Class 9

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Kerala State Syllabus 9th Standard Hindi Solutions Unit 1 Chapter 1 पुल बनी थी माँ (कविता)

पुल बनी थी माँ Textual Questions and Answers

पुल बनी थी माँ आशयग्रहण के प्रश्न

प्रश्ना 1.
‘पुल बनी थी माँ’ से क्या तात्पर्य है?

उत्तर:
पुल दो किनारों को आपस में जोड़ता है। माँ परिवार के हर सदस्य को आपस में जोड़नेवाली कड़ी थी। इसलिए माँ को पुल कहा गया है।

प्रश्ना 2.
‘बुढ़ा रही है माँ’ इसका आशय क्या है?

उत्तर:
प्रस्तुत पंक्ति का आशय यह है कि माँ के शरीर पर बुढ़ापे का असर दिखने लगा। वह शारीरिक और मानसिक रूप से कमज़ोर होने लगी।

प्रश्ना 3.
‘माँ आख़िर माँ ही तो है’ इससे आपने क्या समझा?

उत्तर:
इसका मतलब है कि माँ का मातृत्व बच्चों की कठिनाइयों को अच्छी तरह जानता है। अर्थात् माँ अपने बच्चों के बारे में सबकुछ जानती है।

पुल बनी थी माँ Text Book Activities

पुल बनी थी माँ अभ्यास के प्रश्न

प्रश्ना 1.
बेटों का जीवन बेरोकटोक चलती गाड़ी के समान रहा।

उत्तर:
दौड़ती रहती थी बेधड़क
बिना किसी हरी लाल बत्ती के
हम लोगों की छुक छुक छक छक

प्रश्ना 2.
माँ की देख-भाल की ज़िम्मेदारी बेटों पर आ गई।

उत्तर:
हाथों हाथ रहती माँ
एक दिन हमारे कंधों में आ गई

प्रश्ना 3.
बेटे अपने दायित्व बदलते रहे।

उत्तर:
जब तक जीवित रही माँ।
हम बदलते रहे अपने कंधे

प्रश्ना 4.
माँ के चले जाने से बेटे बेसहारे बने।

उत्तर:
और माँ के कंधों से उतरते ही
उतर गए हमारे कंधे

पुल बनी थी माँ विधात्मक प्रश्न

प्रश्ना 1.
कविता का परिचय देते हुए टिप्पणी लिखें।

उत्तर:
पुल बनी थी माँ : बदलते पारिवारिक संबंधों की आलोचना
कविता ‘पुल बनी थी माँ’ बूढ़े-बुजुर्गों के प्रति उत्तरदायित्वों से विमुख होती जा रही नई पीढ़ी के व्यवहार को दर्शाता है। कविता में माँ के पुल होने और पुल से बोझ बनने की हालत पर चर्चा की गई है।

माँ भाइयों के बीच पुल बनी थी। पुल दो किनारों को आपस में जोड़ता है। माँ परिवार के हर सदस्य को आपस में जोड़नेवाली कड़ी रही। इस माँ रूपी पुल से बच्चों की जिंदगी रूपी रेल गाड़ी बेरोकटोक चलती रही। पिता के चल बसने के बाद भी भाइयों के बीच माँ पुल बनी रही। माँ धीरे-धीरे टूटने लगी। यानी मानसिक रूप से वह धीरे-धीरे कमज़ोर होती गई। उसके शरीर पर बुढ़ापे का असर दिखने लगा। वह शारीरिक रूप से भी
कमज़ोर होने लगी थी। एक ही बात को माँ बार-बार कहने लगी। बच्चे इस आदत को उनके बढ़ते हुए बुढापे की निशानी मानकर जीने लगे। उसकी आवाज़ कमज़ोर होती रही। वह धीरे-धीरे दुर्बल होती रही।

बच्चों के प्रति प्यार और दुलार से रहनेवाली माँ एक दिन बच्चों के आश्रय में आ गईं। धीरे-धीरे बच्चों के सशक्त कंधों में माँ बोझ बन गईं। जब तक बूढ़ी माँ जीवित रही, बच्चे माँ की देखरेख की ज़िम्मेदारी एक दूसरे के कंधों पर डालते रहे। सारी जिंदगी बच्चों के लिए जीनेवाली माँ बुढ़ापे में बच्चों के लिए भार बन गई। पर माँ का मातृत्व बच्चों की इस कठिनाई को सह नहीं पाया। वह स्वयं उनके कंधों से उतर गई मतलब उसका अंतिम प्रयाण हो गया। माँ के अभाव में बच्चे बेसहारे बन गए। कविता में प्रयुक्त शब्द, कथन और मुहावरे- वृषभ कंधा, कंधा बदलना, उतर गए कंधे आदि कविता को और सशक्त बनाया है।

पुल बनी थी माँ Summary in Malayalam and Translation

पुल बनी थी माँ शब्दार्थ

Students rely on Kerala Syllabus 9th Standard Chemistry Textbook Solutions Chapter 1 Structure of Atom Notes Questions and Answers English Medium to help self-study at home.

Kerala SCERT Class 9 Chemistry Chapter 1 Solutions Structure of Atom

Kerala Syllabus Std 9 Chemistry Chapter 1 Structure of Atom Notes Solutions Questions and Answers

Class 9 Chemistry Chapter 1 Let Us Assess Answers Structure of Atom

Question 1.
Some observations related to experiments on cathode rays are given. Write the inference based on each observation.
a. A paddle wheel placed in the path of cathode rays rotates.
b. A shadow is formed if an object is placed in the path of cathode rays.
c. When an electric field is applied perpendicular to the path of cathode rays, the rays deflect towards the positive plate.
Answer:
a. A paddle wheel placed in the path of cathode rays rotates shows that the particles of cathode rays have mass

b. A shadow is formed if an object is placed in the path of cathode rays, indicating that the cathode
rays travel in a straight line.

c. When an electric field is applied perpendicular to the path of cathode rays, the rays deflect towards the positive plate because the cathode rays are composed of negatively charged particles (electrons)

Question 2.
The atomic number of an atom is 16 and mass number is 32.
Answer:
a. How many electrons, protons and neutrons are present in this atom?
b. Write the electron configuration of this atom.
c. Draw the orbit electron configuration of this atom.
Answer:
a) Atomic number of atom = 16
Mass number of atom = 32
Number of protons = Atomic number = 16
No: of electrons = 16
No: of neutrons = Mass number – Atomic number = 32 – 16 = 16

b. Electronic configuration of atom = 2, 8, 6

c. Orbit electronic configuration of the atom

Question 3.
Electrons are present in the K, L and M shells of an atom.
a. Which of these shells has the highest energy?
b. If M shell contains only 3 electrons, write the atomic number of this atom.
c. What is the number of electrons in this atom?
d. If the nucleus of this atom contains 16 neutrons, what is its mass number?
Answer:
a. M shell has maximum energy in K, L, and M shells

b. M shell contains 3 elecrons, indicates that inner shells are filled. i.e., K and L shells contain maximum of 2 and 8 electrons respectively.
Total number of electrons = 2 + 8 + 3 = 13
Atomic number = no: of protons = no: of electrons in a neutral atom = 13

c. No: of electrons in the atom = sum of electrons in each shell = 2 + 8 + 3= 13

d. No: of neutrons = 16
Mass number = No: of protons + No: of neutrons = 13 + 16 = 29

Question 4.
The orbit electron configuartion of an atom is given below.

a. What is the mass number of this atom?
b. Write its electron configuration.
Answer:
a. From the figure
No: of protons = 13
No: of neutrons = 14
Mass number of atom = No: of protons + No: of neutrons = 13 + 14 = 27

b. From the figure
No: of electrons = 13
Electron configuration = 2, 8, 3

Question 5.
The symbols of some elements are given.
\({ }_12^{24} \mathrm{Mg}\), \({ }_6^{12} \mathrm{C}\), \({ }_7^{15} \mathrm{N}\), \({ }_6^{14} \mathrm{C}\), \({ }_11^{24} \mathrm{Na}\)
a. Select a pair of isotopes from the given elements. Write the reason for selecting it.
b. Select a pair of isobars from the given elements.
Answer:
a. \({ }_6^{12} \mathrm{C}\) and \({ }_6^{14} \mathrm{C}\)C are the pair of isotopes. Isotopes have same atomic number and different mass number. They are atoms of same element with different mass number.

b. \({ }_12^{24} \mathrm{C}\)Mg and \({ }_11^{24} \mathrm{C}\)Na are the pair of isobars. Isobars are the atoms of different elements with same mass number and different atomic number.

Question 6.
Match the items in column A & B suitably.

A	B
Plum pudding model	James Chadwick
Planetary model of atom	Goldstein
Canal rays	J.J. Thomson
Neutron	Rutherford

Answer:

A	B
Plum pudding model	J.J. Thomson
Planetary model of atom	Rutherford
Canal rays	Goldstein
Neutron	James Chadwick

Question 7.
The atomic number and mass number of an element are 15 and 31 respectively.
a. What is the number of valence electrons in this atom?
b. How many neutrons are present in this atom?
c. Draw the orbit electron configuration of this atom.
Answer:
a. Atomic number of an atom = No: of protons = No: of electrons = 15
Electron configuration = 2,8,5
Valence electrons No: of electrons in outermost shell = 5

b. Mass number = 31
Mass number = atomic number + No: of nuetrons
No: of neutrons = Mass number – atomic number = 31 – 15=16
c. Orbit electron configuration of the atom

Question 8.
Isotope of an element is used to determine the age of fossils.
a. Which is this isotope?
b. Which are the other two main isotopes of this element?
c. Write the number of neutrons in each isotope.
Answer:
a. \({ }_6^{14} \mathrm{C}\) is the radioactive isotope of carbon used to determine the age of fossils.

b. \({ }_6^{12} \mathrm{C}\) and \({ }_6^{13} \mathrm{C}\) are the other two isotopes of carbon.

c. No: of neutrons in \({ }_6^{14} \mathrm{C}\) isotope = 8
No: of neutrons in \({ }_6^{13} \mathrm{C}\) isotope = 7
No: of neutrons in \({ }_6^{12} \mathrm{C}\) isotope = 6

Extended Activities

Question 1.
Prepare a presentation on scientists connected to the history of atom and their contributions and present it in the classroom.
Answer:
Here are some points for your reference:

• John Dalton – In 1809, found out all substances are made up of very minute particles called atoms.
• Heinrich Geissler – In 1854, developed discharge tubes and vaccum pumps.
• William Crookes – In 1875, passed high voltage electricity through gaes at low pressure in test tube.
• Eugen Goldstein – In 1886, found out canal rays.
• J.J Thomson – In 1906 found out cathode rays. Put forward the plumpudding model of atom
• Henry Becquerel – in 1896 discovered radioactivity.
• Ernest Rutherford – In 1911 conducted gold foil experiment. Found out the nucleus. Put forward the planetary model of atom.
• Neils Bohr – In 1913 put forward the Bohr model of atom.
• James Chadwick – In 1932 discovered neutron.

Question 2.
Prepare a timeline chart on the main events that led to the discovery of different subatomic particles.
Answer:
In 1886 Eugen Goldstein discovered canal rays. From the study of canal rays discovered protons.
Rutherford named the particle proton. In 1906 J.J.Thomson discovered electrons through discharge tube experiments. In 1932 James Chadwick discovered neutrons.

Question 3.
You have learned about isotopes. Find more examples for radio isotopes. Prepare an article on the uses of each radio isotope and publish it in the science magazine. Use word processor for this work.
Answer:

Radioactive isotopes	Uses
Iodine-131	For the study of thyroid glands and treatment
Uranium-235	Used as nuclear fuel
Cobalt-60	For the treatment of cancer
Sodium-24	To find out leakage in industrial pipe lines
Iron-59	To detect anaemia
Carbon-14	To detect the age of fossils
Dueterium	To produce heavy water
Phosphorus-31	As a tracer to find out material transportation in plants

Question 4.
If you get a chance to conduct an interview with Rutherford, what questions would you ask him? Prepare a questionnaire.
Answer:
Here are some questions for your reference:

• Can you remember the teachers who influenced you the most?
• What was your basic subject of interest?
• Which topic did you selected for your research?
• How long did you spend in research?
• Which do you think is your important discovery?

Structure of Atom Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
Analyse table

How do molecules of different substances differ?
Answer:

• Differ in the component elements
• Differ in the ratio of the component elements

Question 2.
What are the important particles in an atom?
Answer:

• Electron
• Proton
• Neutron
  They are called sub-atomic particles.

Question 3.
How was it proved that electrons have mass?
Answer:
When a paddle wheel was placed in the path of cathode rays it rotates. This shows that electrons, the particles in cathode rays have mass.

Question 4.
Cathode rays, cast shadows of opaque objects placed in their path. What can be inferred from this?
Answer:
Cathode rays cast shadows of opaque objects placed in their path, indicating that the cathode rays travel in a straight line.

Question 5.
Some properties of the sub-atomic particles like electron, proton and neutron are given in the table. Complete the following table.

Answer:

Question 6.
Some statements are given. Which of them are related to J.J. Thomson?
a) Proposed the idea of the orbit
b) Conducted discharge tube experiments
c) Discovered neutron
d) Discovered electron
e) Proposed the plum pudding model
Answer:
b) Conducted discharge tube experiments
d) Discovered electron

Question 7.
Prepare a questionnaire about the scientists who conducted research on atomic structure and their contributions. Conduct a quiz programme in your classroom based on this.
Answer:
Here are some questions for your reference

• Who developed discharge tubes and vaccum tubes?
  – Heinrich Geissler
• Who discovered electrons?
  – J.J. Thomson
• Name the scientist who discovered the charge to mass ratio of electrons.
  – J.J. Thomson
• Who discovered canal rays?
  – Eugen Goldstein
• What is the charge of the particles of cathode rays?
  – Negative
• Who found the charge and mass of electron?
  – Robert Millikan
• Scientist who discovered radioactivity?
  – Henry Becquerel
• The Particles that do not deviate in electric and magnetic fields is
  – Neutron

Question 8.
What are the particles in the nucleus of an atom?
Answer:
Protons and neutrons are present in nucleus of an atom.

Question 9.
What is the mass number of an atom having 2 protons and 2 neutrons?
Answer:
No: of neutrons = 2
No: of protons = 2
Mass number of the atom = No: of protons + No: of neutrons = 2 + 2 = 4

Question 10.
Find the number of protons, electrons and neutrons in chlorine and calcium atoms

Answer:
1. Chlorine \({ }_17^{35} \mathrm{C}\)Cl
Atomic number of Cl, Z = 17
Mass number of Cl, A = 35
No: of protons = 17
No: of electrons = 17
Mass number = No: protons + No: of neutrons
∴ No: of neutrons = Mass number – No: of protons A – Z = 35 – 17 18

2. Calcium \({ }_20^{40} \mathrm{C}\)Ca
Atomic number of Ca. Z = 20
Mass number of Ca, A =40.
No: of protons = 20
No: of electrons = 20
Mass number = No: protons + No: of neutrons
∴ No: of neutrons Mass number – No: of protons = A – Z = 40 – 20 = 20

Question 11.
Complete the table given below.

