
(c) Both statements are true. Statement B is the reason for Statement A.
(d) Both statements are true. Statement B is not the reason for Statement A.
Answer:
(d) Both statements are true. Statement B is not the reason for Statement A.
Statement B is true.
The area of a rectangle with sides (x + 5) and (x – 3)
(x + 5)(x – 3) = x² + 5x – 3x – 15 = x² + 2x – 15
A second-degree polynomial Statement A is true.

Question 5.
Write the polynomial x2 + 5x – 84 as the product of two first-degree polynomials.
Answer:
x² + 5x – 84 = (x + a)(x + b) = x² + (a + b)x + ab
⇒ a + b = 5, ab = -84
⇒ (a – b)² = (a + b)² – 4ab
⇒ (a – b)² = 5² – 4 × (-84)
⇒ (a – b)² = 25 + 336
⇒ (a – b)² = 361
⇒ a – b = ±19
Take a – b = 19
a + b = 5, a – b = 19
⇒ a = \(\frac {1}{2}\)(5 + 19) = \(\frac {1}{2}\) × 24 = 12
⇒ b = \(\frac {1}{2}\)(5 – 19) = \(\frac {1}{2}\) × (-14) = -7
Take a – b = -19
a + b = 5, a – b = -19
⇒ a = \(\frac {1}{2}\)(5 + (-19) = \(\frac {1}{2}\) × (-14) = -7
⇒ b = \(\frac {1}{2}\)(5 – (-19)) = \(\frac {1}{2}\) × (5 + 19) = 12
∴ x² + 5x – 84 = (x + 12)(x – 7)

Question 6.
If each side of a square is decreased by 1 metre, its area becomes 49 square metres. What is the side of the original square?
Answer:
Let the side of the original square be x.
(x – 2)² = 49
⇒ x² – 4x + 4 = 49
⇒ x² – 4x – 45 = 0
a = 1, b = -4, c = -45

x is the length of the side of the square, so it cannot be negative.
∴ x = 9

Question 7.
The product of the first term and the third term of an arithmetic sequence with a common difference of 1 is 143. Find the first three terms of the sequence.
Answer:
If first term be x, then third term = x + 2
x(x + 2) = 143
⇒ x² + 2x = 143
⇒ x² + 2x – 143 = 0
a = 1, b = 2, c = -143
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

If the first term x = 11, then the first three terms of the sequence are 11, 12, 13.
If the first term x = -13, then the first three terms of the sequence are -13, -12, -11.

Question 8.
1 added to the product of two consecutive even numbers gives 289. What are the numbers?
Answer:
If one number is x, then the next even number is x + 2.
x(x + 2) + 1 = 289
⇒ x² + 2x + 1 = 289
⇒ x² + 2x + 1 – 289 = 0
⇒ x² + 2x – 288 = 0
a = 1, b = 2, c = -288

If x = 16, then next number x + 2 = 18
If x = -18, then next number x + 2 = -16

Question 9.
The length of the rectangular sheet shown in the figure is 13 centimetres. From this sheet, cut off two square sheets of maximum size. The area of the remaining sheet is 15 square metres.

(a) If the breadth of the sheet is x, then what is the length of the remaining sheet?
(b) Form a second-degree equation and find the length and breadth of the remaining sheet.
Answer:
(a) Length of the remaining sheet = 13 – 2x
(b) Area of the remaining sheet = 15 square metres
x(13 – 2x) = 15
⇒ 13x – 2x² = 15
⇒ 2x² – 13x + 15 = 0
a = 2, b = -13, c = 15

If the breadth of the sheet is x = 5,
then length = 13 – 2x = 13 – 2 × 5 = 3
If breadth of the sheet is x = \(\frac {3}{2}\),
then length = 13 – 2x = 13 – 2 × \(\frac {3}{2}\) = 10

Question 10.
The sum of n natural numbers starting from 1 is \(\frac{n(n+1)}{2}\). Then, the sum of the consecutive natural numbers from 1 up to what number is 325?
Answer:
\(\frac{n(n+1)}{2}\) = 325
⇒ \(\frac{n^2+n}{2}\) = 325
⇒ n² + n = 650
⇒ n² + n – 650 = 0
n² + n – 650 = (n + a)(n + b) = n² + (a + b)n + ab
⇒ a + b = 1, ab = -650
(a – b)² = (a + b)² – 4ab
⇒ (a – b)² = 1² – 4 × (-650)
⇒ (a – b)² = 1 + 2600 = 2601
⇒ a – b = ±51
Take a – b = 51,
a + b = 1, a – b = 51
⇒ a = \(\frac {1}{2}\)(1 + 51) = \(\frac {1}{2}\) × 52 = 26
⇒ b = \(\frac {1}{2}\)(1 – 51) = \(\frac {1}{2}\) × (-50) = -25
Take a – b = -51,
a + b = 1, a – b = -51
⇒ a = \(\frac {1}{2}\)(1 – 51) = \(\frac {1}{2}\) × (-50) = -25
⇒ b = \(\frac {1}{2}\)(1 – (-51)) = \(\frac {1}{2}\) × (1 + 51) = 26
Using this, n² + n – 650 = (n + 26)(n – 25)
n² + n – 650 = 0
⇒ (n + 26)(n – 25) = 0
⇒ n = -26 or n = 25
n is the number of terms, so it cannot be negative.
∴ n = 25
The sum of the consecutive natural numbers from 1 up to 25 is 325.

Question 11.
In the figure, chord AB and CD are extended to meet at P. PB = 14 centimetres, AB = 5 centimetres, and CD = 15 centimetres. Find the length of PC.

Answer:
Let the length of PC be x,
PD = x + 15
PB = 14,
AB = 5
⇒ PA = 14 – 5 = 9
PA × PB = PC × PD
⇒ 9 × 14 = x(x + 15)
⇒ x² + 15x = 126
⇒ x² + 15x – 126 = 0
⇒ a = 1, b = 15, c = -126

Since x is the length of PC, and hence not negative.
So, x = 6
∴ The length of PC = 6 centimetres.

Question 12.
The difference between the perpendicular sides of a right triangle is 10 centimetres, and its area is 72 square centimetres. Then find the length of the perpendicular sides of the triangle.
Answer:
Let the length of one perpendicular side be x,
Then the other side is x + 10
Area = \(\frac {1}{2}\) × x(x + 10) = 72
⇒ x(x + 10) = 144
⇒ x² + 10x = 144
⇒ x² + 10x – 144 = 0
⇒ a = 1, b = 10, c = -144

Since x is the length of a side, it cannot be negative.
∴ x = 8
So, length of perpendicular sides = 8, 18

Question 13.
A rectangle is to be made by bending a 70-centimeter-long rod. The length of the diagonal is 25 centimetres. Then, find the length and breadth of the rectangle.
Answer:
Perimeter of the rectangle = 70 centimetres
Let the length of the rectangle be x,
⇒ 2(length + breadth) = 72
⇒ length + breadth = 36
⇒ x + breadth = 36
⇒ breadth = 36 – x
If the length of the diagonal is 25 centimetres,
x² + (36 – x)² = 25²
⇒ x² + 36² – 72x + x² = 252
⇒ 2x² – 72x + 1296 = 625
⇒ 2x² – 72x + 671 = 0
a = 2, b = -72, c = 671
Hence,

A rectangle diagonal of 25 cm cannot be made.

Question 14.
The terms of an arithmetic sequence with a common difference of 4 are positive numbers. The product of two consecutive terms of the sequence is equal to its sum.
(a) If one term is x, then what is the next term?
(b) Find the terms.
Answer:
(a) Common difference = 4
If one term is x, next term = x + 4
(b) x(x + 4) = x + x + 4
⇒ x² + 4x = 2x + 4
⇒ x² + 4x – 2x – 4 = 0
⇒ x² + 2x – 4 = 0
a = 1, b = 2, c = -4
Hence,

If x = -1 + √5, then next term = x + 4
= -1 + √5 + 4
= 3 + √5
Sequence: -1 + √5, 3 + √5, 7 + √5,…
If x = -1 – √5, then next term = x + 4
= -1 – √5 + 4
= 3 – √5
Sequence: -1 – √5, 3 – √5, 7 – √5,….

Question 15.
When 16 is added to the sum of several terms of the arithmetic sequence 9, 11, 13…, we get 256. How many terms are added?
Answer:
When 16 is added to the sum of several terms of the arithmetic sequence 9, 11, 13,…, we get 256.
That is, the sum of terms of the arithmetic sequence = 256 – 16 = 240
First term (a) = 9
Common difference (d) = 2
x_n = dn + (f – d)
= 2n + (9 – 2)
= 2n + 7
If the algebraic form of an arithmetic sequence is an + b,
then the sum of first n terms of the sequence = a\(\frac {n}{2}\)(n + 1) + nb
= 2\(\frac {n}{2}\)(n + 1) + n × 7
= n(n + 1) + 7n
= n² + n + 7n
= n² + 8n
⇒ n² + 8n = 240
⇒ n² + 8n – 240 = 0
a = 1, b = 8, c = -240

n is the number of terms, so it cannot be negative.
Thus, n = 12
12 terms are added.

Students often refer to SSLC Maths Textbook Solutions and Class 10 Maths Chapter 8 Tangents Important Extra Questions and Answers Kerala State Syllabus to clear their doubts.

SSLC Maths Chapter 8 Tangents Important Questions and Answers

Tangents Class 10 Extra Questions Kerala Syllabus

Tangents Class 10 Kerala Syllabus Extra Questions

Question 1.
Choose the correct answer from the options given below.
(i) The distance between two parallel tangents on a circle is 8 cm. What is the radius of the circle?
(a) 4 cm
(b) 2 cm
(c) 5 cm
(d) 8 cm
Answer:
(a) 4 cm
Since the tangent is perpendicular to the radius, the distance between the parallel tangents is the diameter of the circle.

(ii) The length of the tangent to a circle from an outer point and the radius of the circle are equal to 12 cm. What is the distance from the center to the outer point?
(a) 12 cm
(b) 12√3 cm
(c) 12√2 cm
(d) 6 cm
Answer:
(c) 12√2 cm
Use the property of a 45° – 45° – 90° triangle.

(iii) PA, PB are the tangents from P to the circle. If ∠APB = 40°, then what is the measure of ∠AOB?

(a) 140°
(b) 120°
(c) 150°
(d) 110°
Answer:
(a) 140°
Tangents from the outer point to the circle and radii form a cyclic quadrilateral.

(iv) PA, PB are the tangents from P to the circle. If ∠APB = 40°, then what is the measure of angle ACB?

(a) 50°
(b) 60°
(c) 70°
(d) 40°
Answer:
(c) 70°
\(\frac {1}{2}\) × ∠AOB = ∠ACB

(v) P is a point at a distance of 13 cm from the centre of a circle of radius 5 cm. What is the length of the tangents drawn from P to the circle?
(a) 12 cm
(b) 16 cm
(c) 18 cm
(d) 610 cm
Answer:
(a) 12 cm

(vi) In the figure AB = AC = 12 cm. If AP = 4 cm, then what is the length of BC?

(a) 14 cm
(b) 18 cm
(c) 10 cm
(d) 16 cm
Answer:
(d) 16 cm
BP = CQ = 8
BR = CR = 8
BC = 16 cm

(vii) In the circumcircle of triangle PQR, the tangent at P makes 40° with PQ. What is x?

(a) 40°
(b) 50°
(c) 80°
(d) 20°
Answer:
(a) 40°

(viii) The vertices of a square are on a circle. BP is the tangent to the circle at P. What is the measure of ∠PBC?

(a) 50°
(b) 40°
(c) 45°
(d) 30°
Ans:
(c) 45°

(ix) What is the radius of the incircle of an equilateral triangle with a side of 6 cm?
(a) √3
(b) √2
(c) 1
(d) 3
Answer:
(a) √3

(x) The number of tangents that can be drawn from a point outside a circle to the circle.
(a) 3
(b) 1
(c) 2
(d) 4
Answer:
(c) 2

(xi) From a point, the angle between the tangents drawn to a circle is 90°. Find the angle between the lines joining this point to the points of contact.
(a) 90°
(b) 60°
(c) 100°
(d) 80°
Answer:
(a) 90°

Question 2.
Read the two statements given below.
Statement 1: If the angle between the tangents from an outer point is 90 + x, then the angle between the radii to the point where the tangents touch the circle is 90 – x.
Statement 2: Tangents from the outer point and radii to the points where they touch the circle make a cyclic quadrilateral.
Choose the correct answer from those given below.
(a) Statement 1 is true, and Statement 2 is false.
(b) Statement 1 is false, and Statement 2 is true.
(c) Both statements are true, and Statement 2 is the correct reason for Statement 1.
(d) Both statements are true, but Statement 2 is not the correct reason for Statement 1.
Answer:
(c) Both statements are true, and Statement 2 is the correct reason for Statement 1.

Question 3.
Read the two statements given below.
Statement 1: If the area and perimeter of a triangle are numerically equal, that is, both are the same number, then its inradius is 2 units.
Statement 2: If A is the area, s is half of the perimeter, then the inradius r = \(\frac {A}{s}\)
Choose the correct answer from those given below.
(a) Statement 1 is true, and Statement 2 is false.
(b) Statement 1 is false, and Statement 2 is true.
(c) Both statements are true, and Statement 2 is the correct reason for Statement 1.
(d) Both statements are true, but Statement 2 is not the correct reason for Statement 1.
Answer:
(c) Both statements are true, and Statement 2 is the correct reason for Statement 1.

Question 4.
In the figure, PA is the tangent from P to A on the circle, and OA is the radius. If OP = 18 and ∠OPA = 40° then
[sin 40° = 0.64, cos 40° = 0.76, tan 40° = 0.83]

(a) What is the length of the tangent?
(b) What is the radius of the circle?
Answer:
(a) cos 40° = \(\frac{A P}{O P}\)
⇒ 0.76 = \(\frac{A P}{18}\)
⇒ AP = 0.76 × 18 = 13.68 cm
(b) sin 40° = \(\frac{O A}{O P}=\frac{O A}{18}\)
⇒ OA = 11.52 cm

Question 5.
Angles of triangle OAP are in the ratio 1 : 2 : 3, O is the center of the circle, and PA is the tangent from P. ∠P is the smallest angle, and OP = 12 cm.

(a) What is the ratio of the sides of this triangle?
(b) Find the other two sides of the triangle?
Answer:
(a) Angles are x, 2x, 3x
6x = 180°
x = 30°
Angles are 30°, 60°, 90.
Ratio of the sides is 1 : √3 : 2
(b) OA = 6, PA = 6√3

Question 6.
In the figure, O is the centre of the circle, PA is the tangent, and OP = 24 cm.

(a) What are the angles of a triangle?
(b) What is the radius of the circle?
(c) Find the length of the tangent.
Answer:
(a) ∠POA = 180 – 120 = 60°
∠PAO = 90°
∠OPA = 30°
(b) 12 cm
(c) 12√3 cm

Question 7.
In the figure, a circle touches the sides of a triangle. If AP = 1, BQ = 2 and CR = 3 then

(a) What is the perimeter of the triangle?
(b) What is the area of the triangle?
Answer:
(a) AR = 1, CQ = 3, BP = 2
Perimeter of triangle ABC = 12 cm
(b) The perpendicular sides of the right-angled triangle ABC are 3 and 4 cm.
Area = \(\frac {1}{2}\) × 3 × 4 = 6 sq.cm

Question 8.
In the figure, PA = 12 cm, O is the center of the circle, radius of the circle is 5 cm.

(a) What is the length PB?
(b) Find the area of APBO.
Answer:
(a) PB = 12
(b) Area of APBO = 2 × \(\frac {1}{2}\) × 12 × 5 = 60 sq. unit

Question 9.
In the figure, PA and QB are parallel tangents. Another line PQ touches the circle at R and cuts the parallel tangents.

(a) Draw a rough diagram and join OA, OR, and OB.
(b) Name the equal triangles in the figure.
(c) Find the measure of ∠POQ.
Answer:
(a) Draw the diagram yourself.
(b) PA = PR, OA = OR and OP is common.
Triangle PAO and triangle PRO are equal.
Similarly, ∆QRO and ∆QBO are equal.
(c) If ∠POA = ∠POR = x and ∠QOR = ∠QOB = y then 2x + 2y = 180
⇒ x + y = 90
∴ ∠POQ = 90°

Question 10.
In the figure, AB is a common tangent to the circle. The line PC joins the common point of the circle to the point P. It is also a tangent to the circle.

(a) Name the lines of equal length shown in the figure.
(b) If ∠PAC = x and ∠PBC = y then what is ∠ACB?
(c) Prove that ∆ABC is a right triangle.
Answer:
(a) PA = PC, PB = PC
(b) ∠ACB = x + y
(c) In triangle ABC,
2x + 2y = 180
⇒ x + y = 90
Triangle ABC is a right triangle.

Question 11.
In triangle ABC. O is the centre of the circumcircle. ∠BOC = 140°.

(a) What is the measure of ∠BAC?
(b) If PC is the tangent at C, then what is the measure of ∠BCP?
Answer:
(a) ∠BAC = 70°
(b) 70°

Question 12.
ABCDE is a regular pentagon. AD and BD are its diagonals. The tangents intersect at a point P.

(a) What is the angle measure of ∠E and ∠C?
(b) Find ∠ADE, ∠BDC?
(c) What is ∠PAB, ∠PBA?
(d) What is the angle measure of APB?
Answer:
(a) ∠E = ∠C = \(\frac {540}{5}\) = 108°
(b) ∠ADE = 36°
∠BDC = 36°
(c) ∠ADB = 108 – 72 = 36°
∠PAB = ∠PBA = 36°
(d) ∠P = 180 – 72 = 108°

Question 13.
In the figure, the circle touches the sides of the quadrilateral.
If AB = 12, CD = 8, AD = 7.

(a) What is the relation between the length of its sides?
(b) Find BC?
Answer:
(a) AB + CD = AD + BC
(b) 12 + 8 = 7 + BC
⇒ BC = 13

Question 14.
A circle touches two sides of the triangle ABC. The sides AB and AC are equal.

(a) Which lengths shown in the figure are equal?
(b) Prove that BR = CR.
Answer:
(a) AP = AQ, BP = BR, CQ = CR
(b) AB = AC
AB – PA = AC – AQ
⇒ BP = CQ
BP = CQ
⇒ BR = CR

Question 15.
In the figure, O is the center of the triangle ABC.
∠OBC = 20° and ∠OCB = 30°

(a) What is ∠BOC?
(b) Find ∠A.
Answer:
(a) ∠BOC = 180 – (20 + 30) = 130°
(b) This is the incircle of the triangle.
To draw it, the angles must be bisected.
Given that ∠B = 40° and ∠C = 60°
Therefore ∠A = 80°

Question 16.
AD and BC are the common tangents to two circles. The centres of the circles are M and N, respectively. ∠APB = 40°.

(a) Which lines shown in the figure are of equal length?
(b) Prove that AD = BC.
(c) Calculate ∠APB and ∠CPD.
Answer:
(a) PA = PB, PD = PC
(b) By adding the equalities,
PA + PD = PB + PC
AD = BC
(c) ∠AMB = 180 – 40 = 140°
∠CND = 180 – 40 = 140°

Question 17.
In the first figure, the line PA is the tangent from point P to the circle.
In the second figure, OP is the diameter of the semicircle, and A is a point on the semicircle.

(a) In both figures, what is the value of ∠OAP?
(b) Draw a circle with radius 3 cm. Mark a point 7 cm away from the centre. Draw tangents from point P to the circle.
(c) Measure the length of the tangent.
Answer:
(a) 90°
(b) Draw a circle with centre O.
Mark a point P at a distance of 7 cm from the centre.
Draw the diameter OP.
The circle is intersected at points A and B.
Draw the lines PA and PB; these are tangents.
(c) Measure and write the length of the tangents.

Students often refer to Kerala Syllabus 10th Standard Maths Textbook Solutions Chapter 13 Statistics Questions and Answers Notes Pdf to clear their doubts.

SSLC Maths Chapter 13 Statistics Questions and Answers

Statistics Class 10 Questions and Answers Kerala State Syllabus

SCERT Class 10 Maths Chapter 13 Statistics Solutions

Class 10 Maths Chapter 13 Kerala Syllabus – Not a Correct Average & Another Average

(Textbook Page No. 279)

Question 1.
The distances covered by Ahirath in long jump practice are 6.10, 6.20, 6.18, 6.20, 6.25, 6.21, 6.15, 6.10 in metres. Find the mean and median. Why is it that there is not much difference between these?
Answer:
Mean = \(\frac{6.10+6.20+6.18+6.20+6.25+6.21+6.15+6.10}{8}\) = 6.17
Arranging them in ascending order,
6.10, 6.10, 6.15, 6.18, 6.20, 6.20, 6.21, 6.25
Median = \(\frac{6.18+6.2}{2}\) = 6.19
The numbers on either side of the middle value are arranged roughly in equal amounts above and below it.
Therefore, the mean and the median are approximately equal to each other.
Also, there are no extremely large or extremely small numbers in the group.

Question 2.
The table below gives the rainfall during the first week of June 2025 in various districts of Kerala.

Calculate the mean and median rainfall in Kerala during this week. Why is the mean less than the median?
Answer:
Mean = \(\frac{108.7+89.4+74.8+72+42.6+35.7+66.4+73.5+69.1+50.5+43.6+93.1+39+37.5}{14}\) = 63.99
Arranging them in ascending order,
35.7, 37.5, 39, 42.6, 43.6, 50.5, 66.4, 69.1, 72.0, 73.5, 74.8, 89.4, 93.1, 108.7
Median = \(\frac{66.4+69.1}{2}\) = 67.75
The mean is the average calculated considering all the numbers.
The sum of the values less than the middle number is smaller, and the sum of the values greater than the middle number is larger.
Therefore, the mean is less than the median.

Question 3.
Prove that for a set of numbers in an arithmetic sequence, the mean and median are equal.
Answer:
When the number of terms is odd, there will be one middle term.
The mean obtained by dividing the sum of all the terms by their number is the same as this middle term.
This is a special property of an arithmetic sequence – that is, the mean and the median are the same.
When the number of terms is even, the terms can be paired from the ends so that each pair has the same sum.
The median is half the sum of the two middle terms, and this value is also equal to the mean.

SCERT Class 10 Maths Chapter 13 Solutions – Frequency and Median

(Textbook Page No. 282)

Question 1.
35 households in a neighbourhood are sorted according to their monthly income in the table below.

Calculate the median income.
Answer:

Total number of families = 35 (odd number)
Median = \(\frac{35+1}{2}=\frac{36}{2}\) = 18th term
Therefore the median = Rs. 2000

Question 2.
The table below shows the workers in a factory sorted according to their daily wages:

Calculate the median daily wage.
Answer:

Total number of workers = 30 (even number)
The terms in the middle are the 15th and 16th terms.
From the table, it is clear that the monthly income of both the 15th and 16th workers is 1000.
Therefore the median = Rs. 1000

Question 3.
The table below gives the number of babies born in a hospital during a week, sorted according to their birth weight.

Calculate the median birth-weight.
Answer:

Total number of babies = 70 (even number)
The terms in the middle are the 35th and 36th terms.
From the table, it is clear that the weight of the babies of both the 35th and 36th babies is 3.000 kg.
Therefore the median = 3.000 Kg

Class 10 Maths Kerala Syllabus Chapter 13 Solutions – Classes and Median

(Textbook Page No. 287-288)

Question 1.
The table shows some households sorted according to their usage of electricity:

Calculate the median usage of electricity.
Answer:

Total number of houses = 39
The 20th house lies in the middle, and the median class is 280 – 300.
If 20 units is equally distributed to 10 houses, then one part equals \(\frac {20}{10}\) = 2
It is assumed that the electricity consumption within the median class is in an arithmetic sequence.
The consumption of 17th house is 280 + 1 = 281 units
f = 281, d = 2
Then the 20th house represents the fourth term.
Hence the median = f + 3d = 281 + 6 = 287 units

Question 2.
The table below shows the children in a class sorted according to their marks in the math exam:

Calculate the median mark of the class.
Answer:

The total number of children = 36
The 18th and 19th marks lie in the middle.
Median class = 20 – 30
If the 10 marks are equally distributed among 10 students, then one part = 1 mark.
The 13th mark is = 20 + \(\frac {1}{2}\) = 20\(\frac {1}{2}\)
f = 20.5 and d = 1
x₆ = f + 5d = 20.5 + 5 = 25.5
x₇ = 26.5
∴ Median = 26

Question 3.
The table below gives the details of the income tax paid by the employees in an office in a year:

Calculate the median income tax paid.
Answer:

Here, n = 92
The 46th and 47th incomes lie in the middle, and the median class is Rs. 4000 – 5000
If Rs. 1000 is distributed equally among 20 people, then one part = Rs. 50
The 34th income is 4000 + 25 = 4025
f = 4025, d = 50
x₁₃ = f + 12d
= 4025 + 12 × 50
= 4025 + 600
= 4625
x₁₀ = 4625 + 50 = 4675
Median = \(\frac{4625+4675}{2}\) = 4650

Statistics Class 10 Notes Pdf

Class 10 Maths Chapter 13 Statistics Notes Kerala Syllabus

Introduction
In this unit, we discuss certain measures that are based on numerical data collected from a group. These measures reflect the general characteristics of the group. In your previous classes, you have already learned about the measure called the mean. The mean of a group of numbers is obtained by dividing the sum of the numbers by the total number of values in the group.
The marks obtained by a student in seven examinations are given below.
10, 18, 14, 11, 17, 11, 15
Mean = \(\frac{10+18+14+11+17+11+15}{7}\) = 13.7
Therefore, the mean of the marks is 13.7.
When the marks are arranged in order, the number that comes in the middle is called the median. Since there are seven numbers here, there will definitely be one number exactly in the middle. The order is 10,11, 11,14, 15, 17, 18. The fourth number comes in the middle, so the median is 14. One important thing to note is that when the numbers are arranged in order and form an arithmetic sequence, the mean and the median will be equal.

Median is the middlemost number when the given set of values is arranged in ascending or descending order.

Steps to calculate the median of a given set of values.

• Step 1: Arrange the given set of values in ascending order
• Step 2: Find the total number of values in the set. This is denoted by “n”.
• Step 3: Identify the middle position(s).
• Step 4: Calculate the middle value (That is, the median)

If n is odd, then there will be only one middle value, and it is the median.

If n is even, then there will be two middle values, and the average of these two values is the median.

In statistics, frequency simply means how many times something shows up in a set of data.

How to write a cumulative frequency column in a frequency table?

• In the first row, write the first frequency.
• In the second row, write the sum of the first and second frequencies.
• In the third row, write the sum of the first, second, and third frequencies, and so on.

Steps to calculate the median from a frequency table.

• Step 1: Write the cumulative frequency table.
• Step 2: Find the total number of values in the set. This is denoted by “n”.
• Step 3: Identify the middle position (s).
• Step 4: Calculate the middle value (That is the median!)

Steps to calculate the median from a frequency table with classes.

• Step 1: Write the cumulative frequency table.
• Step 2: Find the total number of values in the set. This is denoted by “n”.
• Step 3: Identify the middle position(s).
• Step 4: Identify the class where the middle value(s) belong. (This class is known as the median class.)
• Step 5: Find out the class width and number of values in this class.
• Step 6:Divide the class width by the number of values in this class and find its half.
• Step 7: Calculate the value corresponding to the first observation in the median class.
• Step 8: Calculate the median value.

Not A Correct Average & Another Average
The purpose of calculating the mean is to reduce a whole collection of numbers to a single number, which gives a general understanding of a situation. But numbers in the collection that are very much less or very much more than others affect the mean a lot.

Median: The median is the middlemost number when the given set of values is arranged in ascending or descending order.

Question 1.
The median of the first n odd numbers?
(a) 2n
(b) n²
(c) 3n
(d) n
Answer:
Sum of the first n odd numbers = n²
It is an arithmetic sequence.
Mean = \(\frac{n^2}{n}\) = n
Median = n

Question 2.
The median of the first n even numbers
(a) n + 1
(b) n
(c) n – 1
(d) 2n + 1
Answer:
Sum = n(n + 1)
Median = \(\frac{n(n+1)}{n}\) = n + 1

Question 3.
The algebraic form of an arithmetic sequence is 3n + 2.
(a) What is the 11th term?
(b) What is the median of the first 21 terms?
Answer:
(a) x₁₁ = 3 × 11 + 2 = 33 + 2 = 35
(b) The median of 21 terms is the 11th term.
Median = 35

Question 4.
A company has 10 workers. Among them, three earn a daily wage of Rs. 500 each, and the remaining workers earn Rs. 800 each.
(a) What is the median daily wage?
(b) How many workers earn less than the median wage?
Answer:
(a) When arranged in ascending order, the middle numbers are the 5th and 6th terms, both of which are 800.
Hence the median = Rs. 800
(b) 3 workers earn less than the median wage.

Question 5.
25 numbers are written in a sequence. They form an arithmetic progression, and the median of the numbers is 36.
(a) What is the 13th number?
(b) What is the sum of the smallest and the largest numbers?
(c) When written in order, what is the sum of the numbers on either side of the middle number?
Answer:
(a) The 25 terms are in an arithmetic progression.
Its middle term is its median.
The middle term = 13th term
Therefore median = 36
(b) The sum of the smallest and the largest number = 2 × 13th term
= 2 × 36
= 72
(c) The sum of the numbers on either side of the middle number is 12th term + 14th term = 2 × 13th term
= 2 x 36
= 72

Question 6.
The temperatures recorded in a town over seven consecutive days are as follows:
26°C, 28°C, 25°C, 30°C, 27°C, 26°C, 25°C
(a) What is the median temperature?
(b) How many days have temperatures higher than the median temperature, and how many have lower temperatures?
(c) How many temperatures are lower than the median temperature?
Answer:
(a) Arrange the temperatures in ascending order.
25, 25, 26, 26, 27, 28, 30
There are 7 terms, and the middle term is the 4th term.
Therefore the median = 4th term = 26° C
(b) Higher temperature = 3 days
Lower temperature = 2 days
(c) Temperature lower than the median = 2 temperatures

Frequency and Median
In statistics, frequency simply means how many times something shows up in a set of data.
The marks obtained by 40 students of a class in a test are shown in the table below.
9 students scored 7 marks.
Similarly, 10 students scored 11 marks, 4 students scored 13 marks, 13 students scored 15 marks, and 4 students scored 19 marks.
This table is already arranged in ascending order of scores.
up to 7 marks, there are 9 students
up to 11 marks, 9 + 10 = 19 students
up to 13 marks, 19 + 4 = 23 students
up to 15 marks, 23 + 13 = 36 students
and up to 19 marks, all 40 students.
Let’s now represent this in a table.

Total score = 40
Here, the middle terms are 20 and 21, and the scores of both are 13.
Therefore the median = 13

Question 1.
The scores obtained by 40 students of a class in a quiz competition are given below.

(a) Using a suitable table, calculate the total marks obtained by the class.
(b) What is the mean of the marks?
(c) Find the median mark.
(d) Calculate the number of students who scored higher than the median mark.
Answer:
(a) Total Score = 4 × 5 + 6 × 10 + 9 × 10 + 10 × 7 + 15 × 8
= 20 + 60 + 90 + 70 + 120
= 360
(b) Mean = \(\frac {360}{40}\) = 9
(c) Consider the table

The total number of students is 40.
Therefore, the 20th and 21st terms lie in the middle.
From the 16th student up to the 25th student, the score is 9.
Hence, the median = 9.
(d) The number of students who scored above the median is 15.

Question 2.
The weights of 12 members of a team are given below:

(a) Prepare a table for calculating the median.
(b) What is the median of the weights?
(c) How many members have having median weight and below?
(d) How many members are there above the median weight?
Answer:
(a)

n = 12 (even)
Therefore 6th and 7th members come in the middle.
From the table, it is clear that a member weighs 70.
Median weight is 70.
(c) 7 members are weighing the median weight.
(d) There are 5 members with a median.

Question 3.
The daily wages of 200 workers in a factory are given below.

(a) Prepare the table for calculating the median.
(b) Find the median wage.
(c) How many workers are getting a median wage and below?
(d) How many workers are getting above the median wage?
Answer:
(a)

(b) n = 200 (even)
So the 100th and 101st wage comes in the middle.
From the table, it is clear that the wages of both workers are 500.
Median is 500.
(c) 134 workers have having daily wage below 500.
(d) There are 66 workers having wages above 500.

Classes and Median
The given data can be divided into different groups, and the number of items in each group can be counted and arranged in a table. Such a table is called a frequency table. A frequency table has two columns – the first column shows the groups, and the second column shows the frequency.

Let us calculate the median from the table given below:
The table shows the ages of employees and the number of people working in an organization.

First, we prepare a table by arranging the given data in ascending order.

There are a total of 46 workers.
The 23rd and 24th workers’ ages fall in the middle, and they belong to the 40 – 45 group.
The 40 – 45 group is called the median class.
It is assumed that the ages in the median class are in arithmetic sequence.
If the 5-year interval of the median class is divided equally for 10 workers, that is, 29 – 19 = 10
Then the share for one person is 5 ÷ 10 = \(\frac {1}{2}\) year.
The age of the 20th worker is 40 + (\(\frac {1}{2}\)) ÷ 2 = 40\(\frac {1}{4}\) years.
If we take the 20th worker’s age as the first term and the common difference as \(\frac {1}{2}\), then the age of the 23rd worker will be the fourth term of that arithmetic sequence.
f = 40\(\frac {1}{4}\), d = \(\frac {1}{2}\),
x₄ = f + 3d
= 40\(\frac {1}{4}\) + 3 × \(\frac {1}{2}\)
= 41\(\frac {3}{4}\)
The age of the 24th worker = \(41 \frac{3}{4}+\frac{1}{2}=42 \frac{1}{4}\) years.
The median is the average of the ages of the 23rd and 24th workers.
Median = \(\left(41 \frac{3}{4}+42 \frac{1}{4}\right) \div 2\) = 42

Students often refer to Kerala Syllabus 10th Standard Maths Textbook Solutions Chapter 12 Solids Questions and Answers Notes Pdf to clear their doubts.

SSLC Maths Chapter 12 Solids Questions and Answers

Solids Class 10 Questions and Answers Kerala State Syllabus

SCERT Class 10 Maths Chapter 12 Solids Solutions

Class 10 Maths Chapter 12 Kerala Syllabus – Pyramids and Area

(Textbook Page No. 259-260)

Question 1.
A square of side 5 centimetres, and four isosceles triangles of base 5 centimetres and height 8 centimetres, are to be put together to make a square pyramid. How many square centimetres of paper are needed?
Answer:
Area of one lateral face = \(\frac {1}{2}\) × 8 × 5 = 20 sq.cm
Lateral surface area = 4 × 20 = 80 sq.cm
Base area = 5² = 25 sq.cm
Area of the paper required to make the square pyramid = 80 + 25 = 105 sq.cm

Question 2.
A toy is in the shape of a square pyramid of base edge 16 centimetres and slant height 10 centimetres. What is the total cost of painting 500 such toys, at 80 rupees per square metre?
Answer:
Area of one lateral face = \(\frac {1}{2}\) × 16 × 10 = 80 sq.cm
Lateral surface area = 4 × 80 = 320 sq. cm
Base area = 16² = 256 sq. cm
Total surface area = 320 + 256 = 576 sq. cm
Area of 500 such toys = 500 × 576
= 288000 sq. cm
= 28.8 sq. cm
(1 metre = 100 cm, 1 sq.m = 100 × 100 = 10000 sq.cm)
Total cost of paining 500 toys = 80 × 28.8 = 2,304 rupees

Question 3.
The lateral faces of a square pyramid are equilateral triangles and the length of a base edge is 30 centimetres. What is its surface area?
Answer:
Length of all edges are 30 cm.
Lateral faces are equilateral triangles.
Height of the triangle = 15√3 cm
Area = \(\frac {1}{2}\) × 30 × 15√3 = 225√3 sq.cm
Lateral surface area = 4 × 225√3 = 900√3 sq.cm
Base area = 30² = 900 sq.cm
Total surface area = 900√3 + 900 sq.cm

Question 4.
The perimeter of the base of a square pyramid is 40 centimetres, and the total length of all its edges is 92 centimetres. Calculate its surface area.
Answer:
Length of base edge = \(\frac {40}{4}\) = 10 cm
Total length of all edges is the sum of the lengths of the 4 base edges and the 4 lateral edges.
40 + 4 × lateral edge = 92
⇒ 4 × lateral edge = 92 – 40 = 52 cm
⇒ Lateral edge = \(\frac {52}{4}\) = 13 cm

Slant height = \(\sqrt{13^2-5^2}\) = 12 cm
Total surface area of the pyramid = 10² + 4 × \(\frac {1}{2}\) × 10 × 12
= 100 + 240
= 340 sq. cm

Question 5.
Can we make a square pyramid with the lateral surface area equal to the base area?
Answer:
Assume that the lateral surface area is equal to the base area.
Then the area of each triangle formed by drawing diagonals of the base is equal to the area of each lateral face.
In such cases, no square pyramids can be formed.

SCERT Class 10 Maths Chapter 12 Solutions – Height and Slant Height

(Textbook Page No. 261-262)

Question 1.
Using a square and four triangles with dimensions as specified in the picture, a pyramid is made. What is the height of this pyramid?

What if the squares and triangles are like this?

Answer:
(a) Height = \(\sqrt{18^2-12^2}\)
= √180
= \(\sqrt{36 \times 5}\)
= 6√5 cm

(b) Base edge = 24 cm
So, base diagonal = 24√2 cm
Half of the base diagonal = 12√2 cm
Height = \(\sqrt{30^2-(12 \sqrt{2})^2}\) = √612 cm

Question 2.
A square pyramid of base edge 10 centimetres and height 12 centimetres is to be made of paper. What should be the dimensions of the triangles?
Answer:
Slant height = \(\sqrt{5^2+12^2}\) = 13
Lateral faces are isosceles triangles. There are 4 triangles.
The base side of each triangle is 10 cm, and the height to that side is 13 cm.
The other two sides are equal.
So, the other sides = \(\sqrt{13^2+\left(\frac{10}{2}\right)^2}\)
= \(\sqrt{13^2+5^2}\)
= √194 cm

Question 3.
Prove that in any square pyramid, the squares of the height, slant height, and lateral edge are in arithmetic sequence.
Answer:
If we subtract the square of height from the square of slant height, we get the square of half of the base edge.
slant height² – height² = \(\left(\frac{\text { base edge }}{2}\right)^2\)
Also, if we subtract the square of the height from the square of the lateral edge, we get the square of half of the base edge.
lateral edge² – height² = \(\left(\frac{\text { base edge }}{2}\right)^2\)
These square differences are the common difference of the arithmetic sequence.
∴ The squares of the height, slant height, and lateral edge are in an arithmetic sequence.

Question 4
A square pyramid is to be made with the triangle shown here as a lateral face. What would be its height? What if the base edge is 40 centimetres instead of 30 centimetres?

Answer:
Half of the base edge = 15 cm
Lateral edge = 25 cm
Slant height = \(\sqrt{25^2-15^2}\)
= √400
= 20 cm
Height of the pyramid = \(\sqrt{20^2-15^2}\)
= √175
= \(\sqrt{25 \times 7}\)
= 5√7 cm

Class 10 Maths Kerala Syllabus Chapter 12 Solutions – Volume of a Pyramid

(Textbook Page No. 263)

Question 1.
What is the volume of a square pyramid of base edge 10 centimetres and slant height 15 centimetres?
Answer:
Form a right triangle joining half of the base edge and the slant height.
Height = \(\sqrt{15^2-5^2}\)
= √200
= 10√2 cm
Volume = \(\frac {1}{3}\) × Base area × Height
= \(\frac {1}{3}\) × 10² × 10√2
= \(\frac{1000 \sqrt{2}}{3}\) cm³

Question 2.
Two square pyramids have the same volume. The base edge of one is half that of the other. How many times is the height of the second pyramid the height of the first?
Answer:
Let the base edge of the first pyramid be a and its height be h.
Then, volume = \(\frac {1}{3}\) × Base area × height
= \(\frac {1}{3}\) × a² × h
Let the base edge of the second pyramid be \(\frac {a}{2}\) and its height be x.
Then, volume = \(\frac{1}{3} \times\left(\frac{a}{2}\right)^2 \times x\)
⇒ \(\frac{1}{3} \times a^2 \times h=\frac{1}{3} \times\left(\frac{a}{2}\right)^2 \times x\)
⇒ a²h = \(\frac{a^2 x}{4}\)
⇒ h = \(\frac {x}{4}\)
⇒ x = 4h
∴ The height of the second pyramid is 4 times the height of the first.
OR
One-third of the square of the base edge multiplied by the height gives the volume.
If the base edge becomes half, then its square becomes one-fourth.
Then, to get the same volume, the height must become 4 times.

Question 3.
The base edges of two square pyramids are in the ratio 1 : 2 and their heights in the ratio 1 : 3. The volume of the first is 180 cubic centimetres. What is the volume of the second?
Answer:
Let base edges be a, 2a, and heights be h, 3h
Volume of the first pyramid = \(\frac {1}{3}\) × a² × h = 180
⇒ a²h = 540
Volume of the second pyramid = \(\frac {1}{3}\) × (2a)² × 3h
= 4a²h
= 4 × 540
= 2160 cm³
OR
The ratio of base edges is 1 : 2.
Therefore, the ratio of the base areas is 1:4.
The ratio of heights is 1 : 3.
Therefore, the ratio between the product of base area and height is 1:12.
That is, the ratio between their volumes is 1 : 12.
The volume of the second pyramid is 12 times the volume of the first pyramid.
∴ Volume of the second pyramid = 12 × 180 = 2160 cm³

Question 4.
All edges of a square pyramid are 18 centimetres. What is its volume?
Answer:
Form a right triangle joining half of the base edge = 9 cm, lateral edge = 18 cm, and slant height.
It is a 30° – 60° – 90° triangle.
Slant height = 9√3 cm
Form a right triangle joining the slant height, half of the base edge, and the height.
Height = \(\sqrt{(9 \sqrt{3})^2-9^2}\) = 9√2 cm
Volume = \(\frac {1}{3}\) × 18² × 9√2 = 972√2 cm³

Question 5.
The slant height of a square pyramid is 25 centimetres, and its surface area is 896 square centimetres. What is its volume?
Answer:
If base edge be x, then x² + 4 × \(\frac {1}{2}\) × x × 25 = 896
⇒ x² + 50x – 896 = 0
x² + 50x – 896 = x² + (a + b)x + ab
⇒ a + b = 50, ab = -896
(a – b)² = (a + b)² – 4ab
⇒ (a – b)² = 50² – 4 × -896 = 6084
⇒ a – b = 78
a + b = 50, a – b = 78
⇒ 2a = 128
⇒ a = 64, b = -14
x² + 50x – 896 = (x + 64)(x – 14) = 0
∴ x = 14 is the base edge.
Height = \(\sqrt{25^2-7^2}\) = 24 cm
Volume = \(\frac {1}{3}\) × 14² × 24 = 1568 cm³

Question 6.
All edges of a square pyramid are of the same length, and its height is 12 centimetres. What is its volume?
Answer:
Let a be the length of all edges.
Then, base edge = lateral edge = a
Form a right triangle joining half of the base edge, the lateral edge, and the slant height.
It is a 30° – 60° – 90° triangle.
Slant height = \(\frac{a \sqrt{3}}{2}\) cm
Form a right triangle joining the slant height, half of the base edge, and the height.
Height = \(\sqrt{\left(\frac{a \sqrt{3}}{2}\right)^2-\left(\frac{a}{2}\right)^2}\) = 12 cm
⇒ \(\sqrt{\frac{2 a^2}{4}}\) = 12
⇒ \(\frac{a}{\sqrt{2}}\) = 12
⇒ a = 12√2 cm
∴ Volume = \(\frac {1}{3}\) × (12√2)² × 12 = 1152 cm³

Question 7.
What is the surface area of a square pyramid of base perimeter 64 centimetres and volume 1280 cubic centimetres?
Answer:
Base edge = \(\frac {64}{4}\) = 16 cm
Base area =16² = 256 cm²
Let h be the height.
\(\frac {1}{3}\) × 256 × h = 1280
⇒ h = \(\frac{1280 \times 3}{256}\)
⇒ h = 15 cm
Slant height = \(\sqrt{15^2+8^2}\) = 17 cm
Total surface area = 16² + 4 × \(\frac {1}{2}\) × 16 × 17 = 800 cm²

SSLC Maths Chapter 12 Questions and Answers – Cone

(Textbook Page No. 265)

Question 1.
What are the radius of the base and slant height of a cone made by rolling up a sector of central angle 60° cut out from a circle of radius 10 centimetres?
Answer:
\(\frac{60}{360}=\frac{1}{6}\)
Base radius = 10 × \(\frac {1}{6}\) = \(\frac {10}{6}\) = \(\frac {5}{3}\) cm
Slant height = 10 cm

Question 2.
What is the central angle of the sector to be used to make a cone of base radius 10 centimetres and slant height 25 centimetres?
Answer:
Slant height of the cone is the radius of the sector, which is the radius of the circle from which the sector is cut off.
So, radius of the cone = 10 cm
\(\frac{10}{25}=\frac{2}{5}\)
Central angle is \(\frac {2}{5}\) part of 360°.
So, central angle = 360 × \(\frac {2}{5}\) = 144°

Question 3.
What is the ratio of the base radius and slant height of a cone made by rolling up a semicircle?
Answer:
Central angle of the semicircle = 180°, which is \(\frac {1}{2}\) part of 360°.
The radius of the sector (semicircle) is the slant height.
Its \(\frac {1}{2}\) part is the radius of the cone.
The ratio of base radius and slant height is 1 : 2.

Kerala Syllabus Class 10 Maths Chapter 12 Solutions – Curved Surface Area

(Textbook Page No. 267)

Question 1.
What is the area of the curved surface of a cone of base radius 12 centimetres and slant height 25 centimetres?
Answer:
Curved surface area = πrl
= π × 12 × 25
= 300π cm²

Question 2.
What is the surface area of a cone of base diameter 30 centimetres and height 20 centimetres?
Answer:
Base radius = 15 cm
Height = 20 cm
Slant height = \(\sqrt{15^2+20^2}\)
= √625
= 25 cm
Surface area = Base area + Curved surface area
= π × 15² + π × 15 × 25
= 600π cm²

Question 3.
A conical firework is of base diameter 10 centimetres and height 12 centimetres. 10000 such fireworks are to be wrapped in coloured paper. The price of the coloured paper is 2 rupees per square metre. What is the total cost?
Answer:
Slant height = \(\sqrt{12^2+5^2}\) = 13 cm
Surface area of a cone = π × 5² + π × 5 × 13 = 90π cm²
Total surface area of 10000 fireworks = 90π × 10000 = 90π m²
Total cost = 90π × 2
= 180π
= 180 × 3.14
= 565.2 rupees

Question 4.
Prove that for a cone made by rolling up a semicircle, the area of the curved surface is twice the base area.
Answer:
If the radius of the semicircle is r, then the slant height of the cone is r.
Base radius = \(\frac {r}{2}\)
Curved surface area = π × r × \(\frac {r}{2}\) = \(\frac{\pi r^2}{2}\)
Base area = \(\pi \times\left(\frac{r}{2}\right)^2=\frac{\pi r^2}{4}\)
The curved surface area \(\frac{\pi r^2}{2}\) is twice the base area \(\frac{\pi r^2}{4}\).

Class 10 Maths Chapter 12 Kerala Syllabus – Volume of a Cone

(Textbook Page No. 268)

Question 1.
The base radius and height of a cylindrical block of wood are 15 centimetres and 40 centimetres. What is the volume of the largest cone that can be carved out of this?
Answer:
Volume = \(\frac {1}{3}\) × π × 15² × 40 = 3000π cm³

Question 2.
The base radius and height of a solid metal cylinder are 12 centimetres and 20 centimetres. By melting it and recasting, how many cones of base radius 4 centimetres and height 5 centimetres can be made?
Answer:
The number of cones that can be made = Volume of the melted solid metal cylinder ÷ Volume of the cone made by recasting
= \(\frac{\pi \times 12^2 \times 20}{\frac{1}{3} \times \pi \times 4^2 \times 5}\)
= 108

Question 3.
A sector of central angle 216° is cut out from a circle of radius 25 centimetres and is rolled up into a cone. What are the base radius and height of the cone? What is its volume?
Answer:
\(\frac{216}{360}=\frac{3}{5}\)
Central angle of the sector is \(\frac {3}{5}\) part of 360°.
The radius of the cone is \(\frac {3}{5}\) part of the radius of the sector.
Radius of the cone = \(\frac {3}{5}\) × 25 = 15 cm
Slant height = 25 cm
Height of the cone = \(\sqrt{25^2-15^2}\) = 20 cm
Volume = \(\frac {1}{3}\) × π × 15² × 20 = 1500π cm³

Question 4.
The base radii of two cones are in the ratio 3 : 5 and their heights are in the ratio 2 : 3. What is the ratio of their volumes?
Answer:
Radii are 3r, 5r, and heights 2h, 3h
Ratio of their volume = \(\frac {1}{3}\) × π × (3r)² × 2h : \(\frac {1}{3}\) × π × (5r)² × 3h
= 18 : 75
= 6 : 25

Question 5.
Two cones have the same volume, and their base radii are in the ratio 4 : 5. What is the ratio of their heights?
Answer:
Let their heights be h₁, h₂
\(\frac{1}{3} \times(4 r)^2 \times h_1=\frac{1}{3} \times(5 r)^2 \times h_2\)
⇒ 16h1 = 25h2
⇒ \(\frac{h_1}{h_2}=\frac{25}{16}\)
∴ Ratio of height = 25 : 16

SCERT Class 10 Maths Chapter 12 Solutions – Sphere

(Textbook Page No. 270)

Question 1.
The surface area of a solid sphere is 120 square centimetres. If it is cut into two halves, what would be the surface area of each hemisphere?
Answer:
4πr² = 120
⇒ πr² = 30
Surface area of hemisphere = 3 × πr²
= 3 × 30
= 90 cm²

Question 2.
The volumes of two spheres are in the ratio 27 : 64. What is the ratio of their radii? And the ratio of their surface areas?
Answer:
Volume of a sphere = \(\frac{4}{3} \pi r^3\)
In volume radius is in the third degree.
27 = 3³, 64 = 4³
So, the ratio of their radii = 3 : 4
Surface area of a sphere is 4πr².
In surface area, radius is in the second degree.
So, ratio of their surface area = 3² : 4² = 9 : 16

Question 3.
The base radius and length of a metal cylinder are 4 centimetres and 10 centimetres. If it is melted and recast into spheres of radius 2 centimetres each, how many spheres can be made?
Answer:
Number of spheres = \(\frac{\text { Volume of cylinder melted }}{\text { Volume of sphere formed by recasting }}\)
= \(\frac{\pi \times 4^2 \times 10}{\frac{4}{3} \pi \times 2^3}\)
= 15

Question 4.
A metal sphere of radius 12 centimetres is melted and recast into 27 small spheres of equal size. What is the radius of each small sphere?
Answer:
Volume of the small sphere = \(\frac{\text { Volume of melted sphere }}{27}\)
= \(\frac{\frac{4}{3} \pi \times 12^3}{27}\)
= \(\frac {4}{3}\) × π × 64
= \(\frac {4}{3}\) × π × 4³
So, the radius of the small sphere is 4 cm.

Question 5.
From a solid sphere of radius 10 centimetres, the largest cone of height 16 centimetres is carved out. What fraction of the volume of the sphere is the volume of the cone?
Answer:

From the figure,
10² = 6² + r²
⇒ r = 8 cm
Volume of the cone = \(\frac {1}{3}\) × π × 8² × 16
Volume of the sphere = \(\frac {4}{3}\) × π × 10³
∴ Ratio of Volumes = \(\frac{\frac{1}{3} \pi \times 8^2 \times 16}{\frac{4}{3} \pi \times 10^3}=\frac{32}{125}\)

Question 6.
A solid sphere is cut into two hemispheres. From one, a square pyramid, and from the other a cone, each of maximum possible size, are carved out. What is the ratio of their volumes?
Answer:
Consider the figure,

The figure shows the base of a square pyramid and a cone.
From the figure,
Base edge of the square pyramid = √2r
Base radius of the cone is r.
The height of both the square pyramid and the cone is equal to the radius r.
Volume of the square pyramid = \(\frac{1}{3} \times(\sqrt{2} r)^2 \times r\)
Volume of the cone = \(\frac {1}{3}\) × π × r² × r
Ratio of volumes = \(\frac{1}{3} \times(\sqrt{2} r)^2 \times r: \frac{1}{3} \pi \times r^2 \times r\) = 2 : π

Solids Class 10 Notes Pdf

Class 10 Maths Chapter 12 Solids Notes Kerala Syllabus

→ Surface area of a square pyramid = base area + area of lateral faces

→ Volume of a square pyramid = \(\frac {1}{3}\) × base area × height

→ We can make a cone by rolling up a sector of a circle.

→ Radius of the sector = Slant height of the cone.

→ Curved surface area of a cone is the area of the sector used to make it.

→ Curved surface area (CSA) = πrl, where, r = base radius of the cone, l = slant height of the cone.

→ Volume of a cone = \(\frac {1}{3}\) × base area × height

→ The surface of a sphere is curved. The distance from the centre of the sphere to its surface is the radius of the sphere.

→ Surface area of a sphere = 4πr²

→ Surface area of a hemisphere = 3πr²

→ Volume of a sphere is \(\frac{4}{3} \pi r^3\)

→ Volume of a hemisphere is \(\frac{2}{3} \pi r^3\)

Introduction
Have you ever wondered how much ice cream your favourite cone can hold or how the ancient Egyptians built such massive pyramids? The answer lies in understanding solids, the three-dimensional shapes that surround us in the real world. In this chapter, we’ll embark on a journey into the fascinating world of 3D, focusing on three specific shapes: cones, square pyramids, and spheres.

Square Pyramids
Just think about the Egyptian pyramids. Each one has four triangle-shaped walls that meet at a point. We will learn how to spot square pyramids and what makes them unique, such as their height and lateral lines. We will also learn how much space they can hold (volume) and how much “skin” they have (surface area).

Cones
Think of your favourite ice cream cone. The top is pointy, and the bottom is round. We’ll learn how to recognise cones, understand their parts, like the slant and bottom, and figure out how much ice cream or juice they can hold.

Spheres
Imagine a ball – it’s perfectly round, no matter where you look at it. Spheres are cool shapes found in many things, like planets and marbles. We’ll learn all about spheres and discover how much space they hold (volume) and how much “skin” they have (surface area). We will examine the distinct characteristics of every form, mastering the skills to recognise them, compute their volume-the quantity of space they occupy and even investigate their surface area, which is the sum of their faces. Get ready to discover the mysteries behind these fascinating 3D wonders as you buckle in!

Solids with a base and a sharp apex opposite to it are called pyramids. Its base must be a polygon. The pyramid is named on the basis of the number of sides of its base. If base is a square, then it is a square pyramid. The sides of the polygon forming the base of a pyramid are called base edges, and the other sides of the triangles are called lateral edges. The topmost point of a pyramid is called its apex.

Pyramids and Area
Solids with a base and a sharp apex opposite to it are called pyramids. Its base must be a polygon. The pyramid is named on the basis of the number of sides of its base. If base is a square, then it is a square pyramid. Lateral faces of the pyramid are triangles. It can be an isosceles or an equilateral triangle. One edge of the triangle is the base edge of the pyramid. The other two edges of the triangle are the lateral edges. A solid with a square base and lateral faces is a pyramid. So, we have to calculate the perimeter and area of a square, and the perimeter and area of a triangle.

We can make square pyramids by cutting paper and pasting the edges.

The first picture shows the picture drawn on a paper to make the square pyramid. Cut and paste through the outer edges to make the square pyramid. The second picture shows the completed pyramid. In both pictures, A is the apex of the pyramid. B is the centre of the base edge, and C is the vertex of the base. The length of AB is the slant height of the pyramid. BC is the half of the base edge, and AC is the length of the lateral edge. In both pictures, ABC is a right triangle.

Question 1.
The figure shows the sketch to make a square pyramid from a square paper. The base edge of the pyramid is 10 cm, and the lateral edge is 12 cm.

(a) What is the slant height?
(b) What is the length of the side of the square paper?
Answer:
13² = 5² + (Slant height)²
Slant height = \(\sqrt{13^2-5^2}\)
= √144
= 12 cm
(b) Base edge = \(\frac {40}{4}\) = 10 cm

Question 2.
Divide a 96 cm long rod into 8 equal parts, join the edges to make a square pyramid.
(a) Calculate the slant height.
(b) Find the area of the paper required to cover its lateral face.
Answer:
(a) Length of an edge = \(\frac {96}{8}\) = 12 cm
Lateral faces are equilateral triangles.
Length of one side is 12 cm.
Slant height = 6√3 cm
(b) Area of one lateral face = \(\frac {1}{2}\) × 12 × 6√3 = 36√3 sq.cm
Total lateral surface area is the sum of the area of four triangles = 4 × 36√3 = 144√3 sq.cm

Question 3.
Four equal triangles are cut off from a segment of a circle with a central angle of 240° and radius 12 cm, as shown in the figure. Make a square pyramid with these triangles as lateral faces.
(a) What is the base edge of the pyramid?
(b) What is the slant height?
(c) Find the lateral surface area.
Answer:
Central angle of a segment = \(\frac {240}{4}\) = 60°
All triangles cut off are equilateral triangles.
(a) Length of base edge = 12 cm
(b) Slant height = 6√3 cm
(c) Lateral surface area = 4 × \(\frac {1}{2}\) × 12 × 6√3 = 144√3 sq.cm

Height and Slant Height
The height of a pyramid is the perpendicular distance from the apex to the base. In the case of a square pyramid, the foot of the perpendicular is the point where the diagonals of the base meet.

Both of the shaded triangles in the picture are right triangles.
Height of the pyramid, half of the base edge, and slant height are joined to form a right triangle.
The height of the pyramid, half of the diagonal, and the lateral edge are joined to form a right triangle.
A right triangle is also formed by joining the slant height, half of the base edge, and the lateral edge.

Question 1.
Base perimeter of a square pyramid is 8 metres, length of the lateral edge is 8 metres.
(a) What is the length of the base edge?
(b) Find the base diagonal.
(c) What is the height of the pyramid?
Answer:
(a) Length of base edge = \(\frac {8}{4}\) = 2 m
(b) Base diagonal = 2√2 cm
(c) Half of base diagonal = √2 m
Height = \(\sqrt{8^2-\sqrt{2}^2}\) = √62 m

Question 2.
The base edge of a square pyramid is double the height. If the base perimeter is 40 cm, then
(a) Find slant height.
(b) Find lateral surface area.
Answer:
(a) Let the height be x, then base edge is 2x.
4 × 2x = 40
⇒ x = 5
Height = 5
⇒ Half of the base edge = 5 cm
Slant height = 5√2 cm
(b) Lateral surface area = 4 × \(\frac {1}{2}\) × 10 × 5√2 = 100√2 sq.cm

Question 3.
The base area of a square pyramid is 100 sq cm, slant height is 13 cm.
(a) What is the length of the base edge?
(b) What is the surface area?
(c) What is the height of the pyramid?
Answer:
(a) Length of base edge = √100 = 10 cm
Slant height = 13 cm
(b) Surface area = 100 + 4 × \(\frac {1}{2}\) × 10 × 13 = 360 sq. cm
(c) Height of the pyramid = \(\sqrt{13^2-5^2}\) = √144 = 12 cm

Question 4.
The total surface area of a square pyramid is 360 sq.cm, slant height is 13 cm.
(a) If base edge is x, form the equation.
(b) What is the base edge?
(c) Find the height of the pyramid.
Answer:
(a) Lateral surface area = 4 × \(\frac {1}{2}\) × x × 13 = 26x
⇒ x² + 26x = 360
⇒ x² + 26x – 360 = 0
(b) x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{-26 \pm \sqrt{676+1440}}{2}\)
= 10
(c) Height = 12 cm

Volume of a Pyramid
The volume of a solid is the measure of the space it occupies. The volume of any prism is equal to the product of the base area and the height. But, the volume of a square pyramid is equal to a third of the base area and the height. That is, the volume of a pyramid is a third of the volume of the prism of the same base area and height.
For example, let the base area of a prism be 100 sq cm and height 30 cm.
Then its volume is 100 × 30 = 3000 cubic centimetres.
The volume of a pyramid with the same base area and height is 1000 cubic centimetres.

Question 1.
The base area of a square pyramid is 100 sq cm, and the slant height is 13 cm.
(a) Find the height of the pyramid.
(b) Find the volume of the pyramid.
Answer:
(a) Length of base edge = √100 = 10 cm
Half of the base edge, height, and slant height together form a right triangle.
Height = \(\sqrt{13^2-5^2}\) = 12 cm
(b) Volume = \(\frac {1}{3}\) × base area × height
= \(\frac {1}{3}\) × 100 × 12
= 400 cm³

Question 2.
From a wooden square prism, a square pyramid of the same base area and height is carved out. Base area 400 sq cm and height 24 cm.
(a) What is the volume of the pyramid?
(b) What is the slant height?
(c) Find the total surface area.
Answer:
(a) Volume = \(\frac {1}{3}\) × base area × height
= \(\frac {1}{3}\) × 400 × 24
= 3200 cm³
(b) Base edge = √400 = 20 cm, Height = 24 cm
Slant height = \(\sqrt{10^2+24^2}\)
= √676
= 26 cm
(c) Total surface area = Base area + 4 × area of triangle in one lateral face
= 400 + 4 × \(\frac {1}{2}\) × 20 × 26
= 400 + 1040
= 1440 cm²

Question 3.
The base perimeter and height of a square prism are 32 cm and 3 cm, respectively. From this, a square pyramid of maximum size is carved out.
(a) What is the length of the base edge of the pyramid?
(b) What is the slant height?
(c) Find the total surface area.
(d) Find the volume.
Answer:
(a) Base perimeter = 32 cm
So, base edge = \(\frac {32}{4}\) = 8 cm
(b) Slant height = \(\sqrt{4^2+3^2}\) = 5 cm
(c) Base area = 8² = 64 cm
Total surface area = Base area + 4 × Area of one isosceles triangle
= 64 + 4 × \(\frac {1}{2}\) × 8 × 5
= 64 + 80
= 144 sq. cm
(d) Volume = \(\frac {1}{3}\) × 64 × 3 = 64 cm³

Cone
Like this, we can make a cone by rolling up a sector of a circle.

Basic measures of a sector are the central angle and the radius of the sector. Using these measures, we can calculate the length of the arc and the area. What part of 360° is the central angle of the sector? That much of the perimeter is the length of the arc. That much of the area is the area of the sector. From a circular plate, cut out a semicircle, rolling up to form a cone. Half of 360°, that is 180°, is the central angle of the sector. The base perimeter of the cone formed by rolling up a semicircle is half of the circumference of the circle.

Question 1.
A circular paper of radius 12 cm is cut into 4 equal sectors. If one of these sectors is rolling up to form a cone.
(a) What is the central angle of one sector?
(b) What is the radius and slant height of the cone?
Answer:
(a) 360 × \(\frac {1}{4}\) = 90°
(b) Slant height of the cone = 12 cm
Radius = 12 × \(\frac {1}{4}\) = 3 cm

Question 2.
From a circular plate, a sector of central angle 120° is cut off. The radius of the sector formed by rolling up the sector is 10 cm.
(a) What is the radius of the circular plate?
(b) What is the slant height of the cone?
Answer:
(a) 30 cm
(b) 30 cm

Question 3.
A circular plate of circumference 36π cm is divided into six equal sectors. One of them rolled up to form a cone.
(a) What is the radius of the plate?
(b) What is the radius of the cone?
Answer:
(a) 2πr = 36π
⇒ r = 18 cm
(b) Radius of the cone = \(\frac {18}{6}\) = 3 cm

Curved Surface Area
A sector rolling up to form a cone, the area of the sector is the lateral surface area of the cone. The arc length of the sector is the base area of the cone. The length of the perpendicular from the apex of the cone to the centre of the base is the height. We can see a right triangle with the slant height of the cone as the hypotenuse, base radius, and height as perpendicular sides.

Question 1.
A circular plate of radius 20 cm is divided into 4 sectors. One of them is rolled up to form a cone.
(a) What is the slant height?
(b) What is the radius of the cone?
(c) What is the height of the cone?
Answer:
(a) 20 cm
(b) Central angle of the sector is \(\frac {1}{4}\) part of 360°.
Radius of the cone is 5 cm.
Height = \(\sqrt{20^2-5^2}\)
= √375
= 5√15 cm

Question 2.
From a square prism of edge length 30 cm and height 20 cm, a cone of maximum size is carved out.
(a) What is the base radius of the cone?
(b) What is the slant height of the cone?
Answer:
(a) Half of the base edge is the radius of the base circle.
Radius = 15 cm
(b) Slant height = \(\sqrt{20^2+15^2}\)
= √625
= 25 cm

Question 3.
A circular plate of radius 12 cm is divided into two sectors. Its central angles are 120° and 240°. Both sectors rolled up to form cones.
(a) What is the common measure for both cones?
(b) What is the radius of the cone formed using a sector of a central angle?
(c) What is the radius of the cone formed using a sector of central angle?
(d) Which cone is highest?
Answer:
(a) Common measure is the slant height.
(b) \(\frac{120}{360}=\frac{1}{3}\)
\(\frac {1}{3}\) × 12 = 4 cm
(c) \(\frac{240}{360}=\frac{2}{3}\)
\(\frac {2}{3}\) × 12 = 8 cm
(d) Slant height of both cones is are same.
Slant height² = radius² + height²
∴ If the radius increases, the height decreases. The second cone is the shortest.

Question 4.
Find the lateral surface area of a cone of base perimeter 24π and height 5 cm.
Answer:
2π × radius = 24π
⇒ radius = 12 cm
Height = 5 cm
Slant height = \(\sqrt{12^2+5^2}\) = 13 cm
Radius of the sector used to form the cone = 13 cm
The area of the sector is \(\frac {12}{13}\) part of the area of the circle. So is the lateral surface area.
∴ Lateral surface area = π × 13² × \(\frac {12}{13}\) = 156π cm²

Question 5.
If the radius of a cone is r and the slant height is l, then prove that the lateral surface area is πrl.
Answer:
A sector rolled up to form a cone.
If the radius of the sector is l (which can be the slant height of the cone), the area of the circle is πl2.
Circumference is 2πl.
The central angle of the sector is x.
\(\frac{x}{360}=\frac{p}{2 \pi l}\)
p is the length of the arc.
Length of the arc, p = 2πl × \(\frac {x}{360}\), which is the base perimeter of the cone.
2πl × \(\frac {x}{360}\) = 2πr, r is the radius of the cone.
From this, lx = 360r
Lateral surface area is the area of the sector.
∴ Lateral surface area = πl² × \(\frac {x}{360}\)
= \(\frac{\pi \times l \times l x}{360}\)
= \(\frac{\pi \times l \times 360 r}{360}\)
= πrl sq. unit

Volume of a Cone
The volume of a cone is the product of base area and height.
The volume of a cone is \(\frac {1}{3}\) part of the volume of a prism with the same base area and height.
That is, the volume of a cone is one-third of the product of base area and height.

Question 1.
The volume of a cylinder is 144π cubic centimetres. What is the volume of a cone with the same base area and height?
Answer:
Volume must be one-third.
Volume = \(\frac {1}{3}\) × 144π = 48π cubic centimetres.

Question 2.
There is a conical vessel of radius 6 cm and slant height 10 cm.
(a) What is the height of this vessel?
(b) Find the volume of the vessel.
(c) How many times is water to be poured to fill a cylindrical vessel of the same radius and twice the height?
Answer:
(a) Height of the vessel = \(\sqrt{10^2-6^2}\) = 8 cm
(b) Volume = \(\frac {1}{3}\) × π × 6² × 8
= 12 × 8 × π
= 96π cubic centimetres.
(c) Volume of the cylindrical vessel with the same radius and twice the height = π × 6² × 16 = 576π cubic centimetres = 96π × 6
Hence, the water must be poured 6 times to fill the cylindrical vessel.

Question 3.
The volume of a cone is 60 cm³. Find the volume of a cone with the same height and half the radius.
Answer:
If the radius becomes half, the volume becomes \(\frac {1}{4}\)th.
Volume = \(\frac {1}{4}\) × 60 = 15 cm³

Question 4.
Radii of two cones are in the ratio 1 : 2 and the ratio of their heights is 2 : 1. Then, find the ratio of their volumes.
Answer:
Radii are r, 2r, and heights are 2h, h.
Then, volume = \(\frac{1}{3} \pi r^2 \times 2 h: \frac{1}{3} \pi \times(2 r)^2 \times h\)
= 2r²h : 4r²h
= 1 : 2

Sphere
The surface of a sphere is curved. The distance from the centre of the sphere to its surface is the radius of the sphere.
Consider a sphere of radius 4 cm. Four circular papers of radius 4 cm can cover the whole surface of the sphere.
If the radius is 4 cm, the surface area of a sphere is 4πr².
The surface area of a sphere is equal to the square of its radius multiplied by 4π.
A solid sphere cut into two equal halves exactly through the middle, we get two hemispheres.
It has one curved surface and one flat circular face.
Curved surface area of a solid sphere is 2πr², and its total surface area is 3πr².
Volume of a sphere is \(\frac{4}{3} \pi r^3\) and volume of a hemisphere is \(\frac{2}{3} \pi r^3\).

Question 1.
Find the radius of a sphere with volume is equal to its surface area. What is the surface area of this sphere?
Answer:
4πr² = \(\frac{4}{3} \pi r^3\)
⇒ r = 3
Surface area = 4π × 3² = 36π sq.unit

Question 2.
If the radius of a sphere becomes twice, what is the change in its volume?
Answer:
To calculate surface area, we have to multiply the square of its radius by 4π.
If the radius becomes twice, the surface area becomes 4 times.
Volume of a sphere is \(\frac{4}{3} \pi r^3\).
So, if the radius becomes twice, then the volume becomes 2³ = 8 times.

Question 3.
How many solid spheres of radius 1 cm can be formed by melting and recasting a solid metal sphere of radius 4 cm?
Answer:
Number of Spheres = \(\frac{\text { Volume of larger sphere }}{\text { Volume of smaller sphere }}\)
= \(\frac{\frac{4}{3} \pi \times 4^3}{\frac{4}{3} \pi \times 1^3}\)
= 64

Question 4.
Find the surface area of the sphere formed by joining two solid hemispheres of surface area 75π cm². What is the radius of the sphere? Find the volume.
Answer:
3πr² = 75π
⇒ r² = 25
⇒ r = 5 cm
Surface area of the sphere = 4π × 5² = 100π cm²
Volume of the sphere = \(\frac {4}{3}\) × π × 5³ = \(\frac{500 \pi}{3}\) cm³

Question 5.
What is the slant height of a cone of maximum size that can be carved out from a solid hemisphere of radius 12 cm? Find surface area and volume.
Answer:

Slant height = 12√2 cm
Surface area = π × 12² + π × 12 × 12√2 = 144π + 144√2π
Volume = \(\frac {1}{3}\) × π × 12² × 12
= 4 × 144π
= 576π cm³

Students rely on Geography Class 10 Notes Kerala Syllabus Chapter 8 Towards Sustainability to help self-study at home.

Class 10 Geography Chapter 8 Important Questions and Answers Towards Sustainability

Kerala Syllabus Class 10 Social Science Geography Chapter 8 Towards Sustainability Important Questions

Towards Sustainability Class 10 Important Questions

Question 1.
Which of the following best defines a resource?
a) Anything that is man-made and used for trade
b) Anything that satisfies human needs, occurs naturally, is technologically feasible, and culturally acceptable
c) Anything found in nature without human use
d) Anything created only for industrial purposes
Answer:
b) Anything that satisfies human needs, occurs naturally, is technologically feasible, and culturally acceptable

Question 2.
Identify the examples of Natural resources from the following.
A. Air, Water
B. Electricity. Machines
C. Soil, Sunlight
D. Vehicles, Buildings

a) A and C are correct
b) A and D are correct
c) B and D are correct
d) A and B are correct
Answer:
a) A and C are correct

Question 3.
Arrange the table suitably.

A. Biotic Resources	i. Minerals, Air
B. Abiotic Resources	ii. Humans, Plants
C. Renewable Resources	iii. It is formed over millions of years and whose reserves decrease with use.
D. Non- renewable Resources	iv. It is considered sustainable as long as they are not overused.

a) A – iii, B – i, C – iv, D – ii
b) A – ii, B – i , C – iv, D – iii
c) A – iv , B – iii , C – ii, D – i
d) A – i, B – iv, C – ii, D – iii
Answer:
b) A – ii, B – i, C – iv, D – iii

Question 4.
Choose the correct answer by reading the given assertion and reason.
Assertion: Based on the origin, resources are classified into two types: Biotic and Abiotic Resources.
Reason: Humans, Plants, and animals are examples of abiotic resources.
a) Assertion is correct, Reason is wrong
b) Both assertion and reason are correct
c) Both assertion and reason are wrong
d) Assertion is wrong, Reason is correct
Answer:
a) Assertion is correct, Reason is wrong

Question 5.
Which of the following is not a basis for classifying resources?
a) Origin
b) Temperature
c) Renewability
d) Ownership
Answer:
b) Temperature

Question 6.
Define Stocks.
Answer:
Materials that can meet human needs but for which humans do not have the appropriate technology to utilize are called stocks.

Question 7.
What is natural resources?
Answer:
Natural resources are substances, objects, or forms of energy that are obtained from nature and are useful to humans.

Question 8.
Find the non-ferrous minerals from the following,
a) Iron
b) Manganese
c) Gold
d) Nickel
Answer:
c) Gold

Question 9.
Assertion (A): Coal is widely used for generating electricity in thermal power plants.
Reason (R): Coal is rich in carbon and is a major source of heat energy when burnt.
a) Both A and R are true, and R is the correct explanation of A.
b) Both A and R are true, but R is not the correct explanation of A.
c) A is true, but R is false.
d) A is false, but R is true.
Answer:
a) Both A and R are true, and R is the correct explanation of A.

Question 10.
Assertion (A): Petroleum is called ‘black gold’.
Reason (R): Petroleum is a naturally occurring liquid mixture that is very valuable and used for various purposes like fuel, medicines, and waterproofing.
a) Both A and R are true, and R is the correct explanation of A.
b) Both A and R are true, but R is not the correct explanation of A.
c) A is true, but R is false.
d) A is false, but R is true.
Answer:
a) Both A and R are true, and R is the correct explanation of A.

Question 11.
Which of the following is correct for non- conventional resources?
A. These resources get depleted and exhausted with use.
B. Eco-friendly
C. Availability is limited
D. Available in plenty

a. A and C are correct
b. A and B are correct
c. B and C are correct
d. C and D are correct
Answer:
a. A and C are correct

Question 12.
Arrange the following suitably.

A. Natural Gas	i. Non-conventional resources
B. Bio energy	ii. Mineral
C. Coal	iii. Conventional resources
D. Aluminum	iv. Fossil Fuel

A. A – iv, B – ii, C – i, D – iii
B. A – i, B – ii, C – iv, D – iii
C. A – iii, B – i, C – iv, D – ii
D. A – ii, B – i, C – iv, D – iii
Answer:
C. A – iii, B – i, C – iv, D – ii

A. Natural Gas	iii. Conventional resources
B. Bio energy	i. Non-conventional resources
C. Coal	iv. Fossil Fuel
D. Aluminum	ii. Mineral

Question 13.
Based on their replenishment capacity or renewability, resources can be classified into two types. Which are they?
Answer:

1. Renewable Resources
2. Non-renewable Resources

Question 14.
Write examples for developed resources and Potential resources.
Answer:

1. Developed Resources: Coal mines, Petroleum wells, Irrigated agricultural land, etc.
2. Potential Resources: Rajasthan and Gujarat have a huge potential for generating energy from wind and solar power, but these sources are currently only marginally used for power generation.

Question 15.
Define Potential Resources and Developed Resources.
Answer:

1. Potential Resources are resources that are found in a region but are not yet fully utilised.
2. Developed Resources are resources whose quantity and quality have been surveyed, assessed, and adequate technology has been developed to utilise them efficiently.

Question 16.
Differentiate between Biotic Resources and Abiotic Resources.
Answer:

1. Biotic Resources: Anything that is part of the biosphere and can be utilised by humans is referred to as a biotic resource.
2. Abiotic Resources: Resources that are made of non-living matter are called abiotic resources.

Question 17.
Complete the table by writing the name of the country by the leading producer of the given Mineral.

Mineral	Leading producer (Country)
Copper	…………. (a) ……………
Iron	…………….(b) ……………..

Answer:
a. Chile
b. Carajas mine (Brazil)

Question 18.
Define the terms energy source and write some examples of energy source.
Answer:
Any substance that can produce heat, move objects, or generate electricity is called an energy source. A substance that releases energy through a chemical process is known as a fuel.
Eg; Solar energy, Coal, Hydel power, petroleum.

Question 19.
a) What is meant by fossil fuel?
b) List the major fossil fuel?
Answer:
a) Fossil fuels belong to the category of non-renewable resources, but they are very important sources of energy. Fossil fuels are formed from the fossilized remains of ancient plants and animals.

b) Coal, petroleum

Question 20.
Write the year in which following report were submitted.
a. Brundtland Commission report: …………………….
b. Earth Summit: ………………………
Answer:
a. 1987
b. 1992

Question 21.
Define the term Sustainable development?
Answer:
Sustainable development is development that meets the needs of the present without compromising the ability of future generations to meet their own needs. We can move forward by engaging in activities that aim to achieve the objectives of sustainable development, protect the Earth, and conserve resources for a better tomorrow.

Question 22.
Write any three examples for Man-made Natural Resources.
Answer:

1. Electricity
2. Vehicles
3. Machines

Question 23.
Write a note on Non-metallic minerals with its example.
Answer:
Non-metallic minerals generally do not have the metallic luster that is typical of metals. They are mined for various industrial purposes. These minerals are used either in their natural state, as raw materials after purification, or in combination with other minerals. Gypsum, limestone, kaolin, and graphite are some of the major non-metallic minerals.

Question 24.
Explain the importance and occurrence of minerals. How are natural resources classified based on their distribution? Give examples.
Answer:
Minerals are naturally occurring substances with definite chemical and physical properties, and they play an important role in people’s welfare, scientific progress, and a country’s economic growth.

They are typically chemical compounds, and many metallic elements can be profitably extracted from mineral-rich rocks known as ores.

Based on distribution, natural resources are classified into:

• Ubiquitous resources – found everywhere (e.g. air, solar energy, water)
• Localised resources – found only in specific areas (e.g. metal ores, fossil fuels)

Question 25.
Why is petroleum called as ‘rock oil’?
Answer:
Petroleum

• It is not known exactly when humans first began using petroleum. In ancient times, people considered the fire produced from natural gas seeping through surface pores to be sacred and worshiped it.
• Asphalt was used to waterproof boats and to regulate indoor temperatures during cold seasons.
• In ancient Egypt, it was also used in pyramid construction and for preserving mummies.
• As it originates from rocks, this fuel is called petroleum, meaning ‘rock oil’.
• The naturally occurring, complex liquid mixture called petroleum is also etymologically referred to as ‘black gold’.

With the Industrial Revolution, petroleum gained importance as it became a primary fuel for machines, replacing coal. Since petroleum became available from the Middle East at very low cost, it has significantly changed human life. Today, petroleum, which remains the primary source of energy, has a profound influence on human life.

• Naturally occurring petroleum is usually black or brown in colour, though it can also appear green or yellow.
• Crude petroleum, a mixture of various components obtained from the Earth, is sent to a refinery to be converted into fuel.
• Petrol, jet fuel, lubricants (used to reduce friction), asphalt, and many other products are extracted from crude oil.
• Thousands of products, including paints, medicines, synthetic fibers, fertilizers, and cosmetics, are produced from petroleum today.

All fossil fuels are nonrenewable energy sources – their reserves will diminish with use and eventually disappear from the Earth forever. Historical records indicate that coal was the first fossil fuel used in the world. It was commonly used as fuel in furnaces to extract metals from their ores. Petroleum began to be used for commercial purposes in the 19th century, and its usage has been rising ever since. Natural gas is another fossil fuel that is widely used today, especially in the domestic sector. If fossil fuels become scarce, all developmental activities will come to a standstill. The uncontrolled use of these non-renewable energy sources could lead to a major energy crisis in the future. It means we need to develop energy sources that can be regenerated. So we must rely bn non-conventional energy sources that are abundantly available and can last for a long time.

Question 26.
How is wind energy converted into electricity?
Answer:
The kinetic energy of the wind is used to produce electricity. The force of the wind spins the blades of a windmill, which is connected to a generator. As the windmill rotates, the generator produces electricity from the mechanical energy.

Question 27.
How is solar energy harnessed and used?
Answer:
The Sun is the primary source of energy, and various techniques are employed to convert solar rays into usable energy. Photovoltaic and solar thermal methods are the primary techniques for harnessing solar energy. The energy obtained is stored in photovoltaic cells and used for various purposes such as heating, lighting, and generating electricity.

Question 28.
Describe the distribution and importance of iron ore in the world.
Answer:
MINERAL RESOURCES

• Minerals are naturally occurring organic or inorganic substances with definite chemical and physical properties.
• They are chemical compounds in which atoms are arranged in a specific manner. Most minerals are
  formed by the combination of two or more elements, while only a few elements occur in their pure (elemental) state in nature.
• The welfare of the people, the development of science and technology, and the economic growth of a country depend, to some extent, on the availability of minerals.
  Natural resources are classified into two types based on their distribution: ubiquitous resources and localised resources. Ubiquitous resources are those found everywhere. Examples: air, solar energy, water etc. Localised resources are those found only in certain parts of the world. Examples: metal ores, fossil fuels, etc.

• The Earth is rich in rocks that contain many minerals with important metallic elements.
  Metallic elements can be extracted from these rocks at a relatively low cost. Ores are Earth materials from which one or more metals can be extracted profitably. A metal can be extracted from more than one type of ore. For example, iron can be extracted from minerals such as hematite and magnetite. Similarly, more than one metal can also be obtained from a single ore deposit. Our Earth is rich in a variety of metals.

The metals we commonly use and their important ores are given below:

When beneficial minerals found as natural occurrences in unusually high concentrations, they can be termed as mineral deposits.

Mineral Deposits
Iron

• Carajas Mine the largest iron ore mine in the world. This mine is located in Brazil, a country in South America.
• The largest reserves of iron ore are found in countries like Australia, Brazil, and Russia.
• India is also one of the major iron-producing countries in the world.
• Iron is a key component in the manufacturing of steel.
• About 98% of the iron ore mined worldwide is used to produce steel.
• Nearly 50 countries are involved in iron ore mining. Steel plays a crucial role in maintaining a strong industrial base.

Question 29.
Why is the conservation of natural resources important? Explain how it can be achieved.
Answer:
We depend on the Earth for all essential resources, but overuse due to population growth and high demand is depleting them rapidly. Natural resources are dwindling, and if this continues, future generations may face shortages. Conservation means protecting the Earth to maintain its self- regenerative capacity and using resources wisely without harming ecosystems or the global climate. To achieve this, we must develop renewable energy sources, such as solar, wind, hydroelectric, and geothermal energy, while reducing the use of fossil fuels, which are limited and polluting.

Question 30.
Explain the uses and distribution of copper. Mention the importance of other useful metals.
Answer:
Early humans used copper for decoration and making tools and weapons. Today, copper is essential in the electrical industry, used for wires, motors, transformers, and generators. Chile is the world’s leading producer of copper, with major mines like Escondida and Collahuasi.
Other important metals include:

• Aluminium – used in wires, vehicles, aeroplanes, cans, and buildings.
• Manganese – important for metal processing and steel production.
• Lead – used in batteries and ceramics.
• Chromium – vital for making stainless steel.

Question 31.
Mention any four differences between conventional and non-conventional energy resources.
Answer:

Conventional (Non-renewable)	Non-Conventional (Renewable)
Limited supply, exhaustible	Available in abundance
Cause pollution	Eco-friendly, clean
Expensive to extract & maintain	Low maintenance cost
Examples: Coal, petroleum	Examples: Solar, wind, tidal

Question 32.
Explain geothermal energy and its importance as an alternative energy source.
Answer:

• Geothermal energy is produced from the heat inside the Earth.
• When magma erupts from the Earth’s interior, the released heat is stored and converted into electricity.
• Thermal energy can also be generated from hot water and steam that come out through geysers.
• It is a clean, renewable, and reliable source of energy, making it an important alternative to fossil fules.

Question 33.
Explain how the excessive use of non-renewable resources affects the environment and economy.
Answer:

• Overuse of non-renewable resources like coal and petroleum causes air, water, and soil pollution.
• It leads to climate change and the depletion of natural reserves.
• Extraction and transport are expensive, increasing economic burden.
• Their scarcity can lead to energy crises, affecting industries, transport, and daily life.
• Therefore, using renewable sources ensures sustainability and economic stability.

Question 34.
Describe coal and explain its formation, characteristics, and major uses.
Answer:
Coal

• Coal is generally the rock formations that originate from plant remains and has a high carbon content. These are carbon-rich sedimentary deposits found in black or brown colour.
• Coal is used as a fuel in metallurgy (the process of extracting metals from their ores and refining them for use).
• It is also used as a chemical raw material in the manufacturing of waxes, ointments, medicines, pesticides, and dyes.
• Coal is a major source of energy for thermal power generation.

Question 35.
Explain the various renewable energy sources and their uses.
Answer:
NON-CONVENTIONAL ENERGY SOURCES

Question 36.
How can resources be classified on the basis of their ownership?
Answer:
RESOURCES BASED ON OWNERSHIP
Resources can also be classified into different types based on their ownership.

Question 37.
How metals can be classified as Metallic and Non- metallic minerals.
Answer:
MINERAL RESOURCES

• Minerals are naturally occurring organic or inorganic substances with definite chemical and physical properties.
• They are chemical compounds in which atoms are arranged in a specific manner. Most minerals are
  formed by the combination of two or more elements, while only a few elements occur in their pure (elemental) state in nature.
• The welfare of the people, the development of science and technology, and the economic growth of a country depend, to some extent, on the availability of minerals.
  Natural resources are classified into two types based on their distribution: ubiquitous resources and localised resources. Ubiquitous resources are those found everywhere. Examples: air, solar energy, water etc. Localised resources are those found only in certain parts of the world. Examples: metal ores, fossil fuels, etc.

• The Earth is rich in rocks that contain many minerals with important metallic elements.
  Metallic elements can be extracted from these rocks at a relatively low cost. Ores are Earth materials from which one or more metals can be extracted profitably. A metal can be extracted from more than one type of ore. For example, iron can be extracted from minerals such as hematite and magnetite. Similarly, more than one metal can also be obtained from a single ore deposit. Our Earth is rich in a variety of metals.

The metals we commonly use and their important ores are given below:

When beneficial minerals found as natural occurrences in unusually high concentrations, they can be termed as mineral deposits.
Mineral Deposits
Iron

• Carajas Mine the largest iron ore mine in the world. This mine is located in Brazil, a country in South America.
• The largest reserves of iron ore are found in countries like Australia, Brazil, and Russia.
• India is also one of the major iron-producing countries in the world.
• Iron is a key component in the manufacturing of steel.
• About 98% of the iron ore mined worldwide is used to produce steel.
• Nearly 50 countries are involved in iron ore mining. Steel plays a crucial role in maintaining a strong industrial base.

Copper

• Copper was often found in a relatively pure state in nature; it was easy to extract and use.
• Early humans used copper for decorative purposes and for making tools and weapons.
• Today, copper has become an indispensable metal in the electrical industry, as it is used in making wires, electric motors, transformers, and generators.
• Chile is the world’s leading producer of copper.

In addition to iron and copper, there are many other metals that are very useful to humans. Aluminium is used for making electrical wires, motor vehicles, airplanes, cans, and buildings. Manganese is indispensable in metal processing and steel production. Lead is used in making batteries and ceramic products. Chromium is important for the production of stainless steel.

Non-metallic minerals: It is generally do not have the metallic luster that is typical of metals. They are mined for various industrial purposes. These minerals are used either in their natural state, as raw materials after purification, or in combination with other minerals. Gypsum, limestone, kaolin, and graphite are some of the major non-metallic minerals.