Class 8

The comprehensive approach in SCERT Class 8 Basic Science Textbook Solutions Chapter 3 Pressure Important Questions ensure conceptual clarity.

Pressure Extra Questions and Answers Class 8 Basic Science Chapter 3 Kerala Syllabus

Pressure Class 8 Important Questions

Question 1.
Which of the following is the unit of atmospheric pressure?
(N, parsec, kg/m³, bar)
Answer:
bar

Question 2.
It is possible to drink juice using a straw because of ……………………
Answer:
Atmospheric pressure

Question 3.
Correct the following. As the number of particles increases, gas pressure decreases.
Answer:
As the number of particles increases, gas pressure increases.

Question 4.
The instrument used to measure atmospheric pressure is a ………………………
Answer:
Barometer.

Question 5.
The pressure exerted on an area of 10 m² by a force of 50 N is ………………….. .
(5 N, 5 N/m², 0.2 N/m², 0.2 N)
Answer:
5 N/m²
P = F/A, F = 50 N, A =10 m²
P = 50 /10 = 5 N/m²

Question 6.
The instrument used to measure liquid pressure is ……………………. .
Answer:
Manometer

Question 7.
Choose the correct equation of pressure.
Answer:
(P = F × A, P = A/F, P = F/A)
Answer:
P = F/A

Question 8.
Observe the diagram

Water flowing down through the holes P, Q and R on a bottle is depicted.
a) Which is the correct figure? i/ii/iii
b) What is the reason for this inference?
Answer:
a) Figure iii is correct

b) As depth of liquid increases pressure increases. So the water flowing from the hole R will have the greatest force and covers more distance.

Question 9.
Observe the picture below. Both beakers contain the same volume of water.

a) Which beaker experiences maximum pressure at the bottom?
b) Justify your answer.
Answer:
a) Beaker A experience maximum pressure at the bottom.

b) The height of liquid column (or depth) is more in beaker A than beaker B. So more pressure is experienced at the bottom of beaker A.

Question 10.
Liquids also exert pressure. Write down any two factors that influence liquid pressure. (Term II, 2022)
Answer:

1. Height of the liquid column
2. Density of the liquid

Question 11.
The weight of the air column over unit area of earth’s surface is atmospheric pressure.
a) What is standard atmospheric pressure?
b) Name the instrument used to measure atmospheric pressure. (Annual 2022)
Answer:
a) Atmospheric pressure at sea level is known as standard atmospheric pressure. It is defined as the weight of a mercury column that is 0.76 m high, with a unit cross sectional area, exerting a pressure of 1 atm.

b) Barometer.

Question 12.
Two identical balloons one filled with water and other with kerosene of equal quantity.

a) Identify the balloon filled with water.
b) Why is the expansion of balloons appear different in figures (a) and (b)? (Annual 2022)
Answer:
a) Figure (b)

b) Water has higher density than kerosene. As density increases, liquid pressure increases. So the balloon which is expanded has water filled in it and the other balloon which is small in size has kerosene filled in it.

Question 13.
Check whether the given statement is correct or wrong. Correct the wrong statement.
(i) Atmospheric pressure increases as height from the Earth’s surface increases.
(ii) The porters in railway stations place a piece of cloth before carrying luggage on head to decrease the area of contact between luggage and head to reduce the pressure felt on their head.
Answer:
(i) Statement is wrong
Corrected statement: Atmospheric pressure decreases as height from the Earth’s surface increases.

(ii) Statement is wrong
Corrected statement: The porters in railway stations place a piece of cloth to increase the area of contact between luggage and head to reduce the pressure felt on their head.

Question 14.
Calculate the force exerted by atmospheric pressure on a table surface of area 3 m². Atmospheric pressure is 101325 Pascal.
Answer:
Force = Pressure × Area
= 1 Pa × 3 m² = 101325 × 3
= 303975 N

Question 15.
The thrust acting per unit area by a liquid is the liquid pressure.

a) Three points A, B and C are marked on a cylindrical glass vessel filled with water as shown in figure. If three identical inflated balloons are placed on these points inside the cylinder, which one will be the largest in size? Justify your answer.
b) Do you expect any change in size to the balloons, if the water is replaced with kerosene? Explain.
Answer:
a) The balloon placed in the position A will be the largest in size. As depth decreases liquid pressure also decreases. The pressure experienced at position A is the least compared to positions B and C and so the balloon with largest size occupies position A.

b) Yes, when kerosene is replaced with water the size of the balloons become comparatively larger than in water because kerosene has low density than water. When density decreases liquid pressure also decreases. So when the pressure experienced is less, the size of balloons (volume of balloons) is comparitively larger.

Question 16.
When area increases, pressure decreases. Find three situations where this principle is applied.
Answer:
a) We can walk easily through a muddy place by placing a plank over it.
b) Ducks walk easily over mud.
c) Constructing dams with wider basements.

Question 17.
Write down reasons for the following statements.
a) Astronauts wear specially designed suits.
b) Sewing needles are made with sharp tips.
c) Connecting the wheels of tanks and other such vehicles using wide chains.
Answer:
a) The suit helps us to maintain the pressure between the dress and the body equal to the atmospheric pressure.

b) Sharp tips have less surface area of contact. As area of contact decreases, pressure increases so that sewing needles can pierce into the materials to be stitched easily.

c) Connecting wheels of tanks and other such vehicles using wide chains helps in making more surface area of contact with the ground to exert less pressure over it. This ensures its smooth motion.

The comprehensive approach in SCERT Class 8 Basic Science Textbook Solutions Chapter 2 Motion and Force Important Questions ensure conceptual clarity.

Motion and Force Extra Questions and Answers Class 8 Basic Science Chapter 2 Kerala Syllabus

Motion and Force Class 8 Important Questions

Question 1.
SI unit of force is ……………………….
Answer:
newton(N)

Question 2.
A child is standing in a field. With respect to the ground, the child is in a state of ……………………. (rest / motion).
Answer:
rest

Question 3.
If a body in motion covers equal distance in equal intervals of time, the body is said to have ………………………….. (uniform speed / non uniform speed).
Answer:
uniform speed

Question 4.
Cheetah is the fastest animal on land. Its speed is given in different ways. Which among these is correct?
a) 25 metre per s
b) 25 m/s
c) 25 m/s²
d) 25 M/S
Answer:
b) 25 m/s

Question 5.
Examine the figure below.

A man starts from A and reaches D through the points B and C. Find the distance travelled from A to D.
Answer:
Distance travelled from A to D = 4 + 3 + 4 = 11 km

Question 6.
Motion of four different objects are stated below. Choose the odd one. Also state the reason for choosing it.
(A train starting from the station, A stone thrown up, The motion of a ball rolling down a hill, The motion of a coconut falling from a coconut tree)
Answer:
A train starting from the station.
Because the other three (stone thrown up, ball rolling down a hill, and coconut falling) move mainly because of gravitational force.

Question 7.
The moving parts of machines experience wear and tear. Why?
Answer:
Due to frictional force

Question 8.
The solid substance used as a lubricant is ………………………
Answer:
Graphite

Question 9.
Bearings are used in motor vehicles. Why?
Answer:
Because rolling friction is less than that of sliding friction.

Question 10.
Define the term streamlining?
Answer:
The method of reducing friction by changing the shape is called streamlining.

Question 11.
“Overspeed of vehicles result in accidents”
Write two messages to be written on an awareness board in your school premises to minimise road accidents.
Answer:
Speed thrills but kills, Don’t mix drink with drive, Better late than never, Wear helmets while driving motor cycles.

Question 12.
The paths of two cars are as shown below. Car A started from P and reached R through Q and car B started from P and reached R through S.

a) What is the total distance covered by the car A?
b) What is the total distance covered by car B?
Answer:
a) Total distance covered by car A = 500 + 600 = 1100 m
b) Total distance covered by car B = 400 + 600 = 1000 m

Question 13.
Analyse the table given below and find which force is felt in each situation.

Situation	Force
A coconut falling from a coconut tree	Gravitational force
Magnet attracts an iron nail	…………a)………….
Hammering a nail	……………b)…………..

Answer:
a) Magnetic force
b) Muscular force

Question 14.
A book on a table is in a state of rest.
a) Identify the reference body for the above statement?
b) Considering the Sun as reference body, the book is in a state of (motion/rest)
Answer:
a) Reference body: Table
b) With respect to Sun, the book is in a state of motion

Question 15.
The conversation between two students standing on the road side about a person inside a moving bus is given below.
Student 1 “That person is at state of rest”
Student 2 “No, he is in motion”
Explain the possibilities of the correctness of these statements based on the concept of a reference body.
Answer:

• Reference body: Bus
  That person with respect to the bus: State of rest
  Student 1 – Statement is correct
• Reference body: Road
  That person with respect to the road: State of motion
  Student 2 – Statement is correct

Question 16.
A person starts from A, moves through B, C and reaches D in 30 s.
a) Calculate the distance.
b) Find out the speed.

Answer:
a) Distance covered from A to D = 10 + 10 + 10 = 30 m
b) Speed = \(\frac{\text { Distance }}{\text { Time }}\) = \(\frac{30 \mathrm{~m}}{30 \mathrm{~s}}\) = 1 m/s

Question 17.
a) Choose the appropriate reference body from the given list so that a passenger sitting in a moving train can said to be in motion, (train, Another passenger next to him in the train, objects outside the train)
b) What is meant by the term ‘reference body’?
Answer:
a) The correct reference body is: Objects outside the train
Because the passenger is moving when seen from outside the train.

b) The object taken to determine the state of motion or the state of rest of a body is called the reference object.

Question 18.
What is frictional force? Point out two disadvantages of friction in our daily life and suggest remedy for overcoming it?
Answer:
When a surface moves or tries to move over another surface, a parallel force is produced between them against their relative motion. This force is called frictional force.

• Difficulty to draw water from the well.
  Remedy – Apply oil in the pulley.
• Chains of bicyles sometimes does not move due to friction.
  Remedy – Apply grease on it.

Question 19.
Figure shows a device used in motor vehicles.

a) Name the device.
b) How much time will it take for a vehicle to cover a distance of 90 km at 60 km/h rate?
Answer:
a) Speedometre

b) Speed = \(\frac{\text { distance travelled }}{\text { time taken }}\)
distance travelled = 90 km
speed = 60 km/h
Time taken = \(\frac{90}{60}\) = 1.5 h

Question 20.
Observe the depiction of the path taken by a person as he moves from A to C through B.

Find out his speed?
Answer:
Speed = \(\frac{\text { distance travelled }}{\text { time taken }}\)
distance travelled = 20 m + 60 m = 80 m
Time taken = 4 s + 6 s = 10 s
Speed = \(\frac{80}{10}\) = 8 m/s

Question 21.
Complete the table suitably.

Measurement	Unit
Time	…………..a)…………..
…………..b)…………..	m
Speed	…………..c)…………..

Answer:
a) s
b) Distance
c) m/s

Question 22.
Classify the following situations into contact and non-contact forces.
a) Applying break in a bicycle.
b) The earth revolving around the sun.
c) The speed of a ball rolling on ground is reduced.
Answer:
Contact forces
a) Applying brake in a bicycle
c) The speed of a ball rolling on ground is reduced.
Non-Contact force
b) The earth revolving around the sun.

Question 23.
The following figure is of a traffic sign board placed near school for road safety.

a) What does the board indicate?
b) A motor bike rider travels 180 km in 2 hours through the road near the school. Calculate his speed. Does he cross the speed limits as per Kerala Road Safety Rules.
c) Which is the device used to find the speed in vehicles.
Answer:
a) Speed limit around this school area is 30 km/h

b) Speed = \(\frac{\text { distance travelled }}{\text { time taken }}\) = \(\frac{180}{2}\) = 90 km/h
Yes, The bike travelled in overspeed,

c) Speedometer

Question 24.
Classify the following situations as contact force and non-contact force.
a) A magnet attracting a nail
b) Pushing a wall
c) An apple falling from an apple tree
Answer:
a) A magnet attracting a nail – Non-contact force
b) Pushing a wall – Contact force
c) An apple falling from an apple tree – Non-contact force

Question 25.
A child starts from A and reaches C through B in 1 s.

a) What is the total distance travelled by the child?
b) Calculate the speed of the child?
Answer:
Total distance travelled by the child = 4 m + 3 m = 7 m
Speed = \(\frac{\text { distance travelled }}{\text { time taken }}\) = \(\frac{7}{1}\) = 7 m/s

Question 26.
Why are airplanes and boats made in special shapes?
Answer:
Airplanes and boats are made in special shapes to reduce friction. Smooth, streamlined designs help them move easily through air and water, making travel faster and easier by decreasing the resistance they face.