Answer:
Atomic number, Z = No: of protons = No: of electrons
Mass Number, A = No: of protons (Z) + No: of neutrons
No: of neutrons = Mass number – Atomic number = A – Z

1. Hydrogen \({ }_1^{1} \mathrm{H}\)
Atomic number of H, Z = 1
Mass number of H, A =1
‘No: of protons = 1
No: of electrons = 1
Mass number = No: protons + No: of neutrons
∴ No: of neutrons = Mass number – No: of protons = A – Z = 1-1=0

2. Lithium \({ }_3^{7} \mathrm{Li}\)
Atomic number of Li, Z = 3
Mass number of Li. A = 7
No: of protons = 3
No: of electrons = 3
Mass number = No: protons + No: of neutrons
∴, No: of neutrons = Mass number – No: of protons = A – Z = 7 – 3 = 4

3. Oxygen \({ }_8^{16} \mathrm{O}\)
Atomic number of O, Z = 8
Mass number of O, A = 16
No: of protons = 8
No: of electrons = 8
Mass number = No: protons + No; of neutrons
∴, No: of neutrons = Mass number – No: of protons = A – Z = 16 – 8 = 8

4. Sodium Na \({ }_11^{23} \mathrm{Na}\)
Atomic number of Na, Z = 11
Mass number of Na, A = 23
No: of protors = 11
No: of electrons = 11
Mass number = No: protons + No: of neutrons
∴, No: of neutrons = Mass number – No: of protons = A – Z = 23 – 11 = 12

5. Neon \({ }_10^{20} \mathrm{Ne}\)
Atomic number of Ne, Z = 10
Mass number of Ne, A = 20
No: of protons = 10
No: of electrons = 10
Mass number = No: protons + No: of neutrons
∴, No: of neutrons = Mass number – No: of protons = A – Z = 20 – 10 = 10

6. Titanium \({ }_22^{48} \mathrm{Ne}\)
Atomic number of Ti, Z = 22
Mass number of Ti, A =48
No: of protons = 22
No: of electrons = 22
Mass number = No: protons + No: of neutrons
∴, No: of neutrons = Mass number – No: of protons = A – Z = 48 – 22 = 26

7. Uranium \({ }_92^{235} \mathrm{U}\)
Atomic number of U, Z = 235
Mass number of U, A = 92
No: of protons = 92
No: of electrons = 92
Mass number = No: protons + No: of neutrons
∴, No: of neutrons = Mass number – No: of protons = A – Z = 235 – 92 = 143

8. Thorium \({ }_90^{232} \mathrm{Th}\)
Atomic number of Th, Z = 90
Mass number of Th, A = 232
No: of protons = 90
No: of electrons = 90
Mass number = No: protons + No: of neutrons
∴, No: of neutrons = Mass number – No: of protons = A – Z = 232 – 90 = 142

9. Zinc \({ }_30^{65} \mathrm{Th}\)
Atomic number of Zn, Z = 30
Mass number of Zn, A = 65
No: of protons = 30
No: of electrons = 30
Mass number = No: protons + No: of neutrons
∴, No: of neutrons = Mass number – No: of protons = A – Z = 65 – 30 = 35

Question 12.
According to the Bohr atom model, where is the electron situated in an atom?
Answer:
Electrons are situated in orbits of an atom.

Question 13.
What are the symbols given to the energy levels 1,2,3, and 4?
Answer:

n	Energy level
1	K
2	M
3	L
4	N

The arrangement of electrons in an atom is done in accordance with certain laws.

Orbit number (n)	Name	Maximum number of electrons that can be accommodated (2n2)
1	K	2 × l2 = 2
2	L	2 × 22 = 8
3	M	2 × 32 = 18
4	N	2 × 42 = 32
5	O	2 × 52 = 50

1. The maximum number of electrons that can be accomodated in an orbit is 2n2, where n is the Table
2. Normally, filling up of electrons in higher energy orbits will take place only after the lower energy orbits are filled.
3. The maximum number of electrons that can be accomodated in the outermost orbit of an atom is 8.

Question 14.
Let us write the electron configuration of some elements. Complete the following table

Answer:

Orbit Electron Configuration-Diagrammatic Representation
The orbit electron configuration of Hydrogen \({ }_{1}^{1} \mathrm{H}\)
No: of electrons in hydrogen = 1

The orbit electron configuration of Boron \({ }_{5}^{10} \mathrm{H}\)
No: of electrons in Boron = 5

Question 15.
Diagrammatically represent the orbit electron configuration of \({ }_{13}^{27} \mathrm{Al}\)
Answer:
The atomic number of aluminium, Z = 13
Mass number of aluminium, A = 27
Neutrons of aluminium = A – Z = 27 – 13 = 14
The orbit electronic’ configuration of aluminium is

Question 16.
The orbit electron configuration of an atom is given.

Analyse the figure and find the following.
(i) Atomic number
(ii) Number of protons
(iii) Number of neutrons
(iv) Mass number
Answer:
(i) Atomic number is the number of protons in an atom.
Atomic number of given atom, Z = 18
(ii) Number of protons (from the figure) = 18
(iii) Number of neutrons (from the figure) = 22
(iv) Mass number of the atom, A = sum of protons and neutron = 18 +22 = 40
(v) Electron configuration = 2,8,8

Question 17.
Write the electron configuration of elements from atomic number 1 to 18 and represent their shell electron configuration.
Answer:

Question 18.
Number of which sub-atomic particle determines the element? (proton/neutron)
Answer:
Proton

Question 19.
See figure given below.

Complete table regarding these atoms.

Answer:

Question 20.
What is the atomic number of these atoms?
Answer:
Atomic number in these atoms is 1.

Question 21.
Which is the element having atomic number 1?
Answer:
Hydrogen is the element with atomic number 1.
Then, all these three are hydrogen atoms.

Question 22.
In the number of which particle do these atoms differ?
Answer:
Neutrons

Question 23.
Are the mass number of these atoms same?
Answer:
Mass number of these atoms are not the same.

Question 24.
Which of them has no neutron in the nucleus?
Answer:
Protium has no nuetron in it.

Question 25.
These are the isotopes of hydrogen. Can you define an isotope?
Answer:
Isotopes are the atoms of the same element with same atomic number and different mass number. Isotopes exhibits the same chemical properties. But they show slight variations in the physical properties. Heavy water is the oxide of deuterium (an isotope of hydrogen) is used in nuclear reactors.

Question 26.
Let us see whether hydrogen alone has isotopes. See the figure given below.

Answer:

\({ }_6^{12} \mathrm{C}\), \({ }_6^{13} \mathrm{C}\), \({ }_6^{14} \mathrm{C}\)C are the three natural isotopes of carbon. Among which the most stable and most abundant is the \({ }_6^{12} \mathrm{C}\) isotope. The amount of ¹³C among the isotopes of carbon is approximately 1.1 %. This isotope is used to study the metabolic processes in plants and animals. ¹⁴C is a radioactive isotope. This is used to determine the age of fossils.

ISOTOPES	USES
Iodine-131	To study the functioning of thyroid gland and its treatment
Uranium-235	Fuel in nuclear reactors
Cobalt-60	Cancer treatment
Sodium-24	To detect the leakage in industrial pipelines
Iron-59	To diagnose Anaemia

Question 27.
Orbit electron configuration of argon (Ar), potassium (K) and calcium (Ca) is given below. Analyse the figure and complete table.


Answer:
From the figure:

Question 28.
What is the pecularity of the mass numbers of these elements?
Answer:
Mass number of the elements Ar, K and Ca are the same.i.e., A = 40

Question 29.
Are the atomic numbers the same?
Answer:
Atomic numbers are not the same for these elements.
Isobars are atoms having the same mass number and different atomic numbers. They are
atoms of different elements in which the number of nucleons (proton  neutron) is equal.
Atoihs having the same number of neutrons are called isotones. Examples: \({ }_6^{14} \mathrm{C}\) and \({ }_7^{15} \mathrm{C}\)

Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 9 Real Numbers Extra Questions and Answers Notes to clear their doubts.

Kerala SCERT Class 9 Maths Chapter 9 Solutions Real Numbers

Real Numbers Class 9 Kerala Syllabus Questions and Answers

Kerala State Syllabus 9th Standard Maths Chapter 9 Real Numbers Solutions Questions and Answers

Class 9 Maths Chapter 9 Kerala Syllabus – Absolute Value

Textual Questions And Answers

Question 1.
Complete the table below:

i) Expand the table by taking some more pairs x, y of numbers. Do you see any relation between \xy\ and |x||y|?
ii) Prove that |xy| = |x||y| for any two numbers x and y.
Answer:

i) If the table is expanded,

From the table, we get that the values of |xy| and |x||y| are equal,

ii) If both the numbers are positive.
Considering the values as, x = 6 and y = 4
Then, |xy| = xy = |x||y|
|(6) (4)| = 24 = |6||4|

If both the numbers are negative.
Considering the values as, x = -6 and y = -4
Then, |xy| – xy = |x||y|
|(-6)(-4)| = 24= |-6||-4|

If one of the number is positive and the another one is negative.
Considering the values as, x = 6 and y = -4
Then, |xy| = xy = |x||y|
|(6)(-4)| = 24= |6||-4|
Hence it is clear that |xy| = |x||y| for any two numbers x and y

Question 2.
Complete the table below:

Expand the table by taking some more pairs x, y of numbers. Do you see any relation between |x| + |y| and |x + y|?
Answer:

If the table is expanded,

From the table it is clear that the values of |x| + |y| is not equal to |x + y|

Class 9 Maths Kerala Syllabus Chapter 9 Solutions – Distances

Textual Questions And Answers

Question 1.
Find x satisfying each of the equations below
i) |x| = 5
ii) |x – 3 | = 2
iii) |x – 2 | = 3
iv) |x + 2 | = 3
Answer:
i) If | x | = 5
Then the values of x will be 5 and -5
That is, x = 5 or x = -5

ii) |x – 3| = 2
If x > 3
Then |x – 3| = x – 3
That means, x – 3 = 2
x = 2 + 3 = 5
x = 5

If x < 3
Then |x – 3| = 3 – x
That means, 3 – x = 2
-x = 2 – 3 = -1
x = 1
If |x – 3| = 2 then the values of x will be x = 1 or x = 5

iii) |x – 2| = 3
If x > 2
Then, |x — 2 | = x – 2
That means, x – 2 = 3
x = 3 + 2 = 5
x = 5

If x < 2
Then |x – 2| = 2 – x
That means, 2 – x = 3
-x = 3 – 2 = 1
x = – 1
If |x – 2 | = 3 then the values of x will be x

iv) |x + 2 | = 3
If x > 2
Then, |x + 2 | = x – (-2)
That means, x – (-2) = 3
x + 2 = 3
x = 3 – 2 = 1
x = 1

If x < 2
Then, |x + 2 | = |x – (-2)| = (-2) – x
That means, (-2) – x = 3
– x = 3 + 2= 5
x = -5
If |x + 2| = 3 then the values of x will be x = 1 or x = -5

Question 2.
Find between which numbers x should lie to satisfy each of the equations below:
i) |x| < 3
ii) |x – 2 | < 1
iii) |x – 11 < 2
iv) |x + 11 < 2
Answer:
i) If |x| = 3
Then the possible values of x will be 3 and -3
But if |x| < 3, then the possible values of x will be in between 3 and -3
That means, —3 < x < 3

ii) |x – 2| < 1
The point x can either be in the left or right of the point 2, (distance from 2 is less than 1).
If x is in the right side of the point 2, whose distance is 1.
Then x – 2 = 1
x = 1 + 2 = 3
x = 3
If x is in the left side of the point 2, whose distance is 1.
Then, 2 – x = 1
– x = 1 – 2 = – 1 .
x = 1
Therefore, |x – 2| < 1, then the possible values of x will be in between 1 and 3
That is, 1 < x < 3.

iii) |x – 1| < 2
The point x can either be in the left or right of the point 1, (distance from 1 is less than 2).
If x is in the right side of the point 1, whose distance is 2.
Then, x – 1 = 2
x = 2 + 1 = 3
x = 3
If x is in the left side of the point 1, whose distance is 2.
Then, 1 – x — 2
-x = 2 – 1 = 1
x = – 1
Therefore, |x – 1| < 2, then the possible values of x will be in between -1 and 3
That is, – 1 < x < 3.

iv) |x + 1| < 2
The point x can either be in the left or right of the point (-1), (distance from (-1) is less than 2).
If x is in the right side of the point (-1), whose distance is 2.
Then, x – (- 1) = 2 – x + 1 = 2
x = 2 – 1 = 1
x = 1
If x is in the left side of the point (-1), whose distance is 2.
Then, – 1 – x = 2
– x = 2 + 1 = 3
x = – 3
Therefore, |x + 1| < 2, then the possible values of x will be in between -3 and 1
That is, – 3 < x < 1.

Question 3.
Find the integers satisfying each of the equations in problem (2).
Answer:
i) |x| < 3
The possible values of x will be in between 3 and -3
Therefore the integer values of x = -2, -1, 0, 1, 2

ii) |x – 2| < 1
The possible values of x will be in between 1 and 3
Therefore the integer values of x – 2

iii) |x – 1| < 2
The possible values of x will be in between -1 and 3
Therefore the integer values of x = 0, 1, 2

iv) |x + 1| < 2
The possible values of x will be in between -3 and 1
Therefore the integer values of x = -2, -1, 0

SCERT Class 9 Maths Chapter 9 Solutions – Midpoint

Intext Questions and Answers

Question 1.
What number x satisfies the equation |x – 2| = |x – 5|
Answer:
Geometrical meaning of the equation |x – 2| = |x – 5| is that the point marked by x on the number line is at the same distance from the points marked by the numbers 1 and 4
That means, the point marked by x is the midpoint of the points marked by 2 and 5
So, x = \(\frac{1}{2}\) × (2 + 5) = \(\frac{7}{2}\)

Textual Questions And Answers

Question 1.
Find the number which mark the midpoint of the points marked by each pair of numbers given below on the number line:
i) 1, -5
ii) -1, -5
iii) –\(\frac{1}{2}\), –\(\frac{1}{3}\)
iv) –\(\frac{1}{3}\), \(\frac{3}{4}\)
v) -2.5, 3.5
vi) 1.3, 8.7
vii) -√2, -√3
viii) -√3, √7
Answer:

Question 2.
Find the number which mark the points dividing the distance between the points marked by 1 and 2 into four equal parts, on the number line.
Answer:
There should be three points which divides the distance between the points marked by 1 and 2 into four equal parts.
To get the three points we have to find the midpoint of
Midpoint of 1 and 2 = \(\frac{1}{2}\)(1 + 2) = \(\frac{3}{2}\)
Midpoint of 1 and \(\frac{3}{2}\) = \(\frac{1}{2}\)(1 + \(\frac{3}{2}\)) = \(\frac{5}{4}\)
Midpoint of \(\frac{3}{2}\) and 2 = \(\frac{1}{2}\)(\(\frac{3}{2}\) + 2) = \(\frac{7}{4}\)

Question 3.
Find x satisfying each of the equations below:
i) |x – 1| = |x – 3|
ii) |x – 3| = |x – 4|
iii) |x + 2| = |x – 5|
iv) |x| = |x + 1|
Answer:
i) x = \(\frac{1}{2}\)(1 + 3) = \(\frac{4}{2}\) = 2
ii) x = \(\frac{1}{2}\)(3 + 4) = \(\frac{7}{2}\)
iii) x = \(\frac{1}{2}\)(-2 + 5) = \(\frac{3}{2}\)
iv) x = \(\frac{1}{2}\)(0 + -1) = \(\frac{-1}{2}\)

Real Numbers Class 9 Extra Questions and Answers Kerala Syllabus

Question 1.
Find the absolute value of the following.
i) 7
ii) -11
iii) π
iv) \(\frac{-1}{2}\)
Answer:
i) |7| = 7
ii) |-11| = 11
iii) |π| = π
iv) |\(\frac{-1}{2}\)| = \(\frac{1}{2}\)

Question 2.
Complete the table below:

Expand the table by taking some more pairs x, y of numbers. Do you see any relation between |x – y| and |x| – |y|?
Answer:

If the table is expanded,

From the table it is clear that the values of |x| – |y| is not equal to|x – y|

Question 3.
Complete the missing columns of the given table.

Find a value for |x| = – x
Answer:

From the above table, the negative numbers are used.
Therefore, x = -2

Question 4.
What is the number x for which
i) |x – 2| = 5
ii) |x + 3| = 2
Answer:
i) |x – 2| = 5
If x < 2,
Then |x – 2 | = 2 – x
That means, 2 – x = 5
– x = 5 – 2 = 3
x = – 3

If x > 2,
Then |x – 2| = x – 2
That means, x – 2 = 5
x = 5 + 2 = 7
x = 7
Therefore|x – 2| = 5, then x = – 3 or x – 7

ii) |x + 3| = 2
If x < 3 ,
Then |x + 3| = |x – (-3) | = – 3 – x
That means, -3 – x = 2
– x = 2 + 3 = 5
x = – 5
Therefore |x + 3| = 2 , then x = – 5 or x = -1

Question 5.
Find between which numbers x should lie to satisfy each of the equations |x – 3| < 5 . Also find the integer values of the equation.
Answer:
|x – 3| < 5
The point x can either be in the left or right of the point 3, (distance from 3 is less than 5).
If x is in the right side of the point 3, whose distance is 5.
Then x — 3 = 5
x = 5 + 3 = 8
x = 8
If x is in the left side of the point 3, whose distance is 5.
Then, 3 – x = 5
– x = 5 – 3 = -2
x = – 2
Therefore, |x – 3| < 5, then the possible values of x will be in between -2 and 8
That is, —2 < x <8.
Therefore the possible integer values of x is = -1, 0, 1, 2, 3, 4, 5, 6, 7

Question 6.
Write down the numbers located 3 units apart from 0 on a number line.
Answer:
The number can either be in the left or right of the point 0,
If the number is in the right side of the point 0, whose distance is 3 units.
Then ,0 + 3 = 3
If the number is in the left side of the point 0, whose distance is 3 units.
Then, 0 – 3 = -3
That means, the numbers located 3 units apart from 0 on a number line is -3 and 3.

Question 7.
On a number line, calculate the number that represents the midpoint of each pair of numbers given below,
i) -3,-7
ii) √5, √6
iii) 4\(\frac{1}{9}\), -2\(\frac{3}{4}\)
iv) -1.8, 0.25
Answer:
i) \(\frac{1}{2}\)(-3 + (-7)) = \(\frac{-10}{2}\) = -5
ii) \(\frac{1}{2}\)(√5 + √6)
iii) \(\frac{1}{2}\left(\frac{37}{2}+\frac{-11}{4}\right)=\frac{1}{2} \times \frac{63}{4}=\frac{63}{8}\)
iv) \(\frac{1}{2}\)(-1.8 + 0.25) = 0.775

Question 8.
i) Find x,if |x – 2| = |x – 6|
ii) Find y, if |y – 3| = |y + 1|,
iii) Then find the value of |x – y|?
Answer:
i) x = \(\frac{2+6}{2}=\frac{8}{2}\) = 4
ii) y = \(\frac{3-1}{2}=\frac{2}{2}\) = 1
iii) |x – y| = |4 – 1| = 3

Question 9.
i) What is the distance between the points representing the numbers 2 and 7 on the number line?
ii) Find the value of x that satisfies the equation |x – 2| = |x – 7|.
Answer:
i) Distance = 7 – 2 = 5
ii) x = \(\frac{7+2}{2}=\frac{9}{2}\) = 4.5

Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 11 Parts of Circles Extra Questions and Answers Notes to clear their doubts.

Kerala SCERT Class 9 Maths Chapter 11 Solutions Parts of Circles

Parts of Circles Class 9 Kerala Syllabus Questions and Answers

Kerala State Syllabus 9th Standard Maths Chapter 11 Parts of Circles Solutions Questions and Answers

Class 9 Maths Chapter 11 Kerala Syllabus – Length And Angle

Textal Questions and Answers

Question 1.
In a circle, the length of an arc of central angle 400 ¡s 3 centimeters. What is the circumference of the circle? And the radius?
Answer:
Central angle = 40°
Length ofan arc = 3 cm
2πr × \(\frac{40}{360}\) = 3π
2r × \(\frac{1}{9}\) = 3
2r = 3 × 9 = 27cm
Circumference = 2πr = 2πr cm
Radius, r = \(\frac{27}{2}\) cm

Question 2.
In a circle, the length of an arc of central angle 25° is 4 centimeters.
i) What is the length of an arc of central angle 75° in this circle?
ii) What is the length of an arc of central angIe 75° in a circle with radius one and a half times this circle?
Answer:
Central angle = 25°
Length of an arc = 4 cm
2πr × \(\frac{25}{360}\) = 4
2πr = 4 × \(\frac{360}{25}\)
=4 × \(\frac{72}{5}=\frac{288}{5}\)

i) Length of an arc of central angle 75° is
2πr × \(\frac{x}{360}=\frac{288}{5} \times \frac{75}{360}\) = 12cm

ii) Length of an arc of central angIe 75° with radius 1\(\frac{1}{2}\)r
2π × \(\frac{3}{2} \mathrm{r} \times \frac{x}{360}=\frac{288}{5} \times \frac{3}{2} \times \frac{75}{360}\) = 18 cm

Question 3.
From a bangle of radius 3 centimeters, a small piece is to be cut off to make a ring of radius centimeter 2
i) What should be the central angle of the piece to be cut off?
ii) The remaining part of the bangle is bent to make a smaller bangle. What is its radius?
Answer:
i) Perimeter of the bangle having radius 3 cm = 2πr = 2 × π × 3 = 6π cm
Perimeter of the ring having radius \(\frac{1}{2}\) cm = 2πr = 2 × π × \(\frac{1}{2}\) = π cm
Since the circumference of the ring and the arc length of the circle are equal Arc length of the bangle = π cm
2πr × \(\frac{x}{360}\) = π
Central angle, x = 360 × \(\frac{1}{6}\) = 60°
Central angle of the piece to be cut off = 60°

ii) Central angle of the remaining part = 360° – 60°= 300°
Arc length of the remaining part = 2πr × \(\frac{300}{360}\) = 5 π cm
Arc length of the remaining part = circumference of the new bangle
Circumference of the new bangle = 5π
2πr = 5π
r = \(\frac{5}{2}\) = 2.5cm

Question 4.
With each vertex of an equilateral triangle as center, an arc of a circle which passes through the other two vertices is drawn to get the given figure. What is its perimeter?

Answer:
Central angle of an equilateral triangle, x = 60°
Arc length = 2πr × \(\frac{x}{360}\)
= 2 × π × 4 × \(\frac{60}{360}\) = \(\frac{4 \pi}{3} cm\) cm
= 3 × \(\frac{4 \pi}{3} cm\) = 4π cm

Question 5.
With each vertex of a regular octagon as center an arc of a circle is drawn and the resulting figure is cut off, as in the pictures below:

What is the perimeter of the figure cut off.’
Answer:
Sum of the interior angle of a regular octagon = (n – 2) × 180°
= (8 – 2) × 180°
= 6 × 180° = 1080°
Angle measure at a vertex = \(\frac{1080^{\circ}}{8}\) = 135°
Centrai angle, x = 135°
Radius of the circular arc = 1 cm
Arc length = 2πr × \(\frac{x}{360}\)
= 2 π × 1 × \(\frac{135}{360}\)
Perimeter of the figure cut off = 8 × Arc length
= 8 × 2 × π × 1 × \(\frac{135}{360}\)
= 6 π cm

Class 9 Maths Kerala Syllabus Chapter 11 Solutions – Angles And Areas

Textual Questions And Answers

Question 1.
i) In a circle of radius 3 centimeters, What is the area of a sector of central angle 120°?
ii) What is the area of a sector of the same central angle in a circle of radius 6 centimeters?
Answer:
i) Radius = 3 cm
Central angle, x = 120°
Area of a sector = πr² × \(\frac{x}{360}\)
= π × 3² × \(\frac{120}{360}\)
= π × 9 × \(\frac{1}{3}\)
= 3 π cm2

ii) Radius = 6 cm
Central angle, x = 120°
Area of a sector = πr² × \(\frac{x}{360}\)
= π × 6² × \(\frac{120}{360}\)
= π × 36 × \(\frac{1}{3}\)
= 12 π cm²

Question 2.
Find the area of green region(shaded region) in the picture below:

Answer:
Central angle, x = 120°
Radius of the larger sector = 3 cm
Area of the larger sector = πr² × \(\frac{x}{360}\)
= π × 9 × \(\frac{120}{360}\)
= 3 π cm²

Radius of the smaller sector = 2 cm
Area of the smaller sector = πr² × \(\frac{x}{360}\)
= π × 4 × \(\frac{120}{360}\)
= 1.33 π cm²

Area of the shaded region = Area of the larger sector – Area of the smaller sector
= 3π – 1.33 π
= 1.67 π cm²

Question 3.
In the picture below, arc of a circle is drawn with each vertex of an equilateral triangle as center and half the side as radius:

Calculate the area of the blue region (shaded region).
Answer:
Area of the shaded region = area of the equilateral triangle – 3 × area of the sector
= \(\frac{\sqrt{3}}{4}\) × 4² – 3 × π × 2² × \(\frac{60}{360}\)
= (4√3 – 2π) cm²

Question 4.
In the picture below, arcs of circles are drawn centered on two opposite vertices of a square and passing through the other two vertices:

What is the area of green region (shaded region)?
Answer:
Area of the shaded region = 2 × area of the sector – area of the square
= 2 × π × 4² × \(\frac{90}{360}\) – 4 × 4
= (8π – 16) cm²

Question 5.
The picture below shows two circles of the same radii, each passing through the center of the other:

Calculate the area of the region common to both.
Answer:

In the figure, triangle ABC and ABD are equilateral triangle.
Area of the region between the two circles = 2 × area of the sector – area of the equilateral triangle
= 2 × π × 2² × \(\frac{60}{360}\) – 2 × \(\frac{\sqrt{3}}{4}\) × 2²
= (\(\frac{8 \pi}{3}\) – 2√3)cm²

Question 6.
The picture below shows semicircles drawn with the sides of a right triangle as diameters:

Prove that the area of the largest semicircle is the sum of the areas of the other two.
Answer:

Area of the semicircle with diameter AB = \(\frac{1}{2}\) × π × \(\left(\frac{A B}{2}\right)^2\)
= \(\frac{1}{2}\) × π × \(\frac{A B^2}{4}=\frac{\pi}{8}\)AB²

Area of the semicircle with diameter BC = \(\frac{1}{2}\) × π × \(\left(\frac{B C}{2}\right)^2\)
= \(\frac{1}{2}\) × π × \(\frac{B C^2}{4}=\frac{\pi}{8}\)BC²

Area of the semicircle with diameter AC = \(\frac{1}{2}\) × π × \(\left(\frac{A C}{2}\right)^2\)
= \(\frac{1}{2}\) × π × \(\frac{A C^2}{4}=\frac{\pi}{8}\)(AC)²

Sum of the areas of smaller semicircles = \(\frac{\pi}{8}\)(AB)² + \(\frac{\pi}{8}\)(BC)²
= \(\frac{\pi}{8}\) x ((AB)² + (BC)²)
= \(\frac{\pi}{8}\)(AC)²

Parts of Circles Class 9 Extra Questions and Answers Kerala Syllabus

Question 1.
In a circle, if the arc length corresponding to a central angle of 40° is 8n centimeters, what will be the arc length for a central angle of 100° in the same circle?
Answer:
Central angle = 40°
Arc length = 8n cm
2πr × \(\frac{40}{360}\) = 8n
2πr = 8π × \(\frac{360}{40}\) = 8π × 9 = 72 π cm
When the central angle is 100°, then the arc length is
2πr × \(\frac{100}{360}\) = 12 π × \(\frac{100}{360}\) = 20 π cm

Question 2.
i) Calculate the area of a circular disc of radius 10 centimeter
ii) It is divided into four parts by drawing two perpendicular diameters .What is the area of each of the sectors obtained?
Answer:
i) Radius = 10 cm
Area of circular disc = πr² = 100π cm²

ii) Area of the sector = πr² × \(\frac{x}{360}\)
= 100π × \(\frac{90}{360}\)
= 25π cm²

Question 3.
In the figure, two arcs are drawn from a point. The distance between the two arcs is 3 cm. The radius of the first arc is 6 cm. Find the difference between the lengths of the two arcs.

Answer:
Arc length having radius 6 cm 2πr × \(\frac{60}{360}=\frac{6 \pi}{3}\)= 2π cm
Arc length having radius 9 cm 2πr × \(\frac{60}{360}=\frac{9 \pi}{3}\) = 3π cm
Difference of arc length = 3π – 2π = π cm

Question 4.
A clock’s minute hand has a length of 10 cm. What is the distance travelled by it between 1:05 PM and 1:40 PM?

Answer:
The minute hand makes a full rotation of 360° in one complete cycle.
In 5 minutes, it rotates \(\frac{360}{12}\) =30°
To rotate from 1:05 to 1:40, it takes 7 × 30 = 210°
Central angle formed by the minute hand = 210°
Distance travelled = 2πr × \(\frac{210}{360}\)
= 2 × π × 10 × \(\frac{210}{360}=\frac{70}{6}\)π cm

Question 5.
A regular hexagon with sides measuring 4 cm has a circle drawn with a diameter equal to half the length of its sides from each vertex. Find the area of the shaded region in the figure?

Answer:
The area of a regular hexagon with sides measuring 4 cm = \(\frac{\sqrt{3}}{4}\) × 4² = 4√3 cm²
The angle at one vertex of the regular hexagon is 60°
The area of the circular segment at one vertex = π × r² × \(\frac{60}{360}\)
= π × r² × \(\frac{60}{360}\)
= \(\frac{2 \pi}{3}\)
The area of the circle at all three vertices = 3 × \(\frac{2 \pi}{3}\) = 2π cm²
Area of the shaded region = area of the regular hexagon – area of the circular segments at the vertices.
= (4√-3 – 2π) cm²

Question 6.
In the diagram, a regular hexagon with sides measuring 2 cm has circular segments with a radius of 1 cm at each vertex. Find the area of the shaded region?

Answer:
The area of a regular hexagon with sides measuring 2 cm = 6 × \(\frac{\sqrt{3}}{4}\) × 22 = 6√3 cm²
Interior angle of the regular hexagon =120°
Exterior angle = 240°
The area of the circular segment at one vertex = π × r² × \(\frac{240}{360}\) = π × 12 × \(\frac{240}{360}\)
= \(\frac{2 \pi}{3}\)
The area of 6 circular segments = 6 × \(\frac{2 \pi}{3}\) = 4π
Area of the shaded portion = area of a regular hexagon with sides measuring 2 cm + area of 6 circular segments
= (6√3 + 4π) cm²

Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 10 Polynomials Extra Questions and Answers Notes to clear their doubts.

Kerala SCERT Class 9 Maths Chapter 10 Solutions Polynomials

Polynomials Class 9 Kerala Syllabus Questions and Answers

Kerala State Syllabus 9th Standard Maths Chapter 10 Polynomials Solutions Questions and Answers

Class 9 Maths Chapter 1 Kerala Syllabus – Algebra of Measurements

Intext Questions And Answers

Question 1.
Let’s start with a question

i) What is the perimeter of the rectangle shown in the figure?
ii) What is the new perimeter if the sides of the rectangle are extended by 1 cm? Also, what is the new perimeter if the sides of the rectangle are extended by 2 cm?
iii) What is the new perimeter if the sides of the rectangle are extended by 3 cm?
iv) What is the new perimeter if the sides of the rectangle are extended by x cm?
Answer:
i) Perimeter = 2( 3 + 2) = 10 cm

ii) If the sides are extended by 1cm ,then
New length = 3 + 1 = 4 cm
New breadth = 2+ 1 = 3 cm
New perimeter = 2(4 + 3) = 14 cm
We can think about this in another way, since all four sides increase by 1 centimeter each, the total increase is 4 centimeters.
Therefore, the new perimeter = 10 + 4 = 14 cm
Similarly, if all four sides increase by 2 centimeters each, the total increase is 8 centimeters.
Therefore, the new perimeter =10 + 8 = 18 centimeters.

iii) If all four sides increase by 3 centimeters each, the total increase is 12 centimeters.
Therefore, the new perimeter =10 + 4 × 3 = 22 cm

iv) If all four sides increase by x centimeters each, the total increase is 4x centimeters.
Therefore, the new perimeter = 10 + 4x centimeters.

Textual Questions And Answers

Question 1.
In all rectangles with one side 1 centimetre less than the other, denote the length of the shorter side as x centimetres.
i) Denote their perimeters as p(x) centimetres and write the relation between x and p(x) as an equation.
ii) Denote their areas as a(x) square centimetres and write the relation between x and a(x) as an equation
iii) Compute p (1), p (2), p (3), p (4), p (5) Do you see any pattern?
iv) Compute a (1), a (2), a (3), a (4), a (5) Do you see any pattern?
Answer:
i) Let x be the shorter side, then the other side will be (x + 1)
Perimeter = 2 (x + x + 1) = 4x + 2 cm
p(x) = 4x + 2

ii) Area = x(x + 1) = x² + x cm²

iii) p(x) = 4x + 2
p(1) = 4 × 1 + 2 = 6
p (2) = 4 × 2 + 2= 10
p (3) = 4 × 3 + 2 = 14
p (4) = 4 × 4 + 2= 18
p (5) = 4 × 5 + 2 = 22
Perimeter is a sequence increasing by 4

iv) a(x) = x² + x a(1) = 1² + 1= 2
a (2) = 2² + 2 = 6
a (3) = 3² + 3 = 12
a (4) = 4² + 4 = 20
a(5) = 5² + 5 = 30
Area is a sequence of consecutive even numbers.

Question 2.
From the four corners of a rectangle, small squares of the same size are cut off and the tabs are raised up to make a box as in the picture below:

i) Denote the length of the sides of the squares as x centimetres and write the lengths of the three edges of the box in terms of x
ii) Denote the volume of the box as v(x) cubic centimetres and write the relation between x and v(x) as an equation
iii) Compute v (\(\frac{1}{2}\)), v (1) and v (1\(\frac{1}{2}\))
Answer:
i) If x cm is the length of the small squares that are cut off, then the length of the maked box by folding it up = 7 – x – x = (7 – 2x) cm
Width = 5 – x – x = (5 – 2x) cm
Height = x cm

ii) V(x)= (7 – 2x)(5 – 2x)(x)
= (35 – 24x + 4x²)(x)
= 4x³ – 24x² + 35x

iii) v(\(\frac{1}{2}\)) = (7 – 2 × \(\frac{1}{2}\))(5 – 2 × \(\frac{1}{2}\))(\(\frac{1}{2}\))
= 6 × 4 × \(\frac{1}{2}\)
=12 cm³

v(1) = (7 – 2)(5 – 2)(1)
= 5 × 3 × 1 = 15 cm³

v(1\(\frac{1}{2}\)) = (7 – 2 × (1\(\frac{1}{2}\))) (5 – 2 × (1\(\frac{1}{2}\)))(1\(\frac{1}{2}\))
= 4 × 2 × \(\frac{3}{2}\)
= 12 cm³

Question 3.
Consider all rectangles that can be made with a rope of length 1 metre. Denote the length of one side as x centimetres and the area enclosed by the rope as a(x) square centimetres.
i) Write the relation between x and a(x) as an equation
ii) Why are a (10) and a (40) the same number?
iii) To get the same number as a(x) when x is taken as two different numbers, what should be the relation between the numbers?
Answer:
i) If one side is x cm ,then the other side is 50 — x cm
Area, a(x) = x (50 – x) = 50x – x² cm²

ii) a(10) = 10(50 – 10) = 10 × 40 = 400
a(40) = 40(50 – 40) = 10 × 40 = 400
10 and 40 are the sides of rectangle, so a (10) and a (40) are same.

iii) Numbers must be sides of rectangle. If we add the two numbers together we get the sum as half the length of the wire (50 cm) used to make it.

Class 9 Maths Kerala Syllabus Chapter 10 Solutions – Special Expressions

Textual Questions And Answers

Question 1.
In each of the problems below, check whether the relation between the specified measurements is a polynomial. Give reasons for your assertions.
i) The relation between the length of the sides of a square park and the area of a 1 metre wide path around it.
ii) The relation between the amount of acid added to a mixture of 3 litres of acid and 7 litres of water, and the change in the percent of acid in the mixture
iii) Two poles of heights 3 metres and 4 metres stand 5 metres apart. A rope is to be stretched from the top of one post to some point on the ground and then stretched to the top of the other pole:

The distance from the foot of one pole to the point on the ground where the rope is fixed, and the total length of the rope.
Answer:
i)

Area of path = Area of large square – Area of small square
= (x + 2)² – x²
= 4x + 4
= 4(x + 1)
This is a polynomial because exponent here is a positive integer.

ii) The solution is \(\frac{3}{10}\) acidic. When x liters of acid are added, the solution’s volume also increases by x litres. The resulting percentage of acid and solution is \(\frac{3+x}{10+x}\) × 100. This expression is not a polynomial because the exponent is not a positive integer.

iii)

Total length of the rope = \(\sqrt{x^2+9}+\sqrt{(5-x)^2+16}\)
This is not a polynomial because it involves square root.

Question 2.
Write each of the following operations as an algebraic expression. Check which of them are polynomials, giving reasons
i) Sum of a number and its reciprocal
ii) Sum of a number and its square root
iii) The product of the sum of a number and its square root, and the difference of the square root and the number
Answer:
Let the number be x
i) x + \(\frac{1}{x}\) It is not a polynomial
ii) x + √x,It is not a polynomial
iii) (x – √x)(x + √x) = x² – x , It is a polynomial

Question 4.
For each of the polynomial p(x) given below, compute p (1) and p (10)
i) p(x) = 2x + 5
ii) p(x) = 3x² + 6x + 1
iii) p(x) = 4x³ + 2x² + 3x + 7
Ans:
i) p(1) = 2 × 1 + 5 = 7
p(10) = 2 × 10 + 5 = 25

ii) p(1) = 3 × 1² + 6 × 1 + 1 = 10
p(10) = 3 × 10² + 6 × 10 + 1 = 361

iii) p(1) = 4 × 1³ + 2 × 1² + 3 × 1 + 7 = 16
p(10) = 4 × 10³ + 2 × 10² + 3 × 10 + 7 = 4237

Question 5.
For each of the polynomial p(x) given below, compute p(0), p(1) and p(-1)
i) p(x) = 3x + 5
ii) p(x) = 5x – 8
iii) p(x) = 3x² + 6x + 1
iv) p(x) = 2x² – 5x + 3
v) p(x) = 4x³ + 2x² + 3x + 7
vi) p(x) = ax³ + bx² + cx + d
Answer:
i) p(0) = 3 × 0 + 5 = 5
p(1) = 3 × 1 + 5 = 8
p(-1) = 3 × -1 + 5 = 2

ii) p(0) = 5 × 0 – 8 = -8
p(1) = 5 × 1 – 8 = -3
p(-1)= 5 × -1 – 8 = -13

iii) p(0) = 3 × 0 + 6 × 0 + 1 = 1
p(1) = 3 × 1² + 6 × 1 + 1 = 10
p(-1)= 3 × (-1)² + 6 × – 1 + 1 = -2

iv) p(0) = 2 × 0 – 5 × 0 + 3 = 3
p(1) = 2 × 1² – 5 × 1 + 3 = 0
p(-1)= 2 × (-1)² – 5 × -1 + 3 = 10

v) p(0) = 4 × O + 2 × O + 3 × O + 7 = 7
p(1) = 4 × 1³ + 2 × 1² + 3 × 1 +7 = 16
p(-1)= 4 × (-1)³ + 2 × (-1)² + 3 × -1 + 7 = 2

vi) p(0) = a × 0 + b × 0 + c × 0 + d = d
p(1) = a × 1³ + b × 1² + c × 1 + d = a + b + c + d
p(-1)= a × (-1)³ + b × (-1)² + c × -1 + d = -a + b- c + d

Polynomials Class 9 Extra Questions and Answers Kerala Syllabus

Question 1.
In a polynomial p(x) = 2x³ + ax² – 7x + b, p (1) = 3, p (2) = 19. Then find the value of ‘a’ and ’b’?
Answer:
p (x) = 2x³ + ax² – 7x + b
We have p(1) = 3
p(1) = 2(1)³ + a(1)² – 7(1) + b
3 = 2 + a – 7 + b
a + b = 8 ………..(1)
p(2) = 19
p (2) = 2(2)³ + a(2)² – 7(2) + b
19= 16 + 4a- 14 + b
4a + b = 17 ……..(2)
Subtracting equation (1) from (2)
we get 3a = 9, a = 3

Put a = 3 in equation (1)
we get b = 8 – 3 = 5

Question 2.
In a polynomial p(x) = 2x³ + 9x² + kx + 3, p (- 2) = p (- 3) Find the value of k.
Answer:
p (x) = 2x³ + 9x² + kx + 3

p (-2) = 2(-2)³ + 9(-2)² + k (-2) + 3
= – 16 + 36 – 2k + 3
= 23-2k

p (-3) = 2(-3)3 + 9(-3)2 + k (-3) + 3
= -54 + 81 – 3k + 3
= 30 – 3k

We have p (-2) = p (-3)
23 – 2k = 30 – 3k
3k – 2k = 30 – 23
k = 7

Question 3.
From the polynomial p(x) = 2x² – 3x + 1 find p (0), p (1) and p (-1).
Answer:
p(x) = 2x² – 3x + 1
p (0) = 2(0)² – 3(0) + 1 = 1
p(1) = 2(1)² – 3(1) + 1 = 0
p (-1) = 2(- 1)² – 3(-1) + 1 = 2 + 3 + 1 = 6

Question 4.
In a polynomial p(x) = 2x³ – 7x² + kx + 20, p(2) = p(3)
i) Find the value of k.
ii) Using the value of k, write the polynomial.
iii) Find p (1).
Answer:
i) p(2) = 2(2)³ – 7(2)² + k(2) + 20
= 16 – 28 + 2k + 20
= 8 + 2k

p (3) = 2(3)3 – 7(3)2 + k(3) + 20
= 54 – 63 + 3k + 20
= 11 + 3k

P(2) = P(3)
8 + 2k = 11 + 3k
3k – 2k = 8 – 11
k = -3

ii) p(x) = 2x³ – 7x² – 3x + 20

iii) p(1) = 2(1)³ – 7(1)² – 3(1) + 20
= 2 – 7 – 3 + 20
= 12

Question 5.
In the polynomial p(x) = 3x² – ax + 1 Find ’a’ satisfying p(2) = 2
Answer:
p(x) = 3x² – ax + 1

p (1) = 3x² – ax + 1
= 3 – a + 1
= 4 – a
Given p(1) = 2, 4-a = 2, a = 4- 2 = 2

Question 6.
i) p(x) = 4x – 5, find p(2)
ii) Write the first degree polynomial q(x) with q (2) = 4 and q (1) = 0
Answer:
i) p(2) = 4(2) – 5 = 8 – 5 = 3
ii) Let q(x) = ax + b
q (2) = 2a + b
4 = 2a + b ….(1)

q (1) = a +b
0 = a + b ….(2)
Solving equations (1) and (2) we get a = 4 and b = -4
Thus q(x) = 4x – 4

Question 7.
p(x) = 2x² + 3x + 5
i) Find p (1), p (0)
ii) Write the second degree polynomial p(x) with p (0) = 2 and p (1) = 5
Answer:
i) p(1) = 2 × 1² + 3 × 1 + 5
= 2 + 3 + 5
= 10

p(0) = 2 × 0² + 3 × 0 + 5 = 5

ii) p(x) = ax³ + bx + c
p(0) = a × 0² + b × 0 + c = c
Given p(0) = 2 c = 2

p(1) = a × 1² + b × 1 + c
= a + b + c

Given p(1) = 5
a + b + c = 5
a + b = 3
Therefore,p(x) = 2x³ + x + 2
The equation a + b = 3 is satisfied for any pair of values (a, b), provided a is non-zero.

Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 12 Prisms Extra Questions and Answers Notes to clear their doubts.

Kerala SCERT Class 9 Maths Chapter 12 Solutions Prisms

Prisms Class 9 Kerala Syllabus Questions and Answers

Kerala State Syllabus 9th Standard Maths Chapter 12 Prisms Solutions Questions and Answers

Class 9 Maths Chapter 12 Kerala Syllabus – Volume

Textual Questions And Answers

Question 1.
The base of a prism is an equilateral triangle of perimeter 18 centimetres and its height is 5 centimetres. Calculate its volume.
Answer:
Given that,
The base perimeter of an equilateral triangle = 18 cm
Therefore the base edge = \(\frac{18}{2}\) = 6 cm
Height = 5 cm
Base area = \(\frac{\sqrt{3}}{4}\) × (side)²
= \(\frac{\sqrt{3}}{4}\) × (6)²
= 9√3 cm²
Volume = Base area × Height
= 9√3 × 5
= 45√3 cm³

Question 2.
The base of a prism is a triangle of sides 13 centimetres, 14 centimetres and 15 centimetres, and its height is 20 centimetres. Calculate its volume.
Answer:
Considering the base edges of a triangular prism as a, b and c where,
a = 13 cm,
b = 14 cm,
c = 15 cm
Height = 20 cm

Considering the semi-perimeter of a triangular prism as S,
S = \(\frac{\text { Sum of the base edges }}{2}=\frac{a+b+c}{2}\)
S = \(\frac{13+14+15}{2}\)
= 21 cm

Base area = \(\sqrt{S(S-a)(S-b)(S-c)}=\sqrt{21(21-13)(21-14)(21-15)}\)
= \(\sqrt{21 \times 8 \times 7 \times 6}\)
= \(\sqrt{7056}\)
= 84 cm²

Volume = Base area × Height
= 84 × 20
= 1680 cm³

Question 3.
There is a hexagonal pit in the school ground to collect rain water. Each side of the hexagon is 2 metres and the depth of the pit is 3 metres. It now contains water one meter deep. How much litres is this? How much more litres can it contain?
Answer:
One side of the hexagon = 2 m
It is given that the depth of the pit is 3 meter and the water level is up to 1 meter.

Base area of a hexagon = 6 × \(\frac{\sqrt{3}}{4}\) × (side)²
= 6 × \(\frac{\sqrt{3}}{4}\) × 2²
= 6√3 m²

Volume of water = Volume of the hexagonal prism
= Base area × height
= 6√3 × 1
= 6 × 1.73 × 1000
= 10380 cubic litre

Depth of the pit = Height of the prism = 3m
Total volume of the hexagonal prism = Base area × height
= 6√3 × 3
= 18√3
= 18 × 1.73 × 1000
= 31140 cubic litre

More liters of water does the prism can hold = 31140 – 10380
= 20760 cubic litre

Question 4.
A hollow prism with base a square of sides 16 centimetres contains water 10 centimetres high. If a cube of edges 8 centimetres is immersed in it, by how much would the water level rise?
Answer:
Base of the hollow prism is a square,
Side = 16 cm
Height =10 cm
Base area = (side)² = 16² = 256 cm²
Volume of the water = Base area × height
= 256 × 10
= 2560 cm³
Edges of the solid cube = 8 cm

Base area = (side)² = 8²
= 64 cm²
Volume of the solid cube = Base area × height
= 64 × 8
= 512 cm³

Let as consider the level of water increased be h,
Volume of the increased water level = Base area × height
=16 × 16 × h
= 256 h cm³

Volume of the increased water level = Volume of the water + Volume of the solid cube
256 h = 2560 + 512
256h = 3072
h = \(\frac{3072}{256}\) = 12
Increased water level = 12 – 10 = 2 cm

Question 5.
A rectangular block of metal has edges of lengths 6 centimetres, 9 centimetres and 15 centimetres. It is melted and recast into identical cubes of sides 3 centimetres. How many cubes would be got?
Answer:
Edges of the rectangular block = 6 cm, 9 cm, 15 cm
Base area of the rectangular block = length × breadth
= 9 × 6
= 54 cm²

Volume of the rectangular block = Base area × height
= 54 × 15
= 810 cm³

Base area of the cube = (side)² = 3² = 9 cm²
Volume of the cube = Base area × height = 9 × 3 = 27 cm³
Number of solid cubes = \(\frac{\text { Volume of the rectangular block }}{\text { Volume of the cube }}=\frac{810}{27}\) = 30

Question 6.
The base of a prism is a square of sides 6 centimetres and its height is 10 centimetres. What is its volume? What is the maximum volume of a triangular prism cut from it?
Answer:
Side of the square = 6 cm
Height = 10 cm
Base area = (side)²
= 6²
= 36 cm²

Volume of the square prism = Base area × height
= 36 × 10
= 360 cm³

To obtain a triangular prism with the maximum volume, the rectangular prism should split through its diagonal.
Base edges of the triangular prism = 6 cm, 6 cm, 10 cm
Base area = \(\frac{1}{2}\)(6 × 6)
= \(\frac{1}{2}\)
= 18 cm²
Volume of the triangular prism = Base area × Third side
= 18 × 10
= 180 cm³

Class 9 Maths Kerala Syllabus Chapter 12 Solutions – Area

Textual Questions And Answers

Question 1.
The base of a prism is an equilateral triangle of perimeter 12 centimetres and its height is 5 centimetres. What is its total surface area?
Answer:
Base perimeter of an equilateral triangle = 12 cm
One side = \(\frac{12}{3}\) = 4 cm
Height = 5 cm
Lateral surface area of the equilateral triangle = Base perimeter × height
= 12 × 5
= 60 cm2
Base area = \(\frac{\sqrt{3}}{4}\) × (side)²
= \(\frac{\sqrt{3}}{4}\) × 4²
= 6.92 cm²
Total surface area = Lateral surface area + 2 (Base area)
= 60 + 2 (6.92)
= 60 + 13.84
= 73.84 cm²

Question 2.
Two identical prisms with right triangles as bases are joined to form a rectangular prism, as in the picture below:

What is the total surface area of this rectangular prism?
Answer:
The sides of the rectangular prism are:
Length = 12 cm
Breadth = 5 cm
Height = 15 cm

Base area = Length × Breadth
= 12 × 5
= 60 cm

Lateral surface area = base perimeter × height
= 2 (length + breadth) × height
= 2(12 + 5) × 15
= 2 × 17 × 15
= 510 cm²

Total surface area = Lateral surface area + 2 ( Base area)
= 510 + 2 (60)
= 510 + 120
= 630 cm²

Question 3.
The base of a prism is a triangle with sides 4 centimetres, 13 centimetres and 15 centimetres and its height is 25 centimetres. Calculate its lateral surface area and total surface area.
Answer:
Considering the sides of the triangular prism be a, b and c
where, a = 4 cm, b = 13 cm, c = 15 cm
Height = 25 cm

Base perimeter = a + b + c
= 4 +13 + 15
= 32 cm

Lateral surface area = base perimeter × height
= 32 × 25
= 800 cm²

Considering the semi-perimeter of a triangular prism as S,
S = \(\frac{a+b+c}{2}=\frac{4+13+15}{2}=\frac{32}{2}\)
= 16 cm

Base area = \(\sqrt{S(S-a)(S-b)(S-c)}\)
= \(\sqrt{16(16-4)(16-13)(16-15)}\)
= \(\sqrt{16 \times 12 \times 3 \times 1}\)
= \(\sqrt{576}\)
= 24 cm²

Total surface area = Lateral surface area + 2 (Base area)
= 800 + 48
= 848 cm²

Question 4.
The lateral surface area of a prism, with base an equilateral triangle, is 120 square centimetres
i) What is the lateral surface area of a prism, with base a rhombus, made by joining two such triangular prisms?
ii) What is the lateral surface area of a prism, with base an isosceles trapezium, made by joining three such triangular prisms?
iii) What is the lateral surface area of a prism, with base a regular hexagon, made by joining six such triangular prisms?
Answer:
Considering the side of a triangular prism be ‘a’ and the height be ‘h’
Lateral surface area =120 cm²
That means, 3 ah = 120
ah = \(\frac{120}{3}\) = 40

i) Number of sides of a rhombus = 4

Lateral surface area = 4 ah = 4 × 40
= 160 cm²

ii) Number of sides of an isosceles trapezium = 5

Lateral surface area = 5 a h
= 5 × 40
= 200 cm²

iii) Number of sides of a regular hexagon = 6

Lateral surface area = 6 a h
= 6 × 40
= 240 cm²

Question 5.
Six sheets of metal, each a square of sides 10 centimetres are joined to make a cube
i) What is its total surface area?
ii) How much water can it contain?
Answer:
Side of the metal sheet and the side of the cube are same
Therefore, side of the cube = 10 cm

i) Total surface area of the cube = 6 × (side)²
= 6 × (10)²
= 600 cm²

ii) Volume of the cube = (side)³
= (10)³
= 1000 cm³

Amount of water does the cube can contain = \(\frac{\text { Volume of the cube }}{1000}=\frac{1000}{1000}\)
= 1 litre

SCERT Class 9 Maths Chapter 12 Solutions – Cylinders

Textual Questions And Answers

Question 1.
The base diameter of a cylindrical tank is 1 meter and its height is 2 meters. How many litres of water can it contain?
Answer:
Base diameter ‘d’ = 1 m
Height = 2 m
Radius r = \(\frac{d}{2}=\frac{1}{2}\) = 0.5 m
Base area = πr² = (0.5)²π – 0.25 π m²

Volume = Base area × Height
= 0.25 π × 2
= 0.5 π m³
1 m³ is equivalent to 1000 liters of water.
That means, the amount of water that the cylindrical tank can contains = volume 1000
= 0.5 π 1000 ×
= 500 π × liters
= 500 × 3.1416
= 1570.8 liters

Question 2.
The base radius of an iron cylinder is 15 centimetres and its height is 32 centimetres. It is melted and recast into a cylinder of base radius 20 centimetres. What is the height of this cylinder?
Answer:
Base radius of an iron cylinder = 15 cm
Height = 32 cm
Base area = πr²
= 225 π cm²

Volume V₁ = Base area × Height
= 225 π × 32
= 7200 π cm³

Iron cylinder is melted and recast into a new cylinder, whose Radius = 20 cm
Considering the height of the new cylinder be ‘h’
Base area = πr² = (20)² π = 400 π cm²

Volume V₂ = Base area × Height
= 400 π h cm³

Volume of the cylinder and the new cylinder are same.
That means, V₁ = V₂
7200 π = 400 π h
h = \(\frac{7200 \pi}{400 \pi}\)
= 18 cm

Question 3.
The base radii of two cylinders of the same height are in the ratio 3 : 4. What is the ratio of their volumes?
Answer:
Lets consider the base radius of the two cylinders be r₁ and r₂
Therefore r₁ : r₂ = 3 : 4
That means, \(\frac{r_1}{r_2}=\frac{3}{4}\)
Considering the height of the cylinder be ‘h’ and
Volume of the cylinder be, v = πr² h

The ratio of their volumes be,
\(\frac{v_1}{v_2}=\frac{\pi r_1^2 h}{\pi r_2^2 h}=\frac{r_1^2}{r_2^2}=\frac{(3)^2}{(4)^2}=\frac{9}{16}\)
That means, v₁ : v₂ = 9 : 16

Question 4.
The base radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 4
i) What is the ratio of their volumes?
ii) The volume of the smaller cylinder is 720 cubic centimeters. What is the volume of the larger one?
Answer:
i) Ratio of the base radius of the two cylinders = 2:3
Ratio of their heights = 5:4

For first cylinder consider,
Radius = 2r and Height = 5h

For the second cylinder consider,
Radius = 3r and Height = 4h

Volume of the first cylinder = Base area × Height
= π (2r)² × 5h
= 4πr² × 5 h
= 20 πr²h cm³

Volume of the second cylinder = Base area × Height
= π (3r)² × 4h
= 9 π r² × 4h
= 36 π r²h cm³

Ratio between their volumes = 20 π r²h : 36 π r²h
= 20 : 36 = 5 : 9

ii) Volume of the smaller cylinder = 720 cm3
Considering the volume of the larger cylinder = x
Therefore the ratio between the volume of the two cylinders = 5 : 9
That means, 5:9 = 720 : x
\(\frac{5}{9}=\frac{720}{x}\)
5x = 720 × 9
5 x = 6480
x = \(\frac{6480}{5}\) = 1296
Volume of the larger cylinder = 1296 cm³

Prisms Class 9 Kerala Syllabus – Curved Surface

Intext Questions And Answers

Question 1.
Rectangular sheets of thick paper, of sides 24 centimetres and 18 centimetres are bent along the longer side to make hollow prisms with bases an equilateral triangle, a square, a regular hexagon, a regular octagon. One of them is rolled into a cylinder. All the prisms have the same lateral surface area and it is also equal to the curved surface area of the cylinder. What about their volumes? Do you note anything special?

What if these shapes are made by bending or rolling along the shorter side of the rectangle? Find out the similarities and differences between these and the first ones, in lateral surface area, curved surface area and volume.
Answer:
Length of the rectangular sheet = 24 cm
Breadth = 18 cm

Bent along the longer side
Base perimeter = 24 cm
Height = 18cm

Lateral surface area
= Base perimeter × Height
= 24 × 18
= 432 cm²

Curved surface area of the cylinder
= Base circumference × Height
= 24 × 18
= 432 cm²

Shorter Side:

Bent along the shorter side
Base perimeter = 18 cm
Height = 24cm

Lateral surface area
= Base perimeter × Height
= 18 × 24
= 432 cm²

Curved surface area of the cylinder
= Base circumference × Height
= 18 × 24
= 432 cm²

Here the lateral surface area and the curved surface areas of the two prisms are equal. But their volumes are different.

1^st Prism:

Equilateral triangle
One side = \(\frac{\text { Base perimeter }}{3}=\frac{24}{3}\) = 8cm
Base area = \(\frac{\sqrt{3}}{4}\) × (side)²
= \(\frac{\sqrt{3}}{4}\) × (8)²
= 16√3 cm²

Volume = Base area × Height
= 16√3 × 18
= 288√3
= 288 × 1.73
= 498.24 cm³

Square prism
One side = \(\frac{\text { Base perimeter }}{4}=\frac{24}{4}\)
= 6cm

Base area = (side)²
= (6)²
= 36 cm²

Volume = Base area × Height
= 36 × 18
= 648 cm³

2^nd Prism:

Equilateral triangle
One side = \(\frac{\text { Base perimeter }}{3}=\frac{18}{3}\) = 6cm
Base area = \(\frac{\sqrt{3}}{4}\) × (side)
= \(\frac{\sqrt{3}}{4}\) × (6)²
= 9√3 cm²

Volume = Base area × Height
= 9√3 × 24 = 216√3
= 216 × 1.73
= 373.68 cm³

Square prism
One side = \(\frac{\text { Base perimeter }}{4}=\frac{18}{4}\) = 4.5 cm
Base area = (side)² = (4.5)²
= 20.25 cm²

Volume = Base area × Height
= 20.25 × 24
=486 cm³

Similarly, we can find the volume of the remaining prisms.
From this we can conclude that, if the base area changes its volume also changes.

Textual Questions And Answers

Question 1.
In a school building there are 12 cylindrical pillars, each of base diameter \(\frac{1}{2}\) metre and height 4 metres.
i) What is the curved surface area of a pillar?
ii) What is the total cost of painting all the pillars at 80 rupees per square metre?
Answer:
i) Number of cylindrical pillars = 12
Base diameter = \(\frac{1}{2}\)m
Height = 4 m
Radius = \(\frac{\text { Base diameter }}{2}=\frac{1 / 2}{2}=\frac{1}{4}\) = 0.25 m
Base circumference = 2 π r
= 2 π × 0.25
= 0.5 π

Curved surface area of the pillar = Base circumference × Height
= 0.5 π × 4
= 2 π cm²

ii) Curved surface area of 12 pillars = 12 × 2π
= 24 π

Cost of painting per square meter = Rs. 80

Total cost for painting the 12 pillars = Curved surface area of 12 pillars × 80
= 24 π × 80
= 1920 π
= 1920 × 3.1416
= 6031.872 Rs

Question 2.
The drum of a road roller has base diameter 80 centimetres and length 1.2 metres:

What is the area of the road levelled when it rolls once?
Answer:
Length of the roller = 1.2 m = 120 cm
Volume = 80 cm
Radius = \(\frac{80}{2}\) = 40 cm

Base circumference = 2 × π × 40
= 80π cm
The area of the road levelled when it rolls once is equivalent to its curved surface area Therefore,
Curved surface area = Base perimeter × Height
= 80 π × 120
= 9600 π
= 9600 × 3.14
= 30,144 cm²
= 3.0144 m²

Question 3.
The curved surface area of a cylinder is equal to its base area. What is the relation between ifs base radius and height?
Answer:
Consider, the radius of the cylinder = r
Height = h
Base circumference = 2π × radius = r

Therefore,
Curved surface area = Base circumference × Height
= 2πr × h
= 2 π r h

Base area = π × (radius)2 = πr²
Given that, the curved surface area of a cylinder is equal to its base area.
That means, 2 π r h = πr²
2 h = r
That means, the radius is twice of its height.

Question 4.
A rectangular sheet of metal with sides 48 centimetres and 25 centimetres is rolled into a cylinder of height 25 centimetres and its ends are closed with exactly fitting circles. What is total surface area of this cylinder?
Answer:
Side of the rectangular sheet = 48 cm
Breadth = 25 cm
Height of the cylinder = 25 cm
Considering radius as ‘r’

The length of the rectangular sheet is equal to its circumference of the base circle.
Circumference of the base circle = 48 cm
2 π r = 48
r = \(\frac{48}{2 \pi}=\frac{48}{2 \times 3.14}\)
= 7.64 cm

From this, radius of the cylinder r = 7.64 cm
Curved surface area = Base circumference × Height
= 2πr × h
=2 × 3.1416 × 7.64 × 25
= 1200 cm²

Base area of a circle = πr²
= 3.14 × (7.64)²
= 183.37 cm²

Base area of two circles = 2 × 183.37
= 366.7 cm²

Total surface area of the cylinder = Curved surface area + Base area of two circles
= 1200 + 366.7
= 1566.7 cm²

Prisms Class 9 Extra Questions and Answers Kerala Syllabus

Question 1.
The base area of a prism is 50 square centimetres and its height is 20 centimetres. What is the volume of the prism?,
Answer:
Base area = 50 cm³
Height = 20 cm
Volume = Base area × height
= 50 × 20
= 1000 cm³

Question 2.
6 square pieces of side 10 centimetres are cut out and joined as shown in the figure. It is then folded to make a prism.

a) Find the surface area of the prism.
b) How many litres of water can it hold?
Answer:
a) one side = 10 cm
Base area of a square = (side)² = (10)²
= 100 cm²

b) Volume = Base area × height = 100 × 10
= 1000 cubic cm
= 1 litre

Question 3.
a) The lateral surface area of an equilateral triangular prism is 90 square centimetres. What is the area of one lateral face?
b) Two such prisms are put together to form a new prism. What is the lateral surface area of the new prism?
c) Six such prisms are put together to form a regular hexagonal prism. What is the lateral surface area of this prism?
Answer:
a) lateral surface area = 90 square cm
Area of one lateral face = \(\frac{90}{3}\) = 30 cm

b) Lateral surface area = 4 × 30 = 120 cm²
[If the lateral faces are joined we get the above answer and if the base faces are joined, Lateral surface area = 6 × 30 = 180sq. cm]

c) Lateral surface area = 6 × 30 = 180 cm²

Question 4
Dimensions of a rectangular box are 50 centimetres, 30 centimetres and 40 centimetres. Find area of the cardboard required for making this box. .
Answer:
Given the dimensions of a rectangular box as,
Length = 50 cm
Breadth = 30 cm
Height = 40 cm
Base perimeter = 2 (50 + 30) = 160 cm

Lateral surface area = Base perimeter × Height
= 160 × 40
= 6400 cm²

Base area = Length × Breadth
= 50 × 30
= 1500 cm²

Total surface area = Lateral surface area + 2 (Base area)
= 6400 + 2 (1500)
= 9400 cm²

Area of the cardboard required = 9400 cm²

Question 5.
If a cylinder having radius 4 centimetre and height 10 centimetre. Then what is the volume of the cylinder. Their is an another cylinder with half the radius and double the height of the first cylinder. What is the relation between the volume of the first cylinder and the second cylinder.
Answer:
Base radius = 4 cm
Height = 10 cm
Base area = πr² = π 4²
= 16 π

Volume of the cylinder = Base area × Height
= 16 π × 10
= 160 π cm³

If the cylinder having half the radius and double the height then,
Base radius = 2 cm

Height = 20 cm
Base area = πr² = π × 2² = 4 π

Volume of the second cylinder = Base area × Height
= 4π × 20
= 80π cm³

That means, volume of the second cylinder is half the volume of the first cylinder.

Question 6.
There are 10 cylindrical pillars of diameter 30 centimetres and height 5 metres in a school varandha. What is the total cost of painting the pillars at the rate of rupees 80 per square metre? (π = 3.14)
Answer:
Number of cylindrical pillars = 10
Base diameter = 30 cm = 0.3 m
Height = 5m
Base radius = \(\frac{\text { Base diameter }}{2}=\frac{0.3}{2}\) = 0.15 m
Base perimeter = 2πr = 2 × π × 0.15 = 0.3 π m

Curved surface area of a pillar = Base perimeter × height
= 0.3 π × 5
= 1.5 π cm²

Total curved surface area of 10 pillar = 10 × 1.5π
= 15π cm²

Total cost = 15π × 80
= 15 × 3.14 × 80
= 3768 Rs

Question 7.
A cylindrical water tank has a radius of 1 metre and a height of 2 metres. How many liters of water can it hold? (1 cubic metre = 1000 liters )
Ans:
Radius of the cylindrical water tank = 1 m
Height = 2 m
Base area = πr² – π 1²
= 1 π m

Volume = Base area × Height
= 1π × 2
= 2 π
— 2 × 3.14
= 6.28 cm³
= 6.28 × 1000 = 6280 liter

Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 7 Negative Numbers Extra Questions and Answers Notes to clear their doubts.

Kerala SCERT Class 9 Maths Chapter 7 Solutions Negative Numbers

Negative Numbers Class 9 Kerala Syllabus Questions and Answers

Kerala State Syllabus 9th Standard Maths Chapter 7 Negative Numbers Solutions Questions and Answers

Class 9 Maths Chapter 7 Kerala Syllabus – Measures And Numbers

Intext Questions And Answers

Question 1.
Solve the following questions.
(i) 6 – 8
Answer:
6 – 8 = -(8 – 6)
= -2

(ii) -6 + 8
Answer:
-6 + 8 = 8 – 6
= 2

(iii) -6 – 8
Answer:
-6 – 8 = -(6 + 8)
= -14

(iv) 2\(\frac{1}{2}\) – 3\(\frac{1}{2}\)
Answer:
2\(\frac{1}{2}\) – 3\(\frac{1}{2}\)
= \(\frac{5}{2}-\frac{7}{2}=-\left(\frac{7}{2}-\frac{5}{2}\right)\)
= \(-\left(\frac{7-5}{2}\right)=-\left(\frac{2}{2}\right)\)
= -1

(v) -2\(\frac{1}{2}\) + 3\(\frac{1}{2}\)
Answer:
-2\(\frac{1}{2}\) + 3\(\frac{1}{2}\)
= \(-\frac{5}{2}+\frac{7}{2}\)
= \(\frac{7}{2}-\frac{5}{2}=\frac{2}{2}\)
= 1

(vi) -2\(\frac{1}{2}\) – 3\(\frac{1}{2}\)
Answer:
-2\(\frac{1}{2}\) – 3\(\frac{1}{2}\)
= \(-\frac{5}{2}-\frac{7}{2}\)
= \(-\left(\frac{5}{2}+\frac{7}{2}\right)=-\left(\frac{12}{2}\right)\)
= -6

Class 9 Maths Kerala Syllabus Chapter 7 Solutions – Position And Number

Intext Question And Answer

Question 1.
Here is a table giving different initial positions and displacements:

Draw pictures to find the final positions and write in the table. Then make another table with the displacements as just numbers, positive or negative and check if in each row, the last number is the sum of the first two numbers.
Answer:
First row:

Second row:

Third row:

Fourth row:

Fifth row:

Sixth row:

Seventh row:

Eighth row:

The given table can be completed as follows;

Initial Position	Displacement	Final Position
7	3 right	10
3	7 right	10
-7	3 right	-4
-3	7 right	4
7	3 left	4
3	7 left	-4
-7	3 left	-10
-3	7 left	-10

Table with the displacements as just numbers, positive or negative;

Initial Position	Displacement	Final Position
7	3	10
3	7	10
-7	3	-4
-3	7	4
7	-3	4
3	-7	-4
-7	. -3	-10
-3	-7	-10

Consider the first row; 7 + 3 = 10
Consider second row; 3 + 7 = 10
Consider third row; -7 + 3 = 3 – 7 = -(7 – 3) = -4
Therefore, in each row, the last number is the sum of the first two numbers.

Textual Questions And Answers

Question 1.
Complete the table below:

Answer:
6 + (-10) = 6 – 10
= -(10 – 6)
= -4

-6 + 10 = 10 – 6
= 4

-6 + (-10) = -6 – 10
= -(6 + 10)
= -16

-6 + 6 = 6 – 6
= 0

6 + (-6) = 6 – 6
= 0

x	y	x + y
6	-10	-4
-6	10	4
-6	-10	-16
-6	6	0
6	-6	0

Question 2.
Find two pairs of numbers x and y satisfying each of the following conditions:
i) x positive, y negative with, x + y = 1
ii) x negative, y positive with, x + y = 1
iii) x positive, y negative with x + y = – 1
iv) negative, y positive with x + y = – 1
Answer:
i) x = 2, y =-1 and x = 5, y = -4
ii) x = -3, y = 4 and x = -6, y = 7
iii) x = 1, y = -2 and x = 3, y = -4
iv) x = -3, y = 2 and x = -5, y = 4

Question 3.
Complete the table below:

Answer:
First row:
(x + y) + z = (2.+ 4) + (-5)
= 6 – 5
= 1
x + (y + z) = 2 + (4 + (-5))
= 2 + (4 – 5)
= 2 + (-1)
= 2 – 1
= 1

Second row:
(x + y) + z = (2 + (-4)) + 5
= (2 – 4) + 5
= -2 + 5
= 5 – 2
= 3

x + (y + z) = 2 + (-4 + 5)
= 2 + (5 – 4)
= 2 + 1
= 3

Third row:
(x + y) + z = (-2 + 4) + (-5)
= (4 – 2) – 5
= 2 – 5
= -(5 – 2)
= -3

x + (y + z) = -2 + (4 + (-5))
= -2 + (4 – 5)
= -2 + -(5 – 4)
= -2 + (-1)
= -2 – 1
= -(2 + 1)
= -3

Fourth row:
(x + y) + z = (2 + (-4)) + (-5)
= (2 – 4) + (-5)
= -2 – 5
= -(2 + 5)
= -7

x + (y + z) = 2 + (-4 + (-5))
= 2 + (-4 – 5)
= 2 + -(4 + 5)
= 2 + (-9)
= 2 – 9
= -(9 – 2)
= -7

Fifth row:
(x + y) + z = (-2 + 4) + 5
= (4 – 2) + 5
= 2 + 5
= 7

x + (y + z) = -2 + (4 + 5)
= -2 + 9
= 9 – 2
= 7

Sixth row:
(x + y) + z = (-2 + (-4)) + 5
= -(2 + 4) + 5
= -6 + 5
= 5 – 6
= -1

x + (y + z) = -2 + (-4 + 5)
= -2 + (5 – 4)
= -2 + 1
= 1 – 2
= -1

Seventh row:
(x + y) + z = (-2 +(-4)) +(-5)
= (-2 – 4) – 5
= -(2 + 4) – 5
= -6 – 5
= -(6 + 5)
= -11

x + (y + z) = -2+ (-4 +(-5))
= -2 + (-4 – 5)
= -2 – 9
= -(2 + 9)
= -11

SCERT Class 9 Maths Chapter 7 Solutions – Displacement

Textual Questions And Answers

Question 1.
Take different numbers, positive and negative, as x, y, z and calculate x – (y – z) and (x – y) + z. Are both the same number in all cases?
Answer:
Let, x = 1, y = -1, z = 2
x – (y – z) = 1 -(-1 – 2)
= 1 – (-3)
= 1 + 3 = 4

(x – y) + z = (1 – (-1)) + 2
= (1 + 1) + 2
= 2 + 2
= 4
Both are same.

Question 2.
Find two pairs of numbers x and y, satisfying each of the conditions below:
i) x positive, y negative, x – y = 1
ii) x negative, y positive, x – y = -1
Answer:

Question 3.
(i) Take four consecutive natural numbers or their negatives as a, b, c, d and calculate
a – b – c + d.
ii) Explain using algebra why it is zero for all such numbers.
iii) What do we get if we calculate a + b – c – d instead of a – b – c + d?
iv) What about a – b + c – d?
Answer:
i) Let, a = 1, b = 2, c = 3, d = 4
a – b – c + d = 1 – 2 – 3 + 4
= -1 – 3 + 4
= -4 + 4
= 0

ii) Let a = x, b = x + 1, c = x + 2, d = a – b – c + d

iii) a + b – c – d = 1 + 2 – 3 – 4
= 3 – 3 – 4
= 0 – 4
= -4

iv) a – b + c – d = 1 – 2 + 3 – 4
= -1 + 3 – 4
= 2 – 4
= -2

Negative Numbers Class 9 Kerala Syllabus – Time And Speed

Textual Questions And Answers

Question 1.
Take different numbers, positive and negative, as x, y, z and calculate (x + y)z and xz + yz. Check if the equation (x + y)z = xz + yz holds in all cases.
Answer:
Let, x = 1, y = 2, z = -1.
(x + y)z = (1 + 2)(-1) = 3 × (-1) = -3.
xz + yz = 1 × (-1) + 2 × (-1)
= -1 + (-2)
= -1 – 2 = -3.

In this case, (x + y)z = xz + yz.
Let, x = -1, y = 3, z = -4.
(x + y)z = (-1 + 3)(-4)
= 2 × (-4)
= -8.

xz + yz = -1 × (-4) + 3 × (-4)
= 4 + (-12)
= 4 – 12
= -8.

In this case, (x + y)z = xz + yz.
∴ (x + y)z = xz + yz holds in all cases.

Question 2.
Prove that the equation (x + y)(u + v) = xu + xv + yu + yv holds if x, y, u, v are replaced by – x, -y, -u, -v.
Answer:
(-x + (-y))(-u + (-v)) = (-x – y)(-u – v)
= -(x + y) x -(u + v)
= (x + y)(u + v)
= xu + xv + yu + yv
Hence, the proof.

Question 3.
In each of the equations below, find y when x is the given number:
i) y = x², x = -1
ii) y = x² + 3x + 2, x = -1
iii) y = x² + 3x + 2, x = -2
iv) y = (x + 1)(x + 2), x = -1
v) y = (x + 1)(x + 2), x = -2
Answer:
i) y = (-1)²
= -1 × -1
= 1

ii) y = (-1)² + 3 × (-1) + 2
= 1 + (-3) + 2
= 1 – 3 + 2
= -2 + 2
= 0

iii) y = (-2)² + 3 × (-2) + 2
= 4 + (-6) + 2
= 4 – 6 + 2
= -2 + 2
= 0

iv) y = (-1 + 1)(-1 + 2)
= 0 × 1
= 0

v) y = (-2 + 1)(-2 + 2)
= -1 × 0
= 0

Question 4.
In the equation y = x² + 4x + 4 take different numbers, positive and negative, a x and calculate y. Why is y positive or zero in all cases?
Answer:
If x = -1, y = (-1)² + 4 × (-1) + 4 = 1 – 4 + 4 = -3 + 4=l.
If x = 1, y = 1² + 4 × 1 + 4 = 1 + 4 + 4 = 9.
If x = 2, y = 2² + 4 × 2 + 4 = 4 + 8 + 4 = 16.
If x = -2, y = (-2)² + 4 × (-2) + 4 = 4 – 8 + 4 = -4 + 4 = 0.
Here, y = x² + 4x + 4 = (x + 2)², whatever the number x, its square is always greater than or equal to zero. That’s why y positive or zero in all cases.

Question 5.
Natural numbers, their negatives and zero can be together called integers.
i) How many pairs of integers (x, y) can you find, satisfying the equation x² + y² = 25?
ii) How many pairs of integers (x, y) can you find satisfying x² – y² = 25?
Answer:
i) 0² + 5² = 25 → (0, 5)
5² + 0² = 25 → (5,0)
0² + (-5)² = 25 → (0, -5) .
(-5)² + 0² = 25 → (-5, 0)
3² + 4² = 25 → (3, 4)
42 + 3² = 25 → (4, 3)
(-3)² + (-4)² = 25 → (-3,-4)
(-4)² + (-3)² = 25 → (-4,-3) ‘
(-3)² + 4² = 25 → (-3, 4)
4² + (-3)² = 25 → (4, -3)
(-4)² + 3² = 25 → (-4, 3)
3² + (4)² = 25 → (3,-4)
12 pairs of integers.

ii) 5² – 0² = 25
(-5)² – o² = 25
13² – 12² = 25
(-13)² – 12² = 25
13² – (-12)² = 25
(-13)² – (-12)² = 25
6 pairs of integers: (5,0), (-5,0), (13,12), (-13,12), (13,-12), (-13,-12)

Question 6.
1 × 2 × 3 × 4 × 5 = 120. What is (- 1)(- 2)(- 3)(- 4)(- 5)?
Answer:
(-1) × (-2) × (-3) × (-4) × (-5) = 2 × (-3) × (-4) × (-5)
= -6 × (-4) × (-5)
= 24 × (-5)
= -120

Question 7.
What is (1 × 2 × 3 × 4 × 5) + [(- 1)(- 2)(- 3)(- 4)(- 5)]?
Answer:
1 × 2 × 3 × 4 × 5 = 120
(-1) × (-2) × (-3) × (-4) × (-5) = -120
(1 × 2 × 3 × 4 × 5) + [(-1) × (-2) × (-3) × (-4) × (-5)]
= 120 + (-120)
= 120 – 120
= 0

Class 9 Maths Negative Numbers Kerala Syllabus – Negative Division

Textual Questions And Answers

Question 1.
Calculate y when x is taken as 2, -2, \(\frac{1}{2}\), –\(\frac{1}{2}\) in the equation y = \(\frac{1}{x}\)
Answer:
If x = 2, y = \(\frac{1}{2}\)
If x = -2, y = \(\frac{1}{-2}\) = -(\(\frac{1}{2}\))
If x = \(\frac{1}{2}\), y = \(\frac{1}{\frac{1}{2}} = 2\) = 2
If x = –\(\frac{1}{2}\), y = \(\frac{1}{\frac{-1}{2}} = 2\) = 2

Question 2.
Calculate y when x = -2 and x = –\(\frac{1}{2}\) in the equation y = \(\frac{1}{x-1}+\frac{1}{x+1}\)
Answer:

Question 3.
Calculate z when x and y are taken as the numbers below in the equation
z = \(\frac{x}{y}-\frac{y}{x}\)
(i) x = 10, y = -5
Answer:
Z = \(\frac{10}{-5}-\frac{-5}{10}\)
= -2 – \(\frac{-1}{2}\)
= -(2 –\(\frac{1}{2}\))
= \(\frac{-3}{2}\)

(ii) x = -10, y = 5
Answer:
Z = \(\frac{-10}{5}-\frac{5}{-10}\)
= – 2 – \(\frac{1}{-2}\)
= -(2 – \(\frac{1}{2}\))
= \(\frac{-3}{2}\)

(iii) x = -10, y = -5
Answer:
Z = \(\frac{-10}{-5}-\frac{-5}{-10}\)
= 2 – \(\frac{1}{2}\)
= \(\frac{3}{2}\)

Negative Numbers Class 9 Extra Questions and Answers Kerala Syllabus

Question 1.
Find the value of z from the equation z = x – y with the given values.
a) x = 7, y = 2
b) x = -3, y = -6
c) x = -8, y = 3
d) x = -4, y = 9
Answer:
a) x = 7, y = 2
z = x – y = 7 – 2 = 5

b) x = -3 y = -6
z = x- y = -3- (-6) = -3 + 6 = 3

c) x = -8 y = 3
z = x – y = -8 – 3 = -11

d) x = -4 y = 9
z = x – y = -4 – 9 = -13

Question 2.
If x = 4 y = -3 and z = 8 find the value of
a) (x + y) + z
b) x + (y + z)
c) xyz
d) (x + y)z
e) xy + xz
Answer:
a) (x + y) + z
= (4 – 3) + 8
= (4 – 3) + 8
= 1 + 8
= 9

b) x + (y + z)
= 4 + (-3 + 8)
=4 + 5
= 9

c) yz = 4 × (-3) × 8
= -12 × 8
= -96

d) (x + y)z = (4 – 3) × 8
= 1 × 8
= 8

e) xy + xz = (4 × (-3)) + (4 × 8)
= -12 + 32
= 20

Question 3.
Compute y = x² + 9x – 5 by taking x as the given number.
a) x = 1
b) x = -3
c) x = 0
Answer:
a) x = 1
y = x² + 9x – 5
= 1² + (9 × 1) – 5
= 1 + 9 – 5
= 10 – 5
= 5

b) x = -3
y = x² + 9x – 5
= (-3)² + (9 × (-3)) – 5
= 9 – 27 – 5
= -23

c) x = 0
y = x² + 9x – 5
= 0² + (9 × 0) – 5
= -5

Question 4.
Find y = x⁴ + x³ + x² + x + 2 if x = -1.
Answer:
y = (-1)⁴4 + (-1)³ + (-1)² + (-1) + 2
= 1 +(-1) + 1 + (-1) +2
= 1 – 1 + 1 -1 + 2
= 2

Question 5.
Check whether the equations are identities. Write the patterns got from each, on taking x = 1, 2,3 and x = – 1, -2, -3
a) – x + (x + 3) = 3
b) (x + 2) – (x + 3) = – 1
c) – x – (x + 1) + 2x + 1 = 0
Answer:
If x = 1
a) -x + (x + 3) = -1 + (1 + 3)
= -1 + 4
= 3

b) (x + 2) – (x + 3) = (1 + 2) – (1 + 3)
= 3 – 4
= -1

c) -x – (x + 1) + 2x + 1 = -1 – (1 + 1) + 2 + 1
= -1 – 2 + 2 + 1
= 0

Similarly, if we take x = 2, 3 we can see all these identities are true always.
Now, if x = -1
a) -x + (x + 3)
= -(-1) + (-1 + 3)
= 1 + 2 = 3

b) (x + 2) – (x + 3) = (-1 + 2) – (-1 + 3)
= 1 – 2
= -1

c) -x – (x + 1) + 2x + 1
= -(-1) – (-1 + 1) – 2 + 1
= 1 – 0 – 1
= 0
Similarly if we take x = -2, -3 we can see all these identities are true always.

Practicing with Kerala Syllabus 9th Standard Question Papers and Answers Malayalam will help students prepare effectively for their upcoming exams.

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Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 8 Circle Measures Extra Questions and Answers Notes to clear their doubts.

Kerala SCERT Class 9 Maths Chapter 8 Solutions Circle Measures

Circle Measures Class 9 Kerala Syllabus Questions and Answers

Kerala State Syllabus 9th Standard Maths Chapter 8 Circle Measures Solutions Questions and Answers

Class 9 Maths Chapter 8 Kerala Syllabus – Diameter And Perimeter

Textual Questions And Answers

Question 1.
The circumference of a circle of diameter 2 metres is measured and found to be 6.28 metres.
i) How do we compute the circumference of a circle of diameter 4 metres, without actually measuring it?
ii) What about the circumference of a circle of diameter 1 metre?
iii) And the circumference of a circle of diameter 3 metre?
Answer:
i) If the diameter is 2 meters, perimeter is 6.28 meters.
If the diameter is 4 meters, perimeter = 6.28 × 2 = 12.56 meters
ii) If the diameter is 1 metre, perimeter = \(\frac{6.28}{2}\) = 3.14 meters
iii) If the diameter is 3 metre, perimeter = \(\frac{6.28 \times 3}{2}\) = 9.42 meters

Question 2.
A piece of wire is bent into a circle of diameter 4 centimetres. If a wire of half the length is bent into a circle, what would be its diameter?
Answer:
The circumferences of circles change in proportion to their diameters. Therefore, if the length of the string is halved, the diameter of the circle formed by the string will also be reduced by half.
This means the diameter of the circle will be 2 centimeters.

Class 9 Maths Kerala Syllabus Chapter 8 Solutions – A New Number

Question 1.
Calculate the circumferences of the circles shown below:

Answer:

Figure 1

AB = 2 cm
In the figure triangle are equilateral triangles, therefore radius OA = 2 cm
Circumference of circle = 2πr = 2 × π × 2 = 4π cm

Figure 2

ABCD is a square AB = BC = 2 cm
AC = \(\sqrt{2^2+2^2}\) = √8 = 2√2 cm
Radius of circle = \(\frac{2 \sqrt{2}}{2}\)= √2 cm
Circumference of circle = 2πr = 2 × π × √2 = 2√2π cm

Figure 3

PR = \(\sqrt{2^2+(1.5)^2}=\sqrt{6.25}\) = 2.5 cm
Radius of circle = \(\frac{2.5}{2}\) = 1.25 cm
Circumference of circle = 2πr = 2 × π × 1.25 = 2.5π cm

Question 2.
In a circle, a chord 4 centimetres away from the centre is 6 centimetres long. What is the circumference of the circle?

Answer:
Length of chord (AB) = 6cm
In the figure, P is the midpoint of AB
AB = 2AP
AP = 3cm
OP = 4cm
r = \(\sqrt{4^2+3^2}=\sqrt{25}\) = 5 cm
Circumference of the circle = 2πr = 2 × π × 5 = 10πcm

Question 3.
The figure below shows an isosceles triangle of base and height 4 centimetres drawn with vertices on a circle.

Calculate the circumference of the circle.
Answer:

Consider the centre of circle as O and triangle as ABC
OC = Radius of circle = r, OD = 4 – r, AD = 2 cm and AO = r
In triangle AOD
(AO)² = (OD)² + (AD)²
r² = 2² + (4 – r)²
r² = 4 + 16 – 8r + r²
8r = 20
Circumference of the circle = 2πr = 2 × π × 2.5 = 5n cm

Question 4.
In each of the pictures below, the centres of the large and small circles are on the same line. In the first and the second pictures, all the small circles have the same diameter. In each of these figures, show that the circumference of the large circle is sum of the circumferences of the small circles.

Answer:
Figure 1

Smaller circles have same diameters.
Consider the diameter as d, circumference of smaller circle = π × diameter = πd
Circumference of two small circles = 2πd
Diameter of the large circle = d + d = 2d
Circumference of the large circle = π × 2d = 2πd
Therefore, Circumference of the large circle is the sum of the circumference of the small circles.

Figure 2

Let d be the diameter of the small circle. .
Sum of circumference of three small circles =3πd
Diameter of the large circle = d + d + d = 3d
Circumference of the large circle = π × 3d
Therefore Circumference of the large circle is the sum of the circumference of the small circles.

Figure 3

In figure diameter of three circles are different, let consider the diameters of small circles are p, q and r.
Circumference of first small circle = πp
Circumference of second small circle = πq
Circumference of third small circle = πr
Sum of circumference of three small circles = πp + πq + πr = π (p + q + r)
Diameter of large circle = p + q + r
Circumference of large circle = π (p + q + r)
Therefore, Circumference of the large circle is the sum of the circumference of the small circles.

Question 5.
In the picture below, the two circles have the same centre. How much more is the circumference < of the larger circle than that of the smaller circle?

Answer:
If r be the radius of the small circle Radius of the large circle = r + 1
Circumference of the small circle = 2πr
Circumference of the large circle = 2π (r + 1) = 2πr + 2π
Therefore, Circumference of the large circle is 2n units more than the circumference of the small circle.

SCERT Class 9 Maths Chapter 8 Solutions – Area

Intext Questions And Answers

Question 1.
What is the relation between the areas of the shaded regions in the two figures?

Answer:
Figure 1
Area = π(2.5)² – π(2)²
= 6.25π – 4π
= 2.25π cm²

Figure 2
r² = (2.5)² – (2)² = 2.25
Area = πr² = 2.25π cm²
Both the area are equal

Textual Questions And Answers

Question 1.
The length of a chord of a circle, 3 centimetres from the centre, is 4 centimetres. What is the area of the circle?
Answer:

Radius = \(\sqrt{3^2+2^2}=\sqrt{9+4}=\sqrt{13}\) cm
Area = πr = π\((\sqrt{13})^2\) = 13π cm²

Question 2.
In each of the pictures below, compute the difference between the area of the circle and the area of the regular polygon, correct up to two decimal places.

Answer:
Figure 1
Radius of the circle = 2cm
Area = πr² = π(2)² = 4π = 3.14 × 4 = 12.56 cm²
Diagonal of the square = 4 cm
One side of a square = \(\frac{4}{\sqrt{2}}\) cm
Area of the square = \(\frac{4}{\sqrt{2}} \times \frac{4}{\sqrt{2}}=\frac{16}{2}\) = 8 cm²
Differences between the areas = 12.56 – 8 = 4.56 cm²

Figure 2
Radius of the circle = 2cm
Area = πr² = π(2)² = 4π = 3.14 × 4 = 12.56 cm²
A regular hexagon is made up of six equilateral triangles.
The side of the equilateral triangle is equal to the radius of the circle.
Side of the equilateral triangle = 2cm
Area of an equilateral triangle = \(\frac{\sqrt{3}}{4}\) × (side)² = \(\frac{\sqrt{3}}{4}\) × (2)² = √3 = 1.73 cm²
Area of a regular hexagon = 6 × Area of an equilateral triangle
= 6 × √3 = 6 × 1.73 = 10.38 cm²
Differences between the areas = 12.56 – 10.38 = 2.18 cm²

Question 3.
In the pictures below, circles are drawn through the vertices of a square and a rectangle.

Calculate the areas of the circles.
Answer:
Figure 1
Diameter of the circle = diagonal of the square
Diameter = \(\sqrt{(3)^2+(3)^2}=\sqrt{2 \times 9}\) = 3√2 cm
Area of the circle = πr² = π × \(\left(\frac{3 \sqrt{2}}{2}\right)^2\) = 4.5 π cm²

Figure 2
Diameter = \(\sqrt{(4)^2+(2)^2}=\sqrt{20}\) cm
Radius = \(\frac{\text { Diameter }}{2}=\frac{\sqrt{20}}{2}\) cm
Area of the circle = π × \(\left(\frac{\sqrt{20}}{2}\right)^2\) = 5 π cm²

Question 4.
Draw a square and draw circles with its vertices as centres and radius as half the side. Draw another square composed of four smaller squares of the same size as the first square, and draw a circle that just fits inside it. Prove that the area of the large circle is the sum of the areas.of the four small circles.

Answer
Figure 1
Radius of one small circle = r cm
Area of one small circle = πr² cm²
Area of the four small circles – 4πr² cm²

Figure 2
Radius of large circle = sum of the radius of two small circles = 2r
Area of the large circle = π(2r)² = 4πr² cm²
The sum of the areas of four small circles is equal to the area of the large circle.

Question 5.
In the pictures below, the squares are of the same size. Prove that the areas of the green regions (shaded regions) in the pictures are equal.

Answer:
Figure 1
Side of the square = 2a
Area of the square = (2a)² = 4a² cm²
A circle is divided into four equal parts and placed at the four comers of a square.
So, radius of the circular segment = a cm
Area of the circular segment = \(\frac{\pi a^2}{4}\)cm²
Area of the circular segment placed at the four comers of the square = 4 × \(\frac{\pi a^2}{4}\) = πa² cm²
Area of the green region (shaded region) = (4 a² – πa²) cm²

Figure 2
Side of the square = 2a
Area of the square = (2 a)² = 4a² cm²
Radius of the circle = a cm
Area of the circle = πr² cm²
Area of the green region (shaded region) = (4 a² – πa²) cm²

Question 6.
Parts of circles are drawn with the vertices of a regular hexagon as centres and the figure below is cut out. Calculate the area of the figure cut out.

Answer:
In the Figure, circular segments are placed at all the six comers of the regular hexagon. The angle at each comer of the regular hexagon isl20°. Combining three circular segments forms a complete circle. The figure contains six such circular segments, which means they can form two complete circles. In other words, the six circular segments are equivalent to two complete circles.

Radius of the circular segment = 1 cm
Area of the circle = πr² = π cm²
Area of two complete circle = 2 × πr² = 2π cm²
Area of the regular hexagon = \(\frac{1}{2}\) × 2 × √3 × 6 = 6√3 cm²
The area of the cut-down portion = Area of the regular hexagon – Area of two complete circle
= (6√3 — 2π) cm²

Question 7.
Parts of a circle are drawn within a square like this. Prove that the area of blue region (shaded region) is half the area of the square.

Answer:
Side of the square = a cm
Area of the square = a² cm²
Area of the semicircle = \(\frac{πa^2}{2}\) cm²
Area of the region found between two quarter circles = [\(\frac{a^2}{2}\) – 2 × \(\frac{1}{4}\) × π × \(\left(\frac{a}{2}\right)^2\)] cm²
Area of the shaded region = area of the semicircle + area of the region found between two quarter circles.
= \(\frac{πa^2}{2}\) + [\(\frac{a^2}{2}\) – 2 × \(\frac{1}{4}\) × π × \(\left(\frac{a}{2}\right)^2\)]
= \(\frac{a^2}{2}\) cm²

Circle Measures Class 9 Extra Questions and Answers Kerala Syllabus

Question 1.
A wire is bent to form a circle with a diameter of 6 centimeters. What will be the diameter of the circle formed by bending a wire that is twice the length of the first one ?
Answer:
The circumferences of circles change in proportion to their diameters. Therefore, if the length of the wire is doubled, the diameter of the circle formed will also get double. This means the diameter of the new circle will be 12 centimeters.

Question 2.
The circumference of a circle with a diameter of 3 meters has been measured to be approximately 9.42 meters.
i) How can we calculate the circumference of a circle with a diameter of 6 meters without measuring it?
ii) Also, what is the circumference of a circle with a diameter of 1.5 meters?
Answer:
i) Since the circumference of a circle with a diameter of 2 meters is 9.42 meters, the circumference of a circle with a diameter of 6 meters will be twice that amount.
Therefore, it will be 2 × 9.42 = 18.84 meters.
ii) Circumference of a circle with diameter 1.5 meters = \(\frac{9.42}{2}\) = 4.7 meters

Question 3.
A wire of 20 centimeters long has been bent to form a circle. If half of that wire is cut and bent into another circle, what will be the diameter of the new circle?
Answer:
Length of the wire = 20 cm
Half the length of the wire = 10 cm
The length of the wire will be equal to the circumference of the circle.
10 = π × diameter
Diameter = \(\frac{10}{\pi}\) cm

Question 4.
Find the length of a chord which is at a distance of 2 cm from the centre of a circle with a circumference of 8ir centimetres.

Answer:
Circumference of a circle = 8π
2πr = 8π
r = 4cm
Considering triangle POR, (OP)² = (OR)² + (PR)²
4² = 2² + (PR)²
PR² = 16 – 4 = 12
PR = √12 = 2√3 cm

Question 5.

i) What Is the area of the circle?
ii) What is the circumference of the circle?
Answer:
The perimeter of the regular hexagon is equal to 6 times the length of one of its sides.
Length of one of its sides = \(\frac{24}{6}\) = 4cm

The regular hexagon is divided into six congruent isosceles triangles.
The length of the side is equal to the radius of the circle.
i) Radius of the circle = 4 cm
ii) Perimeter of the circle = 2πr
= 2 × π × 4
= 8π cm

Question 6.
Calculate the circumference and area of a circle with a radius of 8 centimeters.
Answer:
Radius = 8 cm
Circumference = 2πr = 2 × π × 8 = 16π cm
Area = πr² = π × 8² = 64 π cm²

Question 7.
The diameter of a circle is 10 centimeters. Calculate the area of the circle.
Answer:
Diameter of a circle = 10cm
Radius = \(\frac{10}{2}\) = 5cm
Area of the circle = πr² = π × 5²
= 25π cm²

Question 8.
i) Calculate the perimeter of a wheel of radius 15 cm. What is the distance covered by the wheel in 5 rotations?
ii) What is the distance covered in 5 rotations by another wheel of twice the radius of the circle?
Answer:
i) Radius = 15cm
Perimeter = 2πr = 2 × π × 15 = 30 π cm
Distance travelled when the circle completes 5 rotations = 5 × 30π = 60π cm

ii) Radius = 2 × 15 = 30 cm
Perimeter = 2πr = 2 × π × 30 = 60 π cm
Distance travelled when the circle completes 5 rotations = 5 × 60π = 300π cm

Question 9.
In the diagram, AB is the diameter of the larger semicircle. The segments AC, CD, and DB are the diameters of the smaller semicircles, where AC = 4 centimetres, CD = \(\frac{A C}{2}\), DB = \(\frac{C D}{2}\).What is the area of the shaded region?
Answer:
Area of a circle with diameter AB = πr² = π(3.5)² = 12.25π cm²
Area of a circle with diameter AC = πr² = π(2)² = 4π cm²
Area of a circle with diameter CD = πr² = π(1) = π cm²
Area of a circle with diameter BD = πr² – π(0.5)² = 0.25π cm²
Area of the shaded region = Area of a circle with diameter AB – (Area of a circle with diameter AC + Area of a circle with diameter CD+ Area of a circle with diameter BD)
= 12.25π – (4π + π + 0.25π) = 7π cm²

Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 5 Irrational Multiplication Extra Questions and Answers Notes to clear their doubts.

Kerala SCERT Class 9 Maths Chapter 5 Solutions Irrational Multiplication

Irrational Multiplication Class 9 Kerala Syllabus Questions and Answers

Kerala State Syllabus 9th Standard Maths Chapter 5 Irrational Multiplication Solutions Questions and Answers

Class 9 Maths Chapter 5 Kerala Syllabus – Multiplication

Textual Questions And Answers

Question 1.
We can make a rectangle using four equilateral triangles of the same size by cutting two of them along their heights and rearranging these pieces and the other two whole triangles:

If the sides of all equilateral triangles are 2 centimetres, what is the perimeter and area of the rectangle?
Answer:
If we consider the sides of the triangles, we will get it as follows:

So, the sides of the given rectangle are:

Now,
Perimeter of the rectangle = 2(length + breadth)
= 2(√3 + √3 + 2)
= 2(2√3 + 2)
= 4(√3 + 1)
= 4(1.732 + 1)
= 4 × 2.732
= 10.928 cm

Area of the rectangle = length × breadth
= 2√3 × 2
= 4√3 – 4 × 1.732
= 6.928 cm²

Question 2.
We can make a trapezium by cutting a square and an equilateral triangle with sides twice that of the square, and rearranging the pieces as below:
If the side of the square is 2 centimeters, what is the perimeter and area of the trapezium?

Answer:
Consider one part of the square. Its measures are:

Consider one part of the triangle. Its measures are:

Now, the measures of the trapezium are;

Perimeter of the trapezium = 2√2 + 2 + 2√3 + 2 + 2√2 + 2√3
= 4 + 4√2 + 4√3
= 4(1 + √2 + √3)
= 4(1 + 1.414+ 1.732)
= 4 × 4.146
= 16.584 cm

Area of the trapezium = \(\frac{1}{2}\) × sum of the lengths of the parallel sides x height
= \(\frac{1}{2}\) × (2 + 2√3 +2 + 2√3) × 2
= 4 + 4√3
= 4(1 + √3)
= 4(1 + 1.732)
= 4 × 2.732
= 10.928 cm²

Question 3.
The picture shows the figure formed by joining two squares

Calculate the length of the bottom side of this figure, correct to a centimetre.
Answer:
Side length of the big square = √27
= \(\sqrt{9 \times 3}\)
= √9 × √3
= 3 × √3
= 3 × 1.73
= 5.19 cm
Side length of the small square = √12
= \(\sqrt{4 \times 3}\)
= √4 × √3
= Side length of the big square + Side length of the small square
= 5.19 + 3.46
= 8.65 cm

Question 4.
The figure shows two squares with two corners joined:

Find the length of the slanted line.
Answer:
We know that the diagonal length of a square with side length a is a√2.
Diagonal length of the square with side 3 = 3√2 cm
Diagonal length of the square with side 2 = 2√2 cm
Length of the slanted line
= Diagonal length of the square with side 3 + Diagonal length of the square with side 2
= 3√2 + 2√2
= 5√2
≈ 5 x 1.414
≈ 7.07 cm

Question 5.
Calculate the length of the third side of the right triangle in the picture and also its perimeter.

Answer:
Side length of the big square = \(\sqrt{50}\)
= \(\sqrt{25 \times 2}\)
= \(\sqrt{25}\) × √2
= 5 × √2
≈ 5 × 1.414
≈ 7.07 cm

Side length of the small square = \(\sqrt{18}\)
= \(\sqrt{9 \times 2}\)
= √9 × √2
= 3 × √2
≈ 3 × 1.414
≈ 4.242 cm

Using Pythagoras theorem,
Third side of the right triangle
= \(\sqrt{(\text { side length of the big square })^2-(\text { side length of the small square })^2}\)
= \(\sqrt{(5 \sqrt{2})^2-(3 \sqrt{2})^2}\)
= \(\sqrt{(25 \times 2)-(9 \times 2)}\)
= \(=\sqrt{2(25-9)}\)
= \(=\sqrt{2 \times 16}\)
= √2 × √16
= √2 × 4
= 4√2
Perimeter of the right triangle = 5√2 + 3√2 + 4√2
= 12√2
≈ 12 × 1.414
≈ 16.968 cm

Question 6.
The product of some of the pairs of numbers below are natural numbers or fractions. Find those pairs.
(i) √3, √12
Answer:
√3 × √12 = √3 × \(\sqrt{3 \times 4}\)
= √3 × √3 × √4
= 3 × 2
= 6
Product is a natural number.

(ii) √3, √1.2
Answer:
\(\sqrt{3} \times \sqrt{1.2}=\sqrt{3 \times 1.2}\)
= \(\sqrt{3.6}\)

(iii) √5, √8
Answer:
\(\sqrt{5} \times \sqrt{8}=\sqrt{5 \times 8}\)
= \(\sqrt{40}\)
= \(\sqrt{4 \times 10}\)
= √4 × √10
= 2 × √10
= 2√10

(vi) √0.5, √8
Answer:
√0.5 × √8 = \(\sqrt{0.5 \times 8}\)
= √4
= 2
Product is a natural number.

(v) \(\sqrt{7 \frac{1}{2}}, \sqrt{3 \frac{1}{3}}\)
Answer:
\(\sqrt{7 \frac{1}{2}} \times \sqrt{3 \frac{1}{3}}=\sqrt{\frac{15}{2}} \times \sqrt{\frac{10}{3}}\)
= \(\sqrt{\frac{15}{2} \times \frac{10}{3}}\)
= \(\sqrt{5 \times 5}\)
= \(\sqrt{25}\)
= 5
Product is a fraction.

(vi) \(\sqrt{\frac{2}{5}}, \sqrt{\frac{1}{10}}\)
Answer:
\(\sqrt{\frac{2}{5}} \times \sqrt{\frac{1}{10}}=\sqrt{\frac{2}{5} \times \frac{1}{10}}=\sqrt{\frac{1}{25}}=\frac{1}{5}\)
Product is a natural number.

Class 9 Maths Kerala Syllabus Chapter 5 Solutions – (Division)

Textual Questions And Answers

Question 1.
Prove that (√2 + 1) (√-2 – 1) = 1. Using this:
i. Compute \(\frac{1}{\sqrt{2}-1}\) up to two decimal places.
ii. Compute \(\frac{1}{\sqrt{2}+1}\) up to two decimal places.
Answer:
(√2 + 1)(√2 – 1) = (√2)² – 1²
= 2 – 1
= 1
i. \(\frac{1}{\sqrt{2}-1}=\frac{(\sqrt{2}+1)(\sqrt{2}-1)}{\sqrt{2}-1}\)
= √2 + 1
= 1.41 + 1
= 2.41

ii. \(\frac{1}{\sqrt{2}+1}=\frac{(\sqrt{2}+1)(\sqrt{2}-1)}{\sqrt{2}+1}\)
= √2 – 1
= 1.41 – 1
= 0.41

Question 2.
Compute the lengths of the sides of the equilateral triangle shown below, correct to a millimetre.

Answer:
Consider the triangle given below.

Using Pythagoras theorem;
(2x)² – x² = 4²
4x² – x² = 16
3x² = 16
x² = \(\frac{16}{3}\)
x = \(\sqrt{\frac{16}{3}}\)
= \(\frac{\sqrt{16}}{\sqrt{3}}\)
= \(\frac{4}{1.732}\)
= 2.30 cm

Length of the sides = 2x
= 2 × 2.30
= 4.60 cm

Question 3.
All red triangles in the picture are equilateral and of the same size. What is the ratio of the sides of the outer and inner squares?

Answer:
Let ‘a’ be the side length of the equilaleral triangle
Side length of the outer square = a + a√3 = a( 1 + √3)
Side length of the inner square = a + a√3 – (a + a)
= a + a√3 – 2a
= a√3 – a
= a(√3 – 1)

Ratio of the sides of the outer and inner squares
= \(\frac{\text { Side length of the outer square }}{\text { Side length of the inner square }}\)
= \(\frac{\mathrm{a}(1+\sqrt{3})}{\mathrm{a}(\sqrt{3}-1)}\)
= \(\frac{\sqrt{3}+1}{\sqrt{3}-1}\)
= (√3 + 1) (√3 – 1)

Question 4.
Prove that \(\sqrt{2 \frac{2}{3}}=2 \sqrt{\frac{2}{3}}\) and \(\sqrt{3 \frac{3}{8}}=3 \sqrt{\frac{3}{8}}\) Find other numbers like this.
Answer:

The general form of such numbers can be defined as follows;

Question 5.
Among the pairs of numbers given below, find those for which the quotient of the first by the second is a natural number or a fraction.
(i) √72, √2
(ii) √27, √3
(iii) √125, √50
(iv) √10, √2
(v) √20, √5
(vi) √18, √8
Answer:
(i) \(\frac{\sqrt{72}}{\sqrt{2}}=\frac{\sqrt{36 \times 2}}{\sqrt{2}}=\frac{\sqrt{36} \times \sqrt{2}}{\sqrt{2}}\)
= \(\sqrt{36}\)
= 6
When we divide the first by the second the result is natural number.

(ii) \(\frac{\sqrt{27}}{\sqrt{3}}=\frac{\sqrt{9 \times 3}}{\sqrt{3}}=\frac{\sqrt{9} \times \sqrt{3}}{\sqrt{3}}\)
= √9
= 3
When we divide the first by the second the result is natural number.

(iii) \(\frac{\sqrt{125}}{\sqrt{50}}=\frac{\sqrt{25 \times 5}}{\sqrt{25 \times 2}}\)
= \(\frac{\sqrt{25} \times \sqrt{5}}{\sqrt{25} \times \sqrt{2}}\)
= \(\frac{\sqrt{5}}{\sqrt{2}}\)

(iv) \(\frac{\sqrt{10}}{\sqrt{2}}=\frac{\sqrt{2 \times 5}}{\sqrt{2}}\)
= \(\frac{\sqrt{2} \times \sqrt{5}}{\sqrt{2}}\)
= √5

(v) \(\frac{\sqrt{20}}{\sqrt{5}}=\frac{\sqrt{4 \times 5}}{\sqrt{5}}\)
= \(\frac{\sqrt{4} \times \sqrt{5}}{\sqrt{5}}\)
= √4
= 2
When we divide the first by the second the result is a Natural fraction.

(vi) \(\frac{\sqrt{18}}{\sqrt{8}}=\frac{\sqrt{9 \times 2}}{\sqrt{4 \times 2}}=\frac{\sqrt{9} \times \sqrt{2}}{\sqrt{4} \times \sqrt{2}}\)
= \(\frac{\sqrt{9}}{\sqrt{4}}=\frac{3}{2}\)
When we divide the first by the second the result is a fraction.

SCERT Class 9 Maths Chapter 5 Solutions – Areas of Triangles

Textual Questions And Answers

Question 1.
For each of the lengths below, calculate the area of the equilateral triangle with that as the lengths of the sides:
(i) 10 cm
(ii) 5 cm
(iii) √3 cm
Answer:
(i) Here, a = 10
Area = \(\frac{\sqrt{3}}{4}\) × a
= \(\frac{\sqrt{3}}{4}\) × 10²
= \(\frac{\sqrt{3}}{4}\) × 100
= √3 × 25
~ 1.732 × 25
~ 43.3 cm²

(ii) Here, a = 5
Area = \(\frac{\sqrt{3}}{4}\) × a²
= \(\frac{\sqrt{3}}{4}\) × 5²
= \(\frac{\sqrt{3}}{4}\) × 25
= √3 × 6.25
~ 6.25 × 1.732
~ 10.825 cm²

(iii) Here, a = √3
Area = \(\frac{\sqrt{3}}{4}\) × a²
= \(\frac{\sqrt{3}}{4}\) × (√3)²
= \(\frac{\sqrt{3}}{4}\) × 3
= √3 × 0.75
~ 1.732 × 0.75
~ 1.299 cm²

Question 2.
Calculate the area of the regular hexagon with lengths of the sides 6 centimetres.
Answer:
A regular hexagon can be divided into six equilateral triangles. In such cases; side length of the equilateral triangle = side length of the regular hexagon

So,
Area of the regular hexagon
= sum of the areas of these six equilateral triangles
= 6 × area of one equilateral triangle
= 6 × \(\frac{\sqrt{3}}{4}\) × 6²
= 6 × √3 × 9
= 54 × √3
= 54 × 1.732
= 93.528 cm²

Question 3.
Calculate the perimeter and area of the equilateral triangle with height 12 centimetres.
Answer:
We know that height of an equilateral triangle of side length a is \(\frac{\sqrt{3}}{4}\) × a.
Here, height = 12 cm
ie, \(\frac{\sqrt{3}}{4}\) × a = 12
a = \(=\frac{12 \times 2}{\sqrt{3}}\)
= \(\frac{3 \times 4 \times 2}{\sqrt{3}}\)

Perimeter of the equilateral triangle = 3 × 8√3
= 24√3
≈ 24 × 1.732
≈ 41.568 cm

Area of the equilateral triangle = \(\frac{\sqrt{3}}{4}\) × a²
= \(\frac{\sqrt{3}}{4}\) × (8√3)²
= \(\frac{\sqrt{3}}{4}\) × 64 × 3
= √3 × 16 × 3
≈ 1.732 × 48
≈ 83.136 cm²

Question 4.
Calculate the perimeter and area of the regular hexagon with the distance between parallel sides 6 centimetres.
Answer:

Here,
distance between parallel sides = 2 × height of an equilateral triangle
6 = 2 × \(\frac{\sqrt{3}}{2}\) × a
= √3 × a
a = \(\frac{6}{\sqrt{3}}\)
= \(\frac{3 \times 2}{\sqrt{3}}\)
= \(\frac{\sqrt{3} \times \sqrt{3} \times 2}{\sqrt{3}}\)
= √3 × 2
= 2√3 cm

Perimeter of the regular hexagon = 6 × 2√3
= 12√3
≈ 12 × 1.732
≈ 20.784 cm

Area of the regular hexagon = 6 × area of one equilateral triangle
= 6 × \(\frac{\sqrt{3}}{4}\) × (2√3)²
= 6 × \(\frac{\sqrt{3}}{4}\) × 4 × 3
= 18 × √3
≈ 18 × 1.732
≈ 31.176 cm²

Question 5.
Calculate the height and area of the triangle with sides 8 centimetres, 6 centimetres,6 centimetres.
Answer:
Consider the picture given below;

Using Pythagoras theorem, height
= \(\sqrt{6^2-4^2}\)
= \(\sqrt{36-16}\)
= \(\sqrt{20}\)
= \(\sqrt{4 \times 5}\)
= √4 × √5
= 2 × √5
= 2√5 cm
Area = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 8 × 2√5
= 8√5 cm²

Question 6.
For each of the set of three lengths given below, calculate the area of the triangle with these as the lengths of sides:
(i) 4 cm, 5 cm, 7cm
(ii) 4cm, 13 cm, 15 cm
(iii) 5 cm, 12 cm, 13 cm
Answer:
If the lengths of the sides of a triangle are a, b, c and we take the semi perimeter
s = \(\frac{1}{2}\) (a + b + c), then the area of the triangle is \(\sqrt{s(\mathrm{~s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}\)
(i) Here, a = 4 cm
b = 5 cm
c = 7 cm

(ii) Here, a = 4 cm
b = 13 cm
c = 15 cm

(iii) Here, a = 5 cm
b = 12 cm
c = 13 cm

Irrational Multiplication Class 9 Extra Questions and Answers Kerala Syllabus

Question 1.
Multiply 3√8 and 7√8
Answer:
3√8 × 7√8 = 3 × √8 × 7 × √8
= 3 × 7 × √8 × √8
= 21 × 8
= 168

Question 2.
Write in ascending order.
(i) 3√5 and 4√3
(ii) 2√5, 5√2 and 3√7
Answer:
(i) \(\sqrt{53}=\sqrt{3 \times 3 \times 5}=\sqrt{45}\)
4√3 = \(\sqrt{4 \times 4 \times 3}=\sqrt{48}\)
48 > 45 ⇒ \(\sqrt{48}>\sqrt{45}\)
⇒ 4√3 > 3√5
When arrange in ascending order; 3√5, 4√3

(ii) 2√5 = √2 × 2 × 5 = √20
5√2 = √5 × 5 × 2 = √50
3√7 = √3 × 3 × 7 = √63
20 < 50 < 63 ⇒ √20 <√50 < √63
⇒ 2√5 < 5√2 < 3√7
When arrange in ascending order; 2√5, 5√2, 3√7

Question 3.
x = \(\sqrt{0.5}\), y = \(\sqrt{32}\), z = \(\sqrt{128}\)
(a) Find xy, yz and xz.
Answer:

(b) Find xy + yz + xz
Answer:
xy + yz + xz = 4 + 64 + 8
= 76

(c) Prove that y = 8x
Answer:
8x = 8 x \(\sqrt{0.5}\)
= \(\sqrt{64 \times 0.5}\)
= \(\sqrt{32}\)
= y

Question 4.
√8 can be written as \(\sqrt{4 \times 2}\) = 2√2
a) Write √18, √32 as the product of an integer and √2
Answer:
\(\sqrt{18}=\sqrt{9 \times 2}\) = 3√2
\(\sqrt{32}=\sqrt{16 \times 2}\) = 4√2

b) Write the simplified form of √2 + √8 + √18 – √32
Answer:
\(\sqrt{2}+\sqrt{8}+\sqrt{18}-\sqrt{32}\)
= √2 + 2√2 + 3√2 – 4√2
= 6√2 – 4√2
= 2√2

Question 5.
Find the area of equilateral triangle if perimeter is 18 cm.
Answer:
Perimeter =18
3 × side = 18
side = \(\frac{18}{3}\)
= 6 cm
Area = \(\frac{\sqrt{3}}{4}\) × (side)²
= \(\frac{\sqrt{3}}{4}\) × 6²
= √3 × 9
= 9√3 cm²