Question 27.
Bus A covered 75 m in 5 s. Bus B covered 169 m in 13 s.
a) Which bus covered a greater distance?
b) Which bus has a higher speed?
Answer:
a) Bus B covered greater distance (169 m).
b) Speed of bus A = \(\frac{75 \mathrm{~m}}{5 \mathrm{~s}}\) = 15 m/s
Speed of bus B = \(\frac{169 \mathrm{~m}}{13 \mathrm{~s}}\) = 13 m/s
Bus A has higher speed.

The comprehensive approach in SCERT Class 8 Basic Science Textbook Solutions Chapter 1 Measurement and Units Important Questions ensure conceptual clarity.

Measurement and Units Extra Questions and Answers Class 8 Basic Science Chapter 1 Kerala Syllabus

Measurement and Units Class 8 Important Questions

Question 1.
Find out the odd one from following units and write it down.
a) m
b) g
c) kg
d) mg
Answer:
a) m
All others are the units of mass.

Question 2.
How much time does light take to travel 3 lakh kilometre in vacuum?
(1s, 1/2s, 2s, 3s)
Answer:
1s

Question 3.
Find out the correctly written unit from those given below.
(Volt, pascal, Kelvin, Ampere)
Answer:
pascal

Question 4.
SI unit of force is ………………………
Answer:
newton (N)

Question 5.
Imagine that the distance from school to your friend’s house is 2250 m. State this in kilometres.
Answer:
2250 m = 2250/1000 km = 2.250 km

Question 6.
Find the odd one.
cm³ kg/m³; mm, m³
Answer:
mm, All others are derived units.

Question 7.
Choose the derived unit from the following.
(m, kg, N, s)
Answer:
N

Question 8.
Select the odd one. State reason.
(m², m³, m, kg/m³)
Answer:
m. m is a fundamental unit. Others are derived units.

Question 9.
Identify the relation in first word-pair and complete the second one.
Volume : m³
Density : ………………………
Answer:
kg/m³

Question 10.
Write the appropriate unit for expressing the distance between planets in the solar system.
a) m
b) nm
c) AU
d) km
Answer:
c) AU

Question 11.
Pick out the correct representation of units from the following data.
a) 37.65 m
b) 37 m 65 cm
c) 10 Pa
d) 20 Kg
Answer:
Correct representations are a) 37.65 m and c) 10 Pa

Question 12.
Which of the following does not belong to the group?
(km, kg, mm, cm)
Answer:
kg, others are units of length.

Question 13.
A child measures the total thickness of the papers of his science diary having 200 sheets and finds it to be 2 cm. Find the thickness of one of these papers in SI unit.
Answer:
Thickness of one paper = Height of the pile of papers /No. of papers.
Thickness of one paper = 2 cm/200 = 1/100 cm = 0.01 cm
Thickness of paper in SI unit = 0.01 cm/100 = 0.0001 m

Question 14.
Write down the following units in descending order of their values.
a) 2 cm, 2 m, 2 km, 2 mm
b) 5 g, 5 mg, 5 quintal, 5 kg
Answer:
a) 2 km > 2 m > 2 cm > 2 mm
b) 5 quintal > 5 kg > 5 g > 5 mg

Question 15.
Write down the following units in the ascending order of their values.
a) mm, nm, cm, μm
b) 1 m, 1 cm, 1 km, 1 mm
Answer:
a) nm < μm < mm < cm
b) 1 mm < 1 cm < 1 m < 1 km

Question 16.
Complete the table.
Quantity – SI Units
Length – ………………….
Time – ………………….
Area – ………………….
Density – ………………….
Answer:
Length – metre (m)
Time – second (s)
Area – square metre (m²)
Density – kilogram per cubic metre (kg/m²)

Question 17.
Complete the table.

Unit	Symbol
Newton	(a) …………………..
Kelvin	(b) ……………………
Ampere	(c) ……………………
Metre	(d) …………………..
Kilogram	(e) ……………………
Second	(f) …………………….

Answer:
(a) N
(b) K
(c) A
(d) m
(e) kg
(f) s

Question 18.
Classify the physical quantities given in the box as fundamental quantities and derived quantities.

Length, density, area, mass

Fundamental quantities	Derived quantities

Answer:

Fundamental quantities	Derived quantities
Length, mass	Density, area

Question 19.
1 mm is the least count of a metre scale.
a) What do you mean by least count?
b) Arrange the following units in ascending order of their values.
(μm, km, nm, cm) (Annual 2022)
Answer:
a) The smallest value that can be measured using an instrument is called its least count.
b) nm < μm < cm < km

Question 20.
Based on the fundamental physical quantity length, write down the following questions. (Term 1, 2023)
a) In which unit is the thickness of plastic carry bags expressed?
b) How many kilometres make one astronomical unit?
Answer:
a) micrometre
b) 15 crore kilometre (150 million kilometre)

Question 21.
We know about fundamental units and derived units.
a) What is meant by derived units?
b) Based on which fundamental unit is area expressed?
c) Write down the unit of density.
Answer:
a) Derived units are units that can be stated using fundamental units or that depend on fundamental units.

b) Area (m²) is expressed in terms of the fundamental unit length.

c) Density = mass/volume
The unit of density is kg/m³

Question 22.
Convert the following units into SI units without changing their values.
a) 3500 g
b) 2.5 km
c) 2 h
Answer:
a) 3500 g = 3500/1000 kg = 3.5 kg
b) 2.5 km = 2.5 × 1000 m = 2500 m
c) 2 h = 2 × 60 × 60 s = 7200 s.

Question 23.
State whether the following symbol are right or wrong. Correct them if wrong.
a) m/s/s
b) kilogram per cubic metre
c) 100 n
d) 20 kgs
e) 3M
f) 1 kg 500 g
Answer:
a) Wrong, m/s² is the correct form.
b) Right
c) Wrong. 100 N is the correct form.
d) Wrong. 20 kg is the correct form.
e) Wrong. 3 m is the correct form.
f) Wrong. 1.5 kg or 1500 g is the correct form

Students often refer to Kerala State Syllabus SCERT Class 8 Maths Solutions and Class 8 Maths Chapter 5 Solutions of Equations Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 8 Maths Chapter 5 Solutions Solutions of Equations

Class 8 Kerala Syllabus Maths Solutions Chapter 5 Solutions of Equations Questions and Answers

Solutions of Equations Class 8 Questions and Answers Kerala Syllabus

Adding and Subtracting (Page No. 79)

Question 1.
“Six more marks and I would have gotten a hundred out of a hundred in my math exam”, Rajan thought sadly. How much did he score?
Answer:
The mark is 6 less than 100.
Present mark is 94.

Question 2.
“5 more years, and I would be 18. I can vote”, Lissy calculated. How old is she now?
Answer:
The present age is 5 less than 18.
Her current age is 13 years.

Question 3.
What should be added to 123 to make it 321?
Answer:
Number to be added is 321 – 123 = 198

Question 4.
What should be subtracted from 432 to get 234?
Answer:
Number to be subtracted is 432 – 234 = 198

Textbook Page No. 80

Question 1.
Gopalan bought a bunch of bananas. 7 of the bananas had begun to rot, and he removed them. Now there are 46. How many bananas were there in the bunch?
Answer:
The number of good fruits is 46.
Including spoiled fruits total is 46 + 7 = 53.

Question 2.
Vimala bought some things for 163 rupees, and now she has 337 rupees left. How much did she have at first?
Answer:
The total amount in hand is the sum of the balance amount and the amount for purchasing goods.
It is 337 + 163 = 500 rupees

Question 3.
What number becomes 321 on subtracting 123?
Answer:
The subtraction should be done from the number 123 more than 321.
Number is = 321 + 123 = 444

Multiplication and Division (Page No. 80)

Question 1.
The salary of the manager of an office is five times the salary of a peon. The manager gets 75000 rupees a month. How much does the peon get?
Answer:
Peon’s salary is \(\frac {1}{5}\) of Manger’s salary
Peon’s salary is \(\frac {1}{5}\) × 75000 = 15000 rupees

Question 2.
Some friends went for a trip and decided to divide equally among them the 12500 rupees they had spent. Each had to pay 2500 rupees. How many were there in the group?
Answer:
We have to find out the number of 2500s that make 12500.
This will be the number of members
It is 12500 ÷ 2500
Number of members = \(\frac {12500}{2500}\) = 5

Question 3.
A number multiplied by 12 gives 756. What is the number?
Answer:
12 times a number is 756.
Number is \(\frac {1}{12}\) part of 756
Number = \(\frac {1}{12}\) × 756 = 63

Question 4.
A number divided by 21 gives 756. What is the number?
Answer:
\(\frac {1}{21}\) part of a number is 756.
The number is 21 times 756.
Number = 21 × 756 = 15876

Different Kinds of Change (Page No. 82, 83)

Question 1.
The perimeter of a rectangle is 50 metres and the length of one side is 5 metres. What is the length of the other side?
Answer:
Sum of two adjacent sides is 50 ÷ 2 = 25
One side is 5.
Other side 25 – 5 = 20 cm

Question 2.
When Anita and her friend bought 5 pens together, they got a reduction of 3 rupees in the price. The total cost was 32 rupees. Had they bought the pens individually, how much would it have cost each?
Answer:
If there is no reduction in price, the price of 5 pens is 35 rupees
Price of a pen = 35 ÷ 5 = 7 rupees

Question 3.
In each of the problems below, the result of doing some operations on a number is given. Find each number:
(i) Three added to two times the number gives 101
(ii) Two added to three times the number gives 101
(iii) Three subtracted from two times the number gives 101
(iv) Two subtracted from three times the number gives 101
Answer:
(i) Without adding three, twice the number = 101 – 3 = 98
Number is \(\frac {1}{2}\) part of 98.
Number = \(\frac {1}{2}\) × 98 = 49

(ii) Three times the number 101 – 2 = 99
Number is \(\frac {1}{3}\) of 99.
Number = \(\frac {1}{3}\) × 99 = 33

(iii) Twice the number = 101 + 3 = 104
Number is \(\frac {1}{2}\) of 104.
Number is \(\frac {1}{2}\) × 104 = 52

(iv) 3 times the number is 103.
Number is \(\frac {103}{3}\)

Question 4.
The sum of 6 times a number and 4 is 100. What is the number?
Answer:
If 4 is not added, 6 times the number = 100 – 4 = 96
Number is \(\frac {1}{6}\) of 96.
Number = \(\frac {1}{6}\) × 96 = 16

Question 5.
4 times the sum of a number and 6 gives 100. What is the number?
Answer:
6 added to \(\frac {1}{4}\) part of 100.
6 added to the number is 25
Number is 25 – 6 = 19.

Multiple and Part (Page No. 84)

Question 1.
The sum of 2 times a number and 7 times the same number is 27. What is the number?
Answer:
When 2 times a number and 7 times that number are added, we get 9 times that number.
9 times the number is 27.
Number = \(\frac {27}{9}\) = 3

Question 2.
99 added to 12 times a number gives 21 times the number. What is the number?
Answer:
When 12 times a number is subtracted from 21 times that number, we get 99.
9 times the number is 99.
\(\frac {1}{9}\) of 99 is 11.
Number = \(\frac {1}{9}\) × 99 = 11

Textbook Page No. 85

Question 1.
A third of a piece of rope cut away leaves 10 metres. What was the original length?
Answer:
When \(\frac {1}{3}\) is removed results \(\frac {2}{3}\) of length.
It is 10 m.
Length of rope is 10 × \(\frac {3}{2}\) = 15

Question 2.
Half the milk in a can and a third of the remaining was used up, and now there are 5 litres remaining. How many litres did it originally contain?
Answer:
Half of the amount remains \(\frac {1}{3}\) of this is \(\frac {1}{6}\) of the first amount
Used, \(\frac{1}{2}+\frac{1}{6}\). This is \(\frac {4}{6}\) part
Amount remains \(\frac {2}{6}\) part.
It is 5 L.
Amount in the beginning is 5 times \(\frac {6}{2}\).
It is 5 × \(\frac {6}{2}\) = 15L

Algebraic Methods (Page No. 89)

Question 1.
The perimeter of a rectangle is 80 metres, and its length is one metre more than twice the width. What are its length and width?
Answer:
Sum of length and breadth 40
Width = x, length = 2x + 1
∴ x + 2x + 1 = 40
⇒ 3x + 1 = 40
⇒ 3x = 39
⇒ x = 13
∴ Width = 13, Length = 2 × 13 + 1 = 27 cm

Question 2.
When a hundred-rupee note was changed to twenty-rupee and ten-rupee notes, seven notes were obtained. How many of each?
Answer:
Number of 20 rupee notes x, ten rupee notes 7 – x.
∴ 20x + 10(7 – x) = 100
⇒ 20x + 70 – 10x = 100
⇒ 20x – 10x + 70 = 100
⇒ 10x + 70 = 100
⇒ 10x = 100 – 70
⇒ 10x = 30
⇒ x = 3
∴ Number of 20 rupee notes = 3
∴ Number of 10 rupee notes = 4

Question 3.
The price of a book is 4 rupees more than the price of a pen. The price of a pencil is 2 rupees less than the price of this pen. A person bought 5 books, 2 pens, and 3 pencils and paid 54 rupees for them. What is the price of each?
Answer:
The price of a book and a pencil can be written as follows.
Price of pen = x
price of book = x + 4
price of pencil = x – 2
∴ 5(x + 4) + 2x + 3(x – 2) = 54
⇒ 5x + 20 + 2x + 3x – 6 = 54
⇒ 10x + 14 = 54
⇒ 10x = 40
⇒ x = 4
Price of pen 4 rupees, price of book 8, price of pencil 2.

Question 4.
A square containing four numbers in a calendar is marked, and the sum of these numbers is 80. What are the numbers?
Answer:

∴ x + x + 1 + x + 7 + x + 8 = 80
⇒ 4x + 16 = 80
⇒ 4x = 64
⇒ x = 16
Numbers are 16, 17, 23, 24.

Textbook Page No. 92

Question 1.
The age of Appu’s mother is now 9 times that of Appu. After nine years, it will be three times Appu’s age. What are their ages now?
Answer:
If Appu’s age is x, then the mother’s age is 9x
After 9 years ages are x + 9, 9x + 9
∴ (x + 9) × 3 = 9x + 9
⇒ 3x + 27 = 9x + 9
⇒ 18 = 6x
⇒ x = 3
∴ Appu’s age = 3, mother’s age = 27

Question 2.
A class has the same number of girls and boys. On a day when eight boys were absent, the number of girls was twice the number of boys. What are the number of girls and the number of boys?
Answer:
Let the number of boys be x, number of boys x.
When number of boys is x – 8, x = 2(x – 8)
⇒ x = 2(x – 8)
⇒ x = 16

Question 3.
In a class of girls and boys, 50% of the children are girls. When 10 more girls were admitted to this class, this became 60%. How many children were in the class at first?
Answer:
50% denotes the half.
Let the number of boys and the number of girls be x.
There are 2x students.
When 10 girls are added, the number of girls becomes x + 10
∴ x + 10 = 2x × \(\frac {60}{100}\)
⇒ x + 10 = \(\frac {6x}{5}\)
⇒ x = 50
There are 100 students in the class.

Question 4.
Another problem from folk-math. Some lotus flowers have bloomed in a pond. A flock of birds sat on the flowers to rest. First, one bird sat on each flower, but one bird didn’t have a flower to sit on. Then two birds sat on each flower, and there was one flower extra. How many lotuses and how many birds?
Answer:
Let x be the number of lotus flowers.
Number of birds = x + 1
∴ x + 1 = 2(x – 1)
⇒ x = 3
Number of lotus flowers 3, number of birds 4.

Class 8 Maths Chapter 5 Kerala Syllabus Solutions of Equations Questions and Answers

Class 8 Maths Solutions of Equations Questions and Answers

Question 1.
One third of a number is 10 less than the number. Which of the following is the number?
(a) 30
(b) 15
(c) 10
(d) 27
Answer:
\(\frac {2}{3}\) of the number is 10.
Number is \(\frac {3}{2}\) times 10.
Number = 10 × \(\frac {3}{2}\) = 15

Question 2.
The sum of two numbers is 39, difference is 15. Numbers are
(a) 17, 22
(b) 26, 13
(c) 27, 12
(d) 28, 11
Answer:
x, x + 15
∴ 2x + 15 = 39
⇒ 2x = 4
⇒ x = 12
Numbers = 12, 27

Question 3.
Two sides of a triangle are equal. Small angle is 30°. What is one of the equal angles?
(a) 100°
(b) 75°
(c) 70°
(d) 50°
Answer:
Angle sum = 180°
The smallest is 30°.
The sum of the equal angles is 150°
Equal angle is 75°.

Question 4.
When 3 is added to \(\frac {1}{4}\)th of a number gives 15. Find the number.
Answer:
\(\frac {1}{4}\)th of the number is 12.
Number is 4 × 12 = 48

Question 5.
When \(\frac {2}{3}\) of a number is subtracted from a number, we get 13. What is the number?
Answer:
\(\frac {1}{3}\) of the number is 13.
Number is 3 × 13 = 39

Question 6.
The sum of two numbers is 60. One number is four times the other. Find the number.
Answer:
When the number and four times the number are added, we get five times the number.
Number is \(\frac {1}{5}\) part of 60.
The number is 12.

Question 7.
The sum of two consecutive natural numbers is 63.
(a) If the smaller is x, then what is the other number?
(b) Write the equations and find the numbers.
Answer:
(a) x, x + 1 are the numbers.
(b) 2x + 1 = 63
⇒ 2x = 62
⇒ x = 31
Numbers are 31, 32.

Question 8.
Ammu’s mother is four times her mother’s age. After 20 years, mother’s age will be twice Ammu’s age.
(a) If Ammu’s age is x, then write the equation.
(b) Find the age of Ammu and her mother.
Answer:
(a) Ammu’s age x, Mothers age 4x
(b) 20 years hence
Ammu’s age = x + 20
mothers age = 4x + 20
∴ 4x + 20 = 2(x + 20)
⇒ 4x + 20 = 2x + 40
⇒ x = 10
Ammu’s age is 10, Mother’s age is 40.

Question 9.
A man’s age after 15 years will be twice his age before 15 years.
(a) If the present age is x, then write the equation.
(b) Find the present age?
Answer:
(a) Age after 15 years is x + 15
Age before 15 years is x – 15
∴ 2(x – 15) = x + 15
⇒ 2x – 30 = x + 15
⇒ x = 45
(b) Present age = 45 years

Question 10.
The perimeter of a rectangle is 50 cm. Length is 5 cm more than breadth.
(a) What is length + breadth?
(b) If x is the length, then write the equation.
(c) Find the sides.
Answer:
(a) 25 cm
(b) x + x + 5 = 25
⇒ 2x + 5 = 25
⇒ 2x = 20
⇒ x = 10
(c) Sides are 10 cm, 15 cm

Question 11.
The sum of two numbers is 90. The large number is 15 more than twice the small number. Write the equations and find the numbers.
Answer:
Smaller number = x
Larger number = 2x + 15
∴ x + 2x + 15 = 90
⇒ 3x = 75
⇒ x = 25
Numbers are 25, 65.

Question 12.
The sum of the digits of a two-digit number is 10. On reversing the digits, it is found that the number decreases by 36. Find the number.
Answer:
Digit in ten’s place is x.
Digit in one’s place is 10 – x.
Number is 10x + 10 – x
When the digits are reversed number becomes 10(10 – x) + x
It is given that 10(10 – x) + x = 10x + 10 – x – 36
⇒ 100 – 10x + x = 9x – 26
⇒ 100 – 9x = 9x – 26
⇒ 126 = 18x
⇒ x = 7
∴ Number is 73

Question 13.
Length of a bridge is 60 metres longer than \(\frac {1}{3}\) of its length. What is the length of the bridge?
Answer:
If x is the length of the bridge.
∴ \(\frac {2x}{3}\) = 60
⇒ x = 90

Question 14.
In a class of 52 students, the number of boys is \(\frac {6}{7}\) of the number of girls. How many are boys?
Answer:
Number of boys = x
Number of girls = 52 – x
∴ x = \(\frac {6}{7}\) × (52 – x)
⇒ 7x = 6(52 – x)
⇒ 7x = 312 – 6x
⇒ 13x = 312
⇒ x = 24
∴ Boys = 24, Girls = 28

Question 15.
The sum of three consecutive odd numbers is 63.
(a) If the smallest among them is x, then what are the other numbers?
(b) Find the numbers.
Answer:
(a) x, x + 2, x + 4 are the numbers.
(b) 3x + 6 = 63
⇒ 3x = 57
⇒ x = 19
∴ Numbers are 19, 21, 23.

Class 8 Maths Chapter 5 Notes Kerala Syllabus Solutions of Equations

→ Price of 5 pens and 3 pencils is 34 rupees. The price of 3 pens and 5 pencils is 30 rupees. What is the total price of a pen and a pencil together?

→ 64 rupees are needed for 8 pens and 8 pencils. So 8 rupees is needed for a pen and a pencil.

→ 5 pens and 3 pencils cost 34 rupees. 5 pens 5 pencils cost 40 rupees. Note the increased amount is for 2 pencils.

→ The price of 10 pens and 10 pencils is 80 rupees.

→ Three times \(\frac {1}{3}\) of a number is that number itself. So six times \(\frac {1}{3}\) of that number is two times that number.

→ Eight times \(\frac {1}{4}\) of a number is twice that number.

→ What is the sum of eight times \(\frac {1}{4}\) of a number and six times \(\frac {1}{3}\) of that number?

→ Let x be a number. Three times x is 18. This can be written as an equation. 3x = 18 is the equation \(\frac {1}{3}\) of a number is 18. What is the number?
That means, three times that number is 18.
3x = 18
x = \(\frac {1}{3}\) × 18
x = 6

→ 1 more than two times a number is 5.

→ If 1 is not added, two times the number is 4. The number is 2.

Problems that can be solved by simple logic are common in our daily lives. But some problems cannot be solved by simple logic. In mathematics, we use algebra to solve certain problems. The word algebra is derived from the Arabic word Al-Jabr, which means to join together.

Adding and Subtracting
Finding solutions in the form of equations,
x + a = b, x – a = b
Here, only addition and subtraction are needed to find the solutions. The reverse process is the easiest method.

Worksheet – 1

Question 1.
Babu has 600 rupees with him. He went to a shop and purchased some goods. Now he had 301 rupees left with him. How much did he spend?
Answer:
600 – 301 = 299

Question 2.
If we add 54 to a number, we will get 250. What is the number?
Answer:
250 – 54 = 196

Question 3.
Malini had 78 stamps with her. She gave some of them to her friend. Now she has 46 stamps with her. How many stamps did she give to her friend?
Answer:
78 – 46 = 32

Question 4.
When 164 is subtracted from a number, it makes it 235. What is the number?
Answer:
235 + 164 = 399

Multiplication and Division
Finding solutions in the form \(\frac {x}{a}\) = b and ax = b.
Here, only multiplication and division are needed to find the solution.
The reverse process is the easiest method here also.

Worksheet – 2

Question 1.
The price of 4 shirts is 720 rupees. What is the price of one shirt?
Answer:
720 ÷ 4 = 180

Question 2.
The age of Raju’s father is 9 times that of Raju. If the father’s age is 36 years, what is Raju’s age?
Answer:
36 ÷ 9 = 4

Question 3.
A number when multiplied by 14 gives 1008. What is the number?
Answer:
1008 ÷ 14 = 72

Question 4.
A number when divided by 41 gives 215. What is the number?
Answer:
215 × 41 = 8815

Different Kinds of Change
Here we discuss the changes in the form
ax + b = c, ax – b = c, \(\frac {x}{a}\) + b = c
First, we will adopt the method of reverse process, then only the algebraic method.

For example: One added to three times a number gives 100. What is the number?
Here, the last mathematical calculation done on a number is adding ‘1’.
So first we will subtract ‘1’ from the result 100.
So 100 – 1 = 99.
Then the next calculation is multiplying by three, so we will divide 99 by 3.
So the number = \(\frac {99}{3}\) = 33

Worksheet – 3

Question 1.
The double of 3 less than a number is 97. What is the number?
Answer:
3 less than, so add 3
97 + 3 = 100
Double, so half = \(\frac {100}{2}\) = 50

Question 2.
Two less than 3 times a number is 100. What is the number?
Answer:
2 less, so add 2
100 + 2 = 102
3 times, so divide by 3
\(\frac {102}{3}\) = 34

Question 3.
10 is added to half of a number, giving 100. What is the number?
Answer:
10 added, so subtract 10
100 – 10 = 90
Half of a number, so double it.
90 × 2 = 180

Question 4.
Alia has 150 rupees, which is 8 rupees more than twice what Adhil has. How much has Adhil?
Answer:
8 rupees more, so subtract 8
150 – 8 = 142
twice, so half of 142
\(\frac {142}{2}\) = 71

Multiple and Part
If 3 times a number and 2 times the same number added together make 100. What is the number?
Answer:
3 times any number and 2 times the same number make 5 times the number.
That means 5 times the number is 100.
So the number is \(\frac {1}{5}\) of 100.
Number = \(\frac {100}{5}\) = 20

Algebraic Methods
Now we will discuss the algebraic method of solving these word problems. Here, we will take the unknown number as x, then form an equation. Solving is based on the reverse process itself.

For example, the perimeter of a rectangle is 46 cm. The length is 5 cm more than the breadth. What are its length and breadth?
Here, the length is given as 5 more than the breadth.
Breadth is completely unknown.
So it is better to take breadth as ‘x’, which means length is x + 5
∴ Perimeter = 2(length + breadth)
⇒ 2(x + 5 + x) = 46
⇒ x + 5 + x = 23
⇒ 2x + 5 = 23
⇒ 2x = 23 – 5
⇒ 2x = 18
⇒ x = 9
∴ Length = 9 + 5 = 14

Worksheet – 4

Question 1.
Raju has 75 rupees more than Rani. If both of them together have 725 rupees, what are the amounts with Raju and Rani?
Answer:
Rani has ‘x’ rupees
Raju has 75 + x rupees
∴ x + 75 + x = 725
⇒ 2x + 75 = 725
⇒ 2x = 725 – 75
⇒ 2x = 650
⇒ x = 325
Raju has 325 + 75 = 400 rupees

Question 2.
The second angle of a triangle is double the first. The third angle is 40 degrees less than the first. Find the three angles.
Answer:
First angle = x
Second angle = x + x = 2x
Third angle = x – 40
∴ x + 2x + x – 40 = 180°
⇒ 4x – 40 = 180
⇒ 4x = 180 + 40
⇒ 4x = 220
⇒ x = 55°

Students often refer to Kerala State Syllabus SCERT Class 8 Maths Solutions and Class 8 Maths Chapter 3 Square Identities Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 8 Maths Chapter 3 Solutions Square Identities

Class 8 Kerala Syllabus Maths Solutions Chapter 3 Square Identities Questions and Answers

Square Identities Class 8 Questions and Answers Kerala Syllabus

Squares of Sums (Page No. 40)

Question 1.
Calculate the squares below in your head:
(i) 15²
(ii) 25²
(iii) 33²
(iv) \(\left(5 \frac{1}{2}\right)^2\)
(v) \(\left(10 \frac{1}{2}\right)^2\)
(vi) \(\left(25 \frac{1}{2}\right)^2\)
(vii) \(\left(5 \frac{1}{5}\right)^2\)
Answer:
(i) 15² = 10² + 5² + 2 × 10 × 5 = 225
(ii) 25² = 20² + 5² + 2 × 20 × 5 = 625
(iii) 33² = 30² + 3² + 2 × 30 × 3 = 1089
(iv) \(5 \frac{1^2}{2}=5^2+5+\frac{1}{2^2}=30 \frac{1}{4}\)
(v) \(10 \frac{1}{2}^2=10^2+10+\frac{1}{2^2}=110 \frac{1}{4}\)
(vi) \(25 \frac{1}{2}^2=25^2+25+\frac{1}{2^2}=650 \frac{1}{4}\)
(vii) \(5 \frac{1}{5}^2=5^2+2+\frac{1}{5^2}=27 \frac{1}{25}\)

Question 2.
In the general identity on the square of a sum of two numbers, what special case do we get by taking one of the numbers as 2? Do the calculations below in your head, using this:
(i) What is 22²?
(ii) What is 52²?
(iii) 25² = 625; What is 27²?
Answer:
(x + 1)² = x² + 2 × x × 2 + 2² = x² + 4x + 4
(i) 22² = (20 + 2)²
= 20² + 2 × 20 × 2 + 2²
= 400 + 80 + 4
= 484

(ii) 52² = (50 + 2)²
= 50² + 2 × 50 × 2 + 2²
= 2500 + 200 + 4
= 2704

(iii) 25² = 625
27² = (25 + 2)²
= 25² + 2 × 25 × 2 + 2²
= 625 + 100 + 4
= 729

Question 3.
43² = 1849
(i) What is 44²?
(ii) What is 46²?
Answer:
43² = 1849
(i) 44² = (43 + 1)²
= 43² + 2 × 43 × 1 + 1²
= 1849 + 86 + 1
= 1936

(ii) 46² = (43 + 3)²
= 43² + 2 × 43 × 3 + 3²
= 1849 + 258 + 9
= 2116

Textbook Page No. 45

Question 1.
Calculate the squares of some odd numbers and check the following statement. Explain why they are true:
(i) The square of any odd number is odd.
(ii) The square of any odd number leaves a remainder of 1 on division by 4.
(iii) The square of any odd number leaves a remainder of 1 on division by 8.
Answer:

(i) Let the odd number be, 2n + 1 where n = 0, 1, 2, 3,…..
(2n + 1)² = 4n² + 4n + 1 = 2(2n² + 2n) + 1
∴ The square of an odd number is of the form 2n + 1, which is an odd number.

(ii) (2n + 1)² = 4n² + 4n + 1 = 4(n² + n) + 1
That means the square of an odd number is 1 more than a multiple of 4.
∴ It leaves a remainder of 1 on division by 4.

(iii) (2n + 1)² = 4n(n + 1) + 1
Among the two consecutive numbers n and n + 1, one is always even.
So n(n + 1) is always even.
Thus, 4n(n + 1) is divisible by 8.
Hence, the square is of the form multiple of 8 + 1.
That means it leaves a remainder of 1 on division by 8.

Question 2.
Calculate the remainders on dividing the squares of some natural numbers by 4. Explain why the Statements below are true:
(i) The square of any natural number divided by 4 leaves a remainder of 0 or 1.
(ii) Any natural number that leaves a remainder of 2 or 3 on division by 4 is not a perfect square.
Answer:
The natural numbers on division by 4 are of the form 4n, 4n + 1, 4n + 2, 4n + 3, where n = 0, 1, 2,…
The possible remainders are 0, 1, 2, 3.
(i) Even natural numbers can be written as 2n
Squaring it gives (2n)² = 4n²
Here, the remainder is 0.
Odd natural numbers can be written as 2n + 1
Squaring it gives,
(2n + 1)² = 4n² + 4n + 1 = 4(n² + n) + 1
Here, the remainder is 1.
∴ The square of any natural number divided by 4 leaves a remainder of 0 or 1.

(ii) The square of any natural number divided by 4 leaves a remainder of 0 or 1.
∴ Any natural number leaving a remainder of 2 or 3 when divided by 4 is not a perfect square.

Question 3.
See this pattern:
3 = 2² – 1²
5 = 3² – 2²
7 = 4² – 3²
Check whether some other odd numbers can also be written as the difference of two perfect squares. Explain why all odd numbers greater than 1 can be written like this.
(Hint: Recall the general form of an odd number seen in class 7).
Answer:
All odd numbers greater than 1 can be written as the difference of two perfect squares.
Specifically, the difference of the squares of two consecutive natural numbers.
That is, (n + 1)² – n² = 2n + 1, always an odd number.
So, for any odd number n, it can be written as (n + 1)² – n².

Question 4.
Give reasons for the fact that the square of any number ending in 1 also ends in 1. What about numbers ending in 5? And numbers ending in 6?
Answer:
The last digit of a square is determined by the last digit of the original number.
The numbers ending in 1, 5, or 6 will have squares ending in 1, 5, or 6, respectively.
Because their last digits, when multiplied by themselves, result in the same digit.

Squares of Differences (Page No. 50)

Question 1.
Calculate the squares below in your head:
(i) 29²
(ii) 38²
(iii) 999²
(iv) \(\left(9 \frac{1}{2}\right)^2\)
(v) (9.7)²
Answer:
(i) 29² = (30 – 1)²
= 30² – 2 × 30 × 1 + 1²
= 900 – 60 + 1
= 841

(ii) 38² = (40 – 2)²
= 40² – 2 × 40 × 2 + 2²
= 1600 – 160 + 4
= 1444

(iii) 999² = (1000 – 1)²
= 1000000 – 2000 + 1
= 998001

(iv) \(\left(9 \frac{1}{2}\right)^2=9 \times 10+\frac{1}{4}=90 \frac{1}{4}\)
Or
\(\left(9 \frac{1}{2}\right)^2=\left(10-\frac{1}{2}\right)^2=10^2+\left(\frac{1}{2}\right)^2-10=90 \frac{1}{4}\)

(v) (9.7)² = (10 – 0.3)²
= 10² + (0.3)² – 2 × 10 × 0.3
= 100 + 0.09 – 6
= 94.09

Question 2.
See these computations:
3² – (2 × 3) = 3 = 2² – 1
4² – (2 × 4) = 8 = 3² – 1
5² – (2 × 5) = 15 = 4² – 1
Explain the general principle of these using algebra.
Answer:
3² – (2 × 3) = 3 = 2² – 1
4² – (2 × 4) = 8 = 3² – 1
5² – (2 × 5) = 15 = 4² – 1
…………………………………..
…………………………………..
n² – (2 × n) = (n – 1)² – 1

Textbook Page No. 53, 54

Question 1.
See these computations:
\(\left(\frac{1}{2}\right)^2+\left(1 \frac{1}{2}\right)^2=2 \frac{1}{2}\); 2 = 2 × 1²
\(\left(1 \frac{1}{2}\right)^2+\left(2 \frac{1}{2}\right)^2=8 \frac{1}{2}\); 8 = 2 × 2²
\(\left(2 \frac{1}{2}\right)^2+\left(3 \frac{1}{2}\right)^2=18 \frac{1}{2}\); 18 = 2 × 3²
(i) Write the next two lines following this pattern.
(ii) Explain the general principle of these using algebra.
Answer:

Question 2.
Some natural numbers can be written as the difference of two perfect squares in two different ways. For example
24 = 7² – 5² = 5² – 1²
32 = 9² – 7² = 6² – 2²
40 = 11² – 9² = 7² – 3²
Write the next few multiples of 8 as the difference of two perfect squares in two different ways. Explain algebraically how we can do this for all multiples of 8 starting with 24.
Answer:
The next few lines are:
48 = 13² – 11² = 8² – 4²
56 = 15² – 13² = 9² – 5²
64 = 17² – 15² = 10² – 6²
Algebraic explanation is: 8x = (2x + 1)² – (2x – 1)² where x is 3, 4, 5,……….

Question 3.
See how some numbers are written as the difference of two perfect squares in two different ways:
15 = 8² – 7² = 4² – 1²
21 = 11² – 10² = 5² – 2²
35 = 18² – 17² = 6² – 1²
(i) What are the different ways in which the numbers 15, 21, and 35 can be written as the product of two factors?
(ii) Find two more numbers of this type, which can be written as the difference of two perfect squares in two different ways.
(iii) What general result can we form from this?
Answer:
(i) 15 = (8 – 7) × (8 + 7) = 1 × 15
15 = (4 – 1) × (4 + 1) = 3 × 5
21 = (11 – 10) × (11 + 10) = 1 × 21
21 = (5 – 2) × (5 + 2) = 3 × 7
35 = (18 – 17) × (18 + 17) = 1 × 35
35 = (6 – 1) × (6 + 1) = 5 × 7

(ii) 45 = 9² – 6² = 7² – 2²
65 = 33² – 32² = 9² – 4²

(iii) a² – b² = (a – b)(a + b)

Question 4.
Compute the following products by writing them as the difference of two squares.
(i) 78 × 22
(ii) 301 × 299
(iii) \(2 \frac{1}{3} \times 1 \frac{2}{3}\)
(iv) 10.7 × 9.3
Answer:

Question 5.
Find the largest of each of the following pairs of products, without actual multiplication:
(i) 75 × 25; 76 × 24
(ii) 76 × 24; 74 × 26
(iii) 10.6 × 9.4; 10.4 × 9.6
Answer:

Differences of Squares (Page No. 58, 59, 60)

Question 1.
Do the computations below in your head:
(i) 68² – 32²
(ii) \(\left(3 \frac{1}{2}\right)^2-\left(2 \frac{1}{2}\right)^2\)
(iii) 3.6² – 1.4²
Answer:
(i) 68² – 32² = (68 – 32)(68 + 32)
= 36 × 100
= 3600

(ii) \(\left(3 \frac{1}{2}\right)^2-\left(2 \frac{1}{2}\right)^2=\left(\frac{7}{2}\right)^2-\left(\frac{5}{2}\right)^2=\left(\frac{7-5}{2}\right)\left(\frac{7+5}{2}\right)\)
= 1 × 6
= 6

(iii) 3.6² – 1.4² = (3. 6 – 1.4)(3.6 + 1.4) = 11

Question 2.
Note the pattern in the computations below:
3² – 2² = 5 = 3 + 2
4² – 3² = 7 = 4 + 3
5² – 4² = 9 = 5 + 4
(i) Write the next two computations following these.
(ii) Write the general principle in these as an algebraic equation.
(iii) Write this general principle in ordinary language.
Answer:
(i) 6² – 5² = 11 = 6 + 5
7² – 6² = 13 = 7 + 6

(ii) n² – (n – 1)² = 2n – 1 = n + (n – 1)

(iii) The difference between the square of an integer and the square of the preceding integer is equal to the sum of those two integers.

Question 3.
In a calendar sheet, take a square containing nine numbers and mark the numbers on the left and right, top, and bottom of the middle number:

Calculate the products of the left and right numbers and the top and bottom numbers, and also their difference
8 × 10 = 80
2 × 16 = 32
80 – 32 = 48
Do this for other squares of nine numbers. Explain using algebra why the difference is 48 in all cases.
(Hint: It is convenient to take the middle number as x)
Answer:
Let the middle number be x.
The number to the left is x – 1.
The number to the right is x + 1.
The number on top is x – 7.
The number on the bottom is x + 7.
Since, in a calendar, the numbers in a row differ by 1, and the numbers in a column differ by 7.
Product of left and right numbers be (x – 1)(x + 1) = x² – 1
Product of top and bottom numbers be (x – 7)(x + 7) = x² – 49
Difference = (Product of left and right) – (Product of top and bottom)
= (x² – 1) – (x² – 49)
= x² – 1 – x² + 49
= 48
Therefore, the difference between the product of the left and right numbers and the product of the top and bottom numbers is always 48.

Question 4.
As in the previous problem, mark a square of nine numbers in a calendar. Mark the four numbers in the corners:

Calculate the products of the diagonal pairs of numbers and find their difference:
15 × 3 = 45
1 × 17 = 17
45 – 17 = 28
Do this for other squares of nine numbers. Explain using algebra why the difference is 28 in all cases.
Answer:
Let the middle number be x.
The number to the top left is x – 8.
The number to the top right is x – 6.
The number on the bottom left is x + 6.
The number on the bottom right is x + 8.
Product of first diagonal part = (x – 8)(x + 8) = x² – 64
Product of second diagonal part = (x – 6)(x + 6) = x² – 36
Difference = (x² – 36) – (x² – 64)
= x² – 36 – x² + 64
= 28
Therefore, the difference between the products of the diagonal pairs of numbers is 28.

Question 5.
A square has a perimeter of 20 centimetres. A rectangle has one side 2 centimetres longer and one side 2 centimetres shorter than a side of this square.
(i) What is the perimeter of the rectangle?
(ii) What are the areas of the square and the rectangle?
Answer:
Perimeter of the square = 20 cm
One side = \(\frac {20}{2}\) = 5 cm
(i) One side of the rectangle = 5 + 2 = 7 cm
∴ Other side = 5 – 2 = 3 cm
Perimeter of the rectangle = 2 × (7 + 3)
= 2 × 10
= 20 cm

(ii) Area of the square = side²
= 5²
= 25 cm²
Area of the rectangle = 7 × 3 = 21 cm²

Question 6.
One side of a rectangle is longer than the side of a square, and the other side is equally shorter than the side of the square.
(i) What can we say about the perimeters of the square and the rectangle?
(ii) Which has the larger area, the square or the rectangle?
Answer:
Let the side of the square be x cm
One side of the rectangle = x + a
Other side = x – a (where a > 0)
(i) The perimeter of the square = 4x
Perimeter of the rectangle = 2[(x + a) + (x – a)]
= 2(2x)
= 4x
∴ The perimeter of the rectangle is equal to the perimeter of the square.

(ii) Area of the square = x²
Area of the rectangle = (x + a)(x – a) = x² – a²
Here x² – a² < x²
∴ The square has the larger area.

Class 8 Maths Chapter 3 Kerala Syllabus Square Identities Questions and Answers

Class 8 Maths Square Identities Questions and Answers

Question 1.
Calculate the squares below in your head:
(i) 101²
(ii) 9.2²
Answer:
(i) 101² = 100² + 2 × 100 × 1 + 1²
= 10000 + 200 + 1
= 10201

(ii) 9.2² = (9 + 0.2)²
= 9² + 2 × 9 × 0.2 + 0.2²
= 81 + 3.6 + 0.04
= 84.64

Question 2.
23² = 529
(i) What is 24²?
(ii) What is 26²?
Answer:
23² = 529
(i) 24² = (23 + 1)²
= 23² + 2 × 23 × 1 + 1²
= 529 + 46 + 1
= 576

(ii) 26² = (23 + 3)²
= 23² + 2 × 23 × 3 + 3²
= 529 + 138 + 9
= 676

Question 3.
\(\left(1 \frac{1}{2}\right)^2+\left(3 \frac{1}{2}\right)^2+\left(5 \frac{1}{2}\right)^2\) = _____________
Answer:

Question 4.
Find the value of \(\left(1 \frac{1}{2}\right)^2+\left(2 \frac{1}{2}\right)^2+\left(3 \frac{1}{2}\right)^2+\left(4 \frac{1}{2}\right)^2\)
Answer:

Question 5.
\(\left(19 \frac{1}{2}\right)^2+\left(20 \frac{1}{2}\right)^2\) = _____________
Answer:

Question 6.
101² – 100²
Answer:
101² – 100² = 101 + 100 = 201

Question 7.
99999² – 99998²
Answer:
99999² – 99998² = 99999 + 99998 = 199,997

Question 8.
\(\left(99 \frac{1}{2}\right)^2+\left(100 \frac{1}{2}\right)^2\) = _____________
Answer:

Question 9.
From the following, which of the numbers will be the sum of the digits in the ones and tens place of a square of a number that ends in 5?
(a) 2
(b) 5
(c) 7
(d) 3
Answer:
(c) 7

Question 10.
If the square of a number ending in 5 is of the form 56xy. Then y – x = _____________
Answer:
y – x = 5 – 2 = 3

Question 11.
If the square of a number ending in 5 is of the form 110pq. Then p × q = _____________
Answer:
p × q = 2 × 5 = 10

Question 12.
10² + 2² = 100 + 4 = 104. By using this, write 208 as the sum of the two perfect squares.
Answer:
208 = 2 × 104
= 2(10² + 2²)
= (10 + 2)² + (10 – 2)²
= 12² + 8²

Question 13.
3² + 2² = 9 + 4 = 13.
By using this, write 26 as the sum of the two perfect squares.
Answer:
26 = 2 × 13
= 2(3² + 2²)
= (3 + 2)² + (3 – 2)²
= 5² + 1²

Question 14.
11² + 10² = 121 + 100 = 221.
By using this, write 442 as the sum of the two perfect squares.
Answer:
442 = 2 × 221
= 2(11² + 10²)
= (11 + 10)² + (11 – 10)²
= 21² + 1²

Class 8 Maths Chapter 3 Notes Kerala Syllabus Square Identities

→ The square of the sum of two numbers is equal to the sum of the squares of these numbers and twice their product.

→ (x + y)² = x² + y² + 2xy for all numbers x and y.

→ To get the square of one more than a number, we need only add the square of the number, twice the number, and one.

→ (x + 1)² = x² + 2x + 1 for all numbers x.

→ The square of the difference of two numbers is twice their product subtracted from the sum of their squares.

→ (x – y)² = x² + y² – 2xy for all numbers x and y.

→ The product of any two numbers is the difference of the squares of half their sum and half their difference.

→ The difference of the squares of two numbers is the product of the sum and difference of the numbers.

→ x² – y² = (x + y)(x – y) for all numbers x and y.

→ Sum of digits of a perfect square is either 1 or 4 or 7, or 9.

→ The difference of the squares of two consecutive natural numbers is their sum.

In the first chapter, we discussed the problems related to finding the squares of a number. But in this chapter, ‘Square Identities’, we are going to discuss the algebraic formulas used to simplify expressions that involve the squares of numbers. These identities follow a particular pattern, making it easier to expand without doing lengthy multiplication each time. In this chapter, we are going to discuss squares of sums, squares of differences, and differences of squares.

Squares of Sums
We have seen a method to calculate the squares of many numbers in the section Squares.

How do we find the square of 36?
Answer:

36² = (30 + 6)²
= 30² + (2 × 30 × 6) + 6²
= 900 + 360 + 36
= 1296
Here we wrote 36 as the sum of 30 + 6 to calculate its square.
So the square of 36 can be written as (30 + 6)² = 30² + (2 × 30 × 6) + 6²

In general, we can say that the square of the sum of two numbers is equal to the sum of the squares of these numbers and twice their product.
In algebraic form, we can write it as (x + y)² = x² + y² + 2xy for all numbers x and y.
We can see it geometrically as a result of areas:

Worksheet – 1

Question 1.
1. 42²
2. 56²
3. 73²
4. 84²
5. 93²
Answer:
1. 42² = 40² + 2 × 40 × 2 + 2²
= 1600 + 160 + 4
= 1764

2. 56² = 50² + 2 × 50 × 6 + 6²
= 2500 + 600 + 36
= 3136

3. 73² = 70² + 2 × 70 × 3 + 3²
= 4900 + 420 + 9
= 5329

4. 84² = 80² + 2 × 80 × 4 + 4²
= 6400 + 640 + 16
= 7056

5. 93² = 90² + 2 × 90 × 3 + 3²
= 8100 + 540 + 9
= 8649

To get the square of one more than a number, we need only add the square of the number, twice the number, and one.
In general, (x + 1)² = x² + 2x + 1 for all numbers x

For example, find 31²?
Answer:
31² = 30² + 2 × 30 + 1²
= 900 + 60 + 1
= 961

Find 21²?
Answer:
21² = 20² + 2 × 20 + 1²
= 400 + 40 + 1
= 441

Find 51²?
Answer:
51² = 50² + 2 × 50 + 1²
= 2500 + 100 + 1
= 2601

Find 61²?
Answer:
61² = 60² + 2 × 60 + 1²
= 3600 + 120 + 1
= 3721

Find 71²?
Answer:
71² = 70² + 2 × 70 + 1²
= 4900 + 140 + 1
= 5041

Find 101²?
Answer:
101² = 100² + 2 × 100 + 1²
= 10000 + 200 + 1
= 10201

Suppose take y as \(\frac {1}{2}\), we get
(x + \(\frac {1}{2}\))² = x² + x + \(\frac {1}{4}\) for all numbers x
It can also be written in the form, (x + \(\frac {1}{2}\))² = x(x + 1) + \(\frac {1}{4}\)

For example, Find \(1 \frac{1^2}{2}\)?
Answer:
\(\mathbf{1} \frac{1}{2}^{\mathbf{2}}=1^2+1+\frac{1}{4}=2 \frac{1}{4}\) = 2.25

Find \(9 \frac{1^2}{2}\)?
Answer:
\(9 \frac{1^2}{2}=9 \times 10+\frac{1}{4}=90 \frac{1}{4}\) = 90.25

Find \(10 \frac{1^2}{2}\)?
Answer:
\(10 \frac{1}{2}^2=10 \times 11+\frac{1}{4}=110 \frac{1}{4}\) = 110.25

Find \(19 \frac{1^2}{2}\)?
Answer:
\(19 \frac{\frac{1}{2}^2}{2}=19 \times 20+\frac{1}{4}=380 \frac{1}{4}\) = 380.25

Find \(6 \frac{1^2}{2}\)?
Answer:
\(6 \frac{1}{2}^2=6^2+6+\frac{1}{4}=42 \frac{1}{4}\) = 42.25

Find \(12 \frac{1^2}{2}\)?
Answer:
\(12 \frac{1^2}{2}\) = 12² + 12 + \(\frac {1}{4}\)
= 144 + 12 + \(\frac {1}{4}\)
= 156\(\frac {1}{4}\)
= 42.25

For any non-zero number x, we get
\(\left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2\)

For example, Find \(3 \frac{1}{3}^2\)?
Answer:
\(3 \frac{1}{3}^2=\left(3+\frac{1}{3}\right)^2=3^2+\frac{1}{3^2}+2=9+\frac{1}{9}+2=11 \frac{1}{9}\)

Find \(4 \frac{1}{4}^2\)?
Answer:
\(4 \frac{1}{4}^2=\left(4+\frac{1}{4}\right)^2=4^2+\frac{1}{4^2}+2=16+\frac{1}{16}+2=18 \frac{1}{16}\)

Find \(10 \frac{1}{10}^2\)?
Answer:
\(10 \frac{1}{10}^2=\left(10+\frac{1}{10}\right)^2=10^2+\frac{1}{10^2}+2=100+\frac{1}{100}+2=102 \cdot \frac{1}{100}\)

Worksheet – 2

Question 1.
1. \(\left(5 \frac{1}{5}\right)^2\)
2. \(\left(30 \frac{1}{30}\right)^2\)
3. \(\left(100 \frac{1}{100}\right)^2\)
4. \(\left(3 \frac{1}{2}\right)^2+\left(3 \frac{1}{3}\right)^2\)
5. \(\left(10 \frac{1}{2}\right)^2+\left(10 \frac{1}{3}\right)^2\)
Answer:

Square of a number ending in 5:
The square of any number with the last digit (the digit in the one’s place) 5 has 25 as the last two digits.
The last digit of any number is the remainder on division by 10.
So a number ending in 5, divided by 10, leaves a remainder of 5.
The general form of such a number is 10n + 5.
The square of a number in this form is:
(10n + 5)² = (10n)² + (2 × 10n × 5) + 5²
= 100n² + 100n + 25
= 100(n² + n) + 25
= 100n(n + 1) + 25
The last two digits of 100n(n + 1) are zeroes.
∴ The last two digits of 100n(n + 1) + 25 is 2 and 5

For example: Find 35²?
Answer:
35² = 100 × 3 × 4 + 25 = 1225

Find 55²?
Answer:
55² = 100 × 5 × 6 + 25 = 3025

Remainders on dividing natural numbers by 3:
If we consider any number, it will be a multiple of 3 or the number that leaves a remainder of 1 or 2 on division by 3.
That means, the general form of a number which is not a multiple of 3 is 3n + 1, where n = 0, 1, 2, 3,… or 3n + 2, where n = 0, 1, 2, 3,……….

The square of any non-multiple of 3 leaves a remainder of 1 on division by 3.
That means, (3n + 1)² = 9n² + 2 × 3n × 1 + 1
= 9n² + 6n + 1
= 3(3n² + 2n) + 1
= multiple of 3 + 1
Here, the remainder is 1.
(3n + 2)² = 9n² + 12n + 4
= 3(3n² + 4n) + 4
= 3(3n² + 4n) + 3 + 1
= 3(3n² + 4n + 1) + 1
= multiple of 3 + 1
Here also the remainder is 1.
Any perfect square leaves a remainder of 0 or 1 on division by 3.
Any number that leaves a remainder of 2 on division by 3 is not a perfect square.

Worksheet – 3

Question 1.
Calculate
(i) 45²
(ii) 205²
(iii) 555²
Answer:
(i) 45² = (100 × 4 × 5) + 25
= 2000 + 25
= 2025

(ii) 205² = (100 × 20 × 21) + 25
= 42000 + 25
= 42025

(iii) 555² = (100 × 55 × 56) + 25
= 308000 + 25
= 308025

Squares of Differences
The square of the difference of two numbers is twice their product subtracted from the sum of their squares.
If x and y are any two numbers.
Then, (x – y)² = x² + y² – 2xy for all numbers x and y.

For example: Compute 58².
Answer:
58² = (60 – 2)²
= 60² – 2 × 60 × 2 + 2²
= 3600 – 240 + 4
= 3364

If the equations of squares of sums and differences are added together, we get (x + y)² + (x – y)² = 2(x² + y²)
Therefore, 2(x² + y²) = (x + y)² + (x – y)²
The differences of this two identities is, 4xy = (x + y)² – (x – y)²
That means xy = \(\left(\frac{x+y}{2}\right)^2-\left(\frac{x-y}{2}\right)^2\)

The product of any two numbers is the difference of the squares of half their sum and half their difference.
If all the multiples of 4 starting with 8 can be written as the difference of two perfect squares.

For example: Compute 125 × 75.
Answer:
125 × 75 = \(\left(\frac{125+75}{2}\right)^2-\left(\frac{125-75}{2}\right)^2\)
= 100² – 25²
= 9375

Worksheet – 3

Compute the following.
(i) 112 × 12
(ii) 144 × 14
(iii) 17²
(iv) 39²
Answer:
(i) 112 × 12 = \(\left(\frac{112+12}{2}\right)^2-\left(\frac{112-12}{2}\right)^2\)
= 62² – 50²
= 1344

(ii) 144 × 14 = \(\left(\frac{114+14}{2}\right)^2-\left(\frac{114-14}{2}\right)^2\)
= 79² – 65²
= 2016

(iii) 17² = (20 – 3)²
= 20² – 2 × 20 × 3 + 3²
= 400 – 120 + 9
= 289

(iv) 39² = (40 – 1)²
= 40² – 2 × 40 × 1 + 1²
= 1600 – 80 + 1
= 1521

Differences of Squares
The difference of the squares of two numbers is the product of the sum and difference of the numbers.
x² – y² = (x + y)(x – y) for all numbers x and y.
If we take y as 1, then we get the equation as x² – 1 = (x + 1)(x – 1)

For three consecutive natural numbers, the product of the first and the last is one less than the square of the middle number.

For example: Compute 68² – 32²?
Answer:
68² – 32² = (68 + 32)(68 – 32)
= 100 × 36
= 3600

Worksheet – 4

Question 1.
Calculate
(i) 42² – 17²
(ii) \(\left(10 \frac{1}{2}\right)^2-\left(9 \frac{1}{2}\right)^2\)
Answer:
(i) 42² – 17² = (42 + 17)(42 – 17)
= 59 × 25
= 1475

(ii) \(\left(10 \frac{1}{2}\right)^2-\left(9 \frac{1}{2}\right)^2=\left(10 \frac{1}{2}+9 \frac{1}{2}\right)\left(10 \frac{1}{2}-9 \frac{1}{2}\right)\)
= 20 × 1
= 20

Students often refer to Kerala State Syllabus SCERT Class 8 Maths Solutions and Class 8 Maths Chapter 1 Squares Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 8 Maths Chapter 1 Solutions Squares

Class 8 Kerala Syllabus Maths Solutions Chapter 1 Squares Questions and Answers

Squares Class 8 Questions and Answers Kerala Syllabus

Perfect Squares (Page No. 11)

Question 1.
Calculate the squares given below.
(i) 64²
(ii) 35²
(iii) 47²
(iv) 53²
(v) 88²
Answer:
(i) 64²
64² = 60² + (2 × 60 × 4) + 4²
= 3600 + 480 + 16
= 4096

(ii) 35²
35² = 30² + (2 × 30 × 5) + 5²
= 900 + 300 + 25
= 1225

(iii) 47²
47² = 40² + (2 × 40 × 7) + 7²
= 1600 + 560 + 49
= 2209

(iv) 53²
53² = 50² + (2 × 50 × 3) + 3²
= 2500 + 300 + 9
= 2809

(v) 88²
88² = 80² + (2 × 80 × 8) + 8²
= 6400 + 1280 + 64
= 7744

Decimal Squares (Page No. 13)

Question 1.
Find the squares of the numbers below:
(i) 2.3
(ii) 8.7
(iii) 10.1
(iv) 12.5
(v) 15.7
Answer:
(i) 2.3² = 2² + 2 × 2 × 0.3 + (0.3)²
= 4 + 1.2 + 0.09
= 5.29

(ii) 8.7² = 8² + 2 × 8 × 0.7 + (0.7)²
= 64 + 11.2 + 0.49
= 75.69

(iii) 10.1² = 10² + 2 × 10 × 0.1 + (0.1)²
= 100 + 2 + 0.01
= 102.01

(iv) 12.5² = 12² + 2 × 12 × 0.5 + (0.5)²
= 144 + 12 + 0.25
= 156.25

(v) 15.7² = 15² + 2 × 15 × 0.7 + (0.7)²
= 225 + 21 + 0.49
= 246.49

Class 8 Maths Chapter 1 Kerala Syllabus Squares Questions and Answers

Class 8 Maths Squares Questions and Answers

Calculate the squares given below.

Question 1.
70² = _____
Answer:
70² = 70 × 70
= 7 × 10 × 7 × 10
= 7 × 7 × 10 × 10
= 49 × 100
= 4900

Question 2.
71² = _____
Answer:
71² = 70² + 2 × 70 × 1 + 1²
= 4900 + 140 + 1
= 5041

Question 3.
72² = _____
Answer:
72² = 70² + 2 × 70 × 2 + 2²
= 4900 + 280 + 4
= 5184

Question 4.
75² = _____
Answer:
75² = 7 × 8 | 25 = 5625

Question 5.
8.5² = _____
Answer:
8.5² = 8 × 9 + \(\frac {1}{4}\)
= 72 + 0.25
= 72.25

Question 6.
7.4² = _____
Answer:
7.4² = 7² + 2 × 7 × 0.4 + (0.4)²
= 49 + 5.6 + 0.16
= 54.76

Question 7.
7.62 = _____
Answer:
7.6² = 7² + 2 × 7 × 0.6 + (0.6)²
= 49 + 8.4 + 0.36
= 57.76

Question 8.
75.4² = _____
Answer:
75.4² = 75² + 2 × 75 × 0.4 + (0.4)²
= 5625 + 60 + 0.16
= 5685.16

Question 9.
8.25² = _____
Answer:
8.25² = 8² + \(\frac {8}{2}\) + 0.0625
= 64 + 4 + 0.0625
= 68.0625

Question 10.
10.25² = _____
Answer:
10.25² = 10² + \(\frac {10}{2}\) + 0.0625
= 100 + 5 + 0.0625
= 105.0625

Question 11.
6.25² = _____
Answer:
6.25² = 6² + \(\frac {6}{2}\) + 0.0625
= 36 + 3 + 0.0625
= 39.0625

Question 12.
100.25² = _____
Answer:
100.25² = 100² + \(\frac {100}{2}\) + 0.0625
= 10000 + 50 + 0.0625
= 10050.0625

Class 8 Maths Chapter 1 Notes Kerala Syllabus Squares

→ The digit in the ones place of a perfect square can only be any of these numbers: 0, 1, 4, 5, 6, or 9.

→ If a number ends in 5. The digit in the ones place of its square should be 5, and the digit in its tens place should be 2.

→ The square of an even number is always even.

→ The square of an odd number is always odd.

→ The number of factors of a perfect square is odd.

→ In the first 100 natural numbers, there are 10 perfect squares, and in the first 1000 natural numbers, there are 31 perfect squares.

A perfect square is a number that is obtained by multiplying a natural number by itself. For example, when we multiply 4 by 4, we get 16. So 16 is called the perfect square. Perfect squares are easy to recognize and have many unique patterns and properties. They are always positive, and their square roots are always natural numbers. These numbers appear frequently in mathematics, especially in geometry, like calculating the area of a square, and in algebra. In this chapter, you will learn how to find the square of an integer. And also discuss how to find the square of a decimal number.

Perfect Squares
The squares of natural numbers 1, 2, 3, … are called perfect squares.

Multiplication of two-digit numbers.
11 × 11

In another form,

11 × 11 = 100 + (10 + 10) + 1
That means,
11² = 10² + 2 × 10 + 1² = 121
Similarly, the square of 12
12 × 12

12 × 12 = 100 + 20 + 20 + 4
= 10² + 2 × 20 + 2²
= 144
Without drawing the cells, we can write them like this,
13² = 10² + 2 × 30 + 3² = 169
14² = 10² + 2 × 40 + 4² = 196

Square of 20 = 20 × 20
= 2 × 10 × 2 × 10
= 2 × 2 × 10 × 10
= 4 × 100
= 400

Square of 212

212² = 200² + 2 × 200 × 12 + 12²
= 40000 + 4800 + 144
= 44944

Square of 386
386² = 300² + 2 × 300 × 86 + 86²
= 90000 + 51600 + 7396
= 148996

Worksheet – 1

Calculate the squares given below.
(i) 24
(ii) 56
(iii) 76
(iv) 35
(v) 91
(vi) 101
(vii) 211
(viii) 912
Answer:
(i) 576
(ii) 3136
(iii) 5776
(iv) 1225
(v) 8281
(vi) 10201
(vii) 44521
(viii) 831744

Square of 30 = 30 × 30
= 3 × 10 × 3 × 10
= 9 × 100
= 900

Square of 40 = 4 × 4 × 100
= 16 × 100
= 1600

Square of 25

25 × 25 = 20² + 2 × 20 × 5 + 5²
25² = 400 + 200 + 25
25² = 625

52 × 52 = 50² + 2 × 50 × 2 + 2²
52² = 2500 + 200 + 4
52² = 2704

87 × 87 = 80² + 2 × 80 × 7 + 7²
87² = 6400 + 1120 + 49
87² = 7569

Simplest method to calculate the square of integers that end in 5:
For example:
35² = 30² + 2 × 30 × 5 + 5²
= 900 + 300 + 25
= 1225
In this answer 1225, if we remove the last two digits 25 we get the remaining as 12.
12 | 25
The 12 is 3 × 4.
We can check this in other numbers also.
For example:
4 | 5² = 20 (4 × 5) | 25
6 | 5² = 6 × 7 | 25 = 4225
9 | 5² = 9 × 10 | 25 = 9025
105² = 10 × 11 | 25 = 11025
295² = 29 × 30 | 25 = 87025
1005² = 100 × 101 | 25 = 1010025

The square of the integer ending in 5 is the multiple of the number of tens (the number obtained after removing 5 from it), with the next natural number, and joining the square of 5 with this.

Decimal Squares
Square of 15 is 15² = 225
Then 1.5 × 1.5 = \(\frac{15}{10} \times \frac{15}{10}\)
= \(\frac {225}{100}\)
= 2.25

If the length of the side of a square is 2.3 metres. Find its area.
Answer:
Area = side × side
= 2.3 × 2.3
= \(\frac{23^2}{100}\)
= \(\frac {529}{100}\)
= 5.29 sq.m
This multiplication can also be considered as a rectangle math problem

3.7² = 3.7 × 3.7
= 3² + 2 × 3 × 0.7 + (0.7)²
= 9 + 4.2 + 0.49
= 13.69

4.3² = 4.3 × 4.3
= 4² + 2 × 4 × 0.3 + (0.3)²
= 16 + 2.4 + 0.09
= 18.49

Worksheet – 2

Calculate the squares given below.
Question 1.
6.8²
Answer:
6.8² = 6.8 × 6.8
= 6² + 2 × 6 × 0.8 + (0.8)²
= 36 + 9.6 + 0.64
= 46.24

Question 2.
12.3²
Answer:
12.3² = 12.3 × 12.3
= 12² + 2 × 12 × 0.3 + (0.3)²
= 144 + 7.2 + 0.09
= 151.29

Question 3.
30.5²
Answer:
30.5² = 30.5 × 30.5
= 30² + 2 × 30 × 0.5 + (0.5)²
= 900 + 30 + 0.25
= 930.25

Simplest method to calculate the square of a number with the decimal part 0.5:
3.5² = 3.5 × 3.5
= 3² + 2 × 3 × 0.5 + (0.5)²
= 9 + 3 + 0.25
= 12.25

7.5² = 7² + 2 × 7 × 0.5 + (0.5)²
= 72 + 7 + 0.25
= 56.25

10.5² = 10² + 10 + 0.25 = 110.25

50.5² = 50² + 50 + 0.25 = 2550.25

We can see this in another form,
3.5² = 3 × 4 + 0.25 = 12.25
7.5² = 7 × 8 + 0.25 = 56.25
10.5² = 10 × 11 + 0.25 = 110.25

If the decimal part is 0.25:
For example:
2.252 = 22 + 2 × 2 × 0.25 + (0.25)2
= 4 + 1 + 0.0625
= 5.0625

3.252 = 32 + 2 × 3 × 0.25 + (0.25)2
= 9 + 1.5 + 0.0625
= 10.5625

If we consider the right side first,
Square of the integer + half of the integer + square of 0.25

Worksheet – 3

Question 1.
Find the square of 12.25.
Answer:
12.25 = Square of 12 + half of 12 + 0.0625
= 144 + 6 + 0.0625
= 150.0625

Question 2.
Square of 30.25.
Answer:
30.25 = Square of 30 + half of 30 + 0.0625
= 900 + 15 + 0.0625
= 915.0625

Students often refer to Kerala State Syllabus SCERT Class 8 Maths Solutions and Class 8 Maths Chapter 2 Equal Triangles Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 8 Maths Chapter 2 Solutions Equal Triangles

Class 8 Kerala Syllabus Maths Solutions Chapter 2 Equal Triangles Questions and Answers

Equal Triangles Class 8 Questions and Answers Kerala Syllabus

Sides and Angles (Page No. 18, 19)

Question 1.
In each pair of triangles below, find the angles of the second triangle equal to the angles of the first triangle, and write these pairs.

Answer:
(i) ∠M = ∠Z, ∠N = ∠X, ∠L = ∠Y
(ii) ∠A = ∠R, ∠C = ∠Q, ∠B = ∠P

Question 2.
In the triangles shown below,
AB = QR, BC = RP, CA = PQ

Calculate angle ∠C of triangle ABC and all angles of triangle PQR.
Answer:
In triangle ABC,
∠C =180 – (40 + 60)
= 180 – 100
= 80°
In triangle PQR,
∠P = ∠C = 80°
∠Q = ∠A = 40°
∠R = ∠B = 60°

Question 3.
In the triangle below,
AB = QR, BC = PQ, CA = RP

Calculate the other two angles of each triangle.
Answer:
AB = QR ⇒ ∠C = ∠P = 50°
CA = RP ⇒ ∠B = ∠Q = 70°
BC = PQ ⇒ ∠A = ∠R = 60°

Question 4.
The diagonal of a quadrilateral splits it into two triangles as shown below:

Do these triangles have the same angles? Why?
Answer:
The diagonal is common to both triangles.
Sides of upper triangles are equal to the sides of lower triangles.
Angles opposite to equal sides are equal.

Question 5.
In the quadrilateral ABCD shown below,
AB = AD, CB = CD

Calculate all the angles of the quadrilateral.
Answer:
AD = AB, CD = CB,
AC is a common Sides of triangle ABC are equal to the sides of triangle ADC
Angles opposite to equal sides are equal.
∠ACD = 50°, ∠BAC = 30°, ∠D = 100°, ∠B = 100°
Angles are ∠A = 60°, ∠B = 100°, ∠C = 100°, ∠D = 100°

Two Angles (Page No. 24)

Question 1.
In each pair of triangles below, find the sides of the triangle on the right equal to the sides of the triangle on the left:

Answer:
(i) BC = PQ, AC = PR, ∠C = ∠P
(ii) MN = XZ, ∠Z = ∠M, LM = YZ

Question 2.
In the picture, the top two sides and the bottom side of a pentagon, with equal angles and equal sides, are extended to form a triangle:

(i) Are the sides of the small triangle on the left equal to the sides of the small triangle on the right? Why?
(ii) Are the left and right sides of the large triangle equal? Why?
Answer:
(i) Look at the picture

Sides and angles of ABCDE are equal.
Since ∠A = ∠B, then ∠PAE = ∠QBC
∠PEA = ∠QCB
In triangle PAE, side AE and angles at the ends, side BC and angles at the ends in triangle QBC are equal.
So triangles are equal.

(ii) In triangle DPQ, ∠P = ∠Q.
Sides opposite to equal angles are equal.
PD = QD

Question 3.
The sides of a triangle are equal to the sides of another triangle.
(i) Is the height from each side of one triangle to the opposite vertex equal to the height from the equal side of the other triangle to its opposite vertex? Why?
(ii) Are the areas of the two triangles equal? Why?
Answer:
(i) In the figure

AB = PQ, AC = PR, BC = QR
AM is the altitude to the side BA.
∠AMB = 90°,
PN is the altitude to the side QR.
∠PNQ = 90°
In triangle AMB and triangle PNQ,
∠B = ∠Q, ∠M = ∠N
⇒ ∠A = ∠P
Side AB of triangle AMB and angles at the ends are equal to side PQ of triangle PNQ and angles at the ends.
Triangles are equal.
Sides opposite to equal angles are equal.
So AM = PN

(ii) In triangle ABC, side BC and altitude to BC are equal to side QR and altitude to QR are equal.
Areas are equal.

Two Sides (Page No. 28, 29)

Question 1.
The green lines in the picture below are parallel and of the same length; one is drawn from the end of the horizontal blue, and the other is drawn from the midpoint of the blue line:

(i) Are the lengths of the red lines in the picture equal? Why?
(ii) Are the red lines parallel?
Answer:

(i) AD = BE, AB = BC, ∠DAB = ∠EBC
In the triangles ABD and BCE, two sides and the angle between them are equal.
Sides opposite to equal angles are equal: BD = CE

(ii) Angle ABD, angle BCE are equal.
Side BD and CE are parallel.

Question 2.
Is the quadrilateral below a parallelogram? Why?

Answer:
∠BAC = ∠DCA, AB = CD
Line AB and CD are equal and parallel.
In triangle ABC and in triangle ADC line AC is common.
Two sides and the angle between them are equal.
AD = BC, ∠DAC = ∠BCA
Opposite sides of ABCD are equal and parallel.
ABCD is a parallelogram.

Question 3.
In the figure below, the lines AB and CD are parallel. M is the midpoint of AB.

(i) Are the quadrilaterals AMCD and MBCD parallelograms? Why?
(ii) Calculate all the angles of the triangles AMD, MBC, and DCM.
Answer:
(i) Opposite sides are equal and parallel
AM = CD = 3, AM is parallel to CD
AMCD is a parallelogram. MBCD is also a parallelogram.

(ii) ∠MCD = 40°, ∠MDC = 60°,
∠ADM = 140 – 60 = 80°
∠AMD = 180 – (40 + 80) = 60°
∠MCB = 120 – 40 = 80°
∠AMC = 140°
∠DMC = 180 – (60 + 40) = 80°

Question 4.
In the picture, O is the centre of the circle and A, B, C, D are points on the circle:

Are the lines AB and CD equal? Why?
Answer:
In triangle OAB and triangle OCD
OA = OC, OB = OD.
The angle between them is 80°.
AB = CD

Question 5.
In the picture, O is the centre of the circle and A, B, and C are points on the circle. ∠AOB and ∠AOC are equal:

Are the lines AB and AC equal? Why?
Answer:
In triangle AOB and triangle AOC, OA is the common side.
OB = OC.
Including angles are equal.
Triangles are equal.
Sides opposite to equal angles are equal.
AB = AC

Isosceles Triangles (Page No. 32)

Question 1.
Find the other angles in each of the isosceles triangles below?

Answer:
(a) AB = AC
⇒ ∠B = ∠C = 30°
∠A = 180 – 60 = 120°

(b) PQ = PR
⇒ ∠Q = ∠R = \(\frac {140}{2}\) = 70°

(c) LM = MN
⇒ ∠L = 45°, ∠N = 45°

(d) XY = YZ
⇒ ∠X = ∠Z = 40°

Textbook Page No. 35, 36

Question 1.
One angle of an isosceles triangle is 120°. What are the other two angles?
Answer:
One angle is 120°.
This will be the angle between equal sides.
So the other two angles are opposite equal sides.
Each is equal to 30°.

Question 2.
The picture shows a triangle drawn by joining the centre of a circle and two points on the circle:

Calculate the other two angles of the triangle.
Answer:
OA and OB are radii, equal.
So angles opposite to radii are equal to 60°.
All angles are 60°.

Question 3.
The picture shows the triangle drawn by joining three points on a circle. Two of the angles formed by joining these points to the centre of the circle are also given:

(i) Calculate the third angle at the centre.
(ii) Calculate all the angles of the large (green) triangle in the circle.
Answer:
(i) In triangle OAB,
OA = OB.
The angles opposite to these sides are 35°.
∠OBA = ∠OAB = 35°
In triangle OAC,
OA = OC.
Sides opposite to these sides are equal.
Each angle is 30°.
In triangle OBC,
∠BOC = 360 – 230 = 130°
∠OBC = ∠OCB = 25°
(ii) ∠A = 65°, ∠B = 60°, ∠C = 55°

Question 4.
Calculate the areas of each of the triangles below:

Answer:
(i) Draw AM perpendicular to BC.
Triangle AMB is a right triangle.
MB = CM = 4 cm
AM² = 5² – 4² = 9
⇒ AM = 3
Area of triangle = \(\frac {1}{2}\) × 8 × 3 = 12 cm

(ii) Draw PN perpendicular to QR.
Triangle PNQ is a right triangle.
NQ = RN = 3 cm
PN² = 5² – 3² = 9
⇒ PN = 4
Area of triangle = \(\frac {1}{2}\) × 6 × 4 = 12 cm

Question 5.
Given that one angle of an isosceles triangle is 70°. What can we say about the other angles?
Answer:
Two possibilities arise
(a) If two angles are 70°, then third angle will be 40°.
Angles are 70°, 70°, 40°.

(b) If one angle is 70° and the other two angles are x, then
70 + 2x = 180
⇒ 2x = 110
⇒ x = 55°
Possible triangles
70°, 70°, 40°
70°, 55°, 55°

Question 6.
How many non-equal isosceles triangles can be drawn with one angle 70° and one side 8 centimetres?
Answer:
(a) One side is 8 cm, and the angles at the ends are 40°. So the third angle will be 70°.
(b) One side is 8 cm, and the angles at the ends are 70°.
(c) Two sides are 8 cm, and the angle between them is 40°. The third angle will be 70°.
(d) Two sides are 8 cm, and the angle between them is 70°.

Class 8 Maths Chapter 2 Kerala Syllabus Equal Triangles Questions and Answers

Class 8 Maths Equal Triangles Questions and Answers

Question 1.
Four statements are given below. Identify the true statement.
(a) Equal triangles have equal area.
(b) Equiangular triangles (Triangles having equal angles) are equal triangles.
(c) Equality: one angle and any two sides of two triangles make them equal.
(d) One side and an angle at the ends can fix a triangle.
(a) (a) and (b) are true
(b) (a) and (c) are true
(c) (b) and (c) are true
(d) (a) and (d) are true
Answer:
(d) (a) and (d) are true
Equality means being identical or having the same shape and size.

Question 2.
Read the following statements and identify the correct option.
p1: A diagonal divides a parallelogram into two equal triangles.
p2: The line joining a corner to the midpoint of the opposite side of any triangle makes the triangle into two equal triangles.
(a) p1 and p2 are true
(b) p1 and p2 are false
(c) p1 is true p2 is false
(d) p1 is false p2 is true
Answer:
(c) p1 is true p2 is false

Question 3.
The base and altitude to the base of an isosceles triangle are 10 cm and 12 cm. The length of the equal sides
(a) 13 cm
(b) 15 cm
(c) 10 cm
(d) 8 cm
Answer:
(a) 13 cm
Half of the base, altitude, and one of the equal sides make a right triangle.

Question 4.
The angle around the centre of a circle is divided into three equal parts, each 120°, by drawing radii. The ends of the radii are joined to make a triangle. The triangle so formed is
(a) Right triangle
(b) Isosceles triangle
(c) Equilateral triangle
(d) None of these
Answer:
(c) Equilateral triangle
Triangles formed around the centre of the circle are equal, one angle is 120°, the Other two angles are 30°. So the angles of the triangle will be 60°.

Question 5.
Two angles of a triangle are equal, and the third is half of one of the equal angles. The small angle of the triangle is
(a) 30°
(b) 40°
(c) 36°
(d) 70°
Answer:
Divide 180° into five equal parts.
Each part is 36°.
Angles of the triangle are 72°, 72°, 36°.

Question 6.
In the diagram, AB = AD, BC = DC.

(a) If ∠BAC = 20°, then what is ∠A?
(b) Name pairs of equal triangles in the figure.
Answer:
(a) Triangle BAC and triangle DAC are equal.
∠BAC = ∠DAC = 20°

(b) Triangle AOB and triangle AOD are equal.
Triangle BOC and triangle DOC are equal.
Triangle ABC and triangle ADC are equal.
Triangle ABD and triangle CBD are equal.

Question 7.
In the figure, AB = AC, BD = CD.
(a) Name two equal triangles in the figure.
(b) If ∠BAC = 40°, ∠ABD = 30°, then what are the angles of triangle ADB?

Answer:
(a) AB = AC, BD = CD, AD is common.
Triangle ABD and triangle ADC are equal.

(b) In triangle ADB,
∠A = 20°, ∠B = 30°, ∠D = 130°

Question 8.
In the figure, AB = CF, EF = BD, ∠and DBC = ∠AFE.

(a) Are the triangles AFE and triangle BDC equal? How do you know this?
(b) What are the equal sides and equal angles of these triangles?
Answer:
(a) AB = CF
∴ AB + BF = CF + BF
AF = CB,
BD = EF,
∠DBC = ∠AFE
Triangles are equal.

(b) CD = AE, EF = BD, ∠DBC = ∠AFE

Question 9.
In the figure, AB = AC, E is the midpoint of AB, and F is the midpoint of AC.

(a) What are the equal triangles in the figure?
(b) If BF = 12 cm, then what is CE?
Answer:
(a) AB = AC
⇒ ∠B = ∠C
BE = CF, BC is common.
Triangle BEC and triangle CFB are equal.

(b) Sides opposite to equal angles are equal.
CE = 12 cm

Question 10.
In the figure, PQ = PR = 10 cm, ∠QPS = ∠RPS.

(a) What are the equal triangles in the figure?
(b) If QS = 18 cm, then what is SR?
(c) Find the perimeter of PQRS.
Answer:
(a) PQ = PR
PS is common.
∠QPS = ∠RPS
Triangles PQR and triangle PRS have equal angles and sides.
(b) 18 cm
(c) 18 + 10 + 18 + 10 = 56 cm

Question 11.
In the figure, AB = AC. The lines BO and CO divide angle B and angle C equally, intersect at O.

(a) Prove that OB = OC.
(b) Does AO divide the angle A into two equal parts?
Answer:
(a) BO divides ∠B equally, and CO divides ∠C equally.
∠OBC = ∠OCB
⇒ BO = CO

(b) Sides of triangle OAB are equal to sides of triangle AOC.
Triangles are equal.
∠BAO = ∠CAO
OA is the bisector of A.

Question 12.
In the figure, D is the midpoint of BC, DL is perpendicular to AB, DM is perpendicular to AC, and DL = DM.

(a) Prove that triangle BLD and triangle CMD have the same side and shape.
(b) Is ∠B = ∠C?
(c) If AB = 12 cm, then what is AC?
Answer:
(a) BD = CD = x, DL = DM = y
Triangle BLD and triangle CMD are right triangles
BD² = BL² + DL², CD² = DM² + CM²
Since D is the midpoint, BD = CD
BL² + DL² = DM² + CM², BL² = CM²
⇒ BL = CM
Three sides of triangle BLD are equal to three sides of triangle CMD.
So these triangles are equal.

(b) Sides opposite to equal angles are equal,
∠B = ∠C

(c) Since ∠B = ∠C we can write AB = AC.
⇒ AB = AC = 12 cm

Class 8 Maths Chapter 2 Notes Kerala Syllabus Equal Triangles

→ If all sides of a triangle are equal to the sides of another triangle, then the triangles are equal.

→ Angles opposite to equal sides are also equal.

→ If one side and angles at the ends of that side are equal to one side and angles at the ends of another triangle, then the triangles are equal. Sides opposite to equal angles are equal. Angles opposite to equal sides are also equal.

→ A quadrilateral having opposite sides parallel and equal is called a parallelogram.

→ If two lines have equal length and are parallel, then joining ends on the same side makes a parallelogram.

→ If two sides and the angle between them in a triangle are equal to two sides and including angle of another triangle, then the triangles are equal.

→ Sides opposite to equal angles are equal.

→ Angles opposite to equal sides are equal.

→ Triangles having two sides equal are called isosceles triangles. Angles opposite to equal sides are equal.

→ Triangles having all sides equal are called equilateral triangles. All angles of an equilateral triangle are 60° each.

→ In an isosceles triangle perpendicular from the apex forming equal angles to the opposite side divides the triangle into two equal angles.

Triangles having the same shape and size are called equal triangles. The geometric conditions for becoming two triangles equal and related concepts are discussed in this unit. To fix a triangle, three measurements are essential. Three sides or one side and angles at the ends, or two sides and including angles are these measurements.

Triangles can be classified into isosceles triangles, equilateral triangles, and scalene triangles. If all sides are equal, then the triangle is called an equilateral triangle. Triangles having two sides equal are called isosceles triangles. If all sides are different, it will be a scalene triangle.

Sides and Angles
If the lengths of the sides of two triangles are the same, then their angles are also the same.
Look at these triangles:

Here, the length of the sides of the two triangles is the same, and their angles are also the same.
That is, each angle in triangle ABC is equal to one of the angles in triangle PQR. Here,

• The longest side is 6 centimetres
• Medium side is 5 centimetres
• Shortest side is 3 centimetres

Thus, in each triangle,

• The largest angle is opposite the 6-centimetre side
• Medium angle is opposite the 5-centimetre side
• The smallest angle is opposite the 3-centimetre side

Therefor the,

• Largest angle is, ∠A = ∠R
• Medium angle is, ∠B = ∠P
• Smallest angle is, ∠C = ∠Q

If the lengths of the sides of two triangles are the same, then the angles opposite to sides of equal length are also equal.

Worksheet – 1

Question 1.
In each pair of triangles below, find all pairs of matching angles and write them down.

Answer:
(i) ∠A = ∠R, ∠B = ∠P, ∠C = ∠Q
(ii) ∠L = ∠Y, ∠N = ∠X, ∠M = ∠Z

Worksheet – 2

Question 1.
Two triangles are given below.

(a) Are the triangles equal? Why?
(b) If the triangles are equal, which are the matching angles?
Answer:
(a) Yes, these two triangles are equal.
Because three sides of ∆ABC are equal to three sides of ∆PQR.
ie. AB = PQ = 8 cm
BC = PR = 6 cm
AC = RQ = 4 cm
(b) ∠C = ∠R, ∠A = ∠Q, ∠B = ∠P

Two Angles
If the angles of a triangle are the same as the angles of another triangle, the lengths of their sides may not be the same.

If in two triangles, the length of one side and the two angles at its ends are the same, then their third angles and the lengths of the other two sides are also the same.

If two angles of a triangle are the same as two angles of another triangle, then the third angles are also the same.

In two triangles, even if all the angles and one side are the same, the other two sides may or may not be the same.
A parallelogram is a quadrilateral with each pair of opposite sides parallel.

The opposite sides of any parallelogram are equal.
The opposite angles of any parallelogram are equal.

Worksheet – 3

Question 1.
In each pair of triangles below, And the pairs of sides and write their names.


Answer:
(i) AB = RQ
BC = QP
AC = RP
Because the two triangles are equal, the sides opposite to the equal angles are also equal.
(ii) LN = YX, MN = ZX, LM = YZ

Two Sides
If in two triangles, the lengths of two sides and the angle between them are the same, then the length of the third side and the remaining two angles are also the same.
If two sides of a quadrilateral are equal and parallel, then the quadrilateral is a parallelogram.

Equal Triangles: If two sides and the angle between them in a triangle are equal to two sides and the included angle of another triangle, then the triangles are equal. Sides opposite to equal angles are equal. Angles opposite to equal sides are equal.

Worksheet – 4

Question 1.
In the figure below, AC and BE are parallel lines.

(i) Are the lengths of BC and DE equal? Why?
(ii) Are the lengths of BC and DE equal? Why?
Answer:
AC = BE
AB = BD
Since AC is parallel to BE
∠BAC = ∠DBE [corresponding angles]
So ∆ABC and ∆BDE are equal.
(i) Since the two triangles are equal, BC = DE
(ii) Since ∠ABC = ∠BDE
BC is parallel to DE.

Question 2.
Is ACBD in the figure a parallelogram? Why?

Answer:
AC = BD = 6 cm
AB = BA
∠BAC = ∠ABD = 35°
Also, AC is parallel to BD
So ∆BAC and ∆ABD are equal.
So AD = BC and parallel.
∴ So ACBD is a parallelogram.

Worksheet – 5

Question 1.
Are the angles of ∆ABC and ∆ABD equal in the figure below? Why?

Answer:
Yes, because ∆ABC and ∆ABD are equal triangles.
AC = AD, BC = BD, also AB = AB itself.
So ∠C = ∠D, ∠CAB = ∠DAB,
∠CBA = ∠DBA (∵ In equal triangles, corresponding angles are equal)

Isosceles Triangles
A triangle with two of its sides equal is called an isosceles triangle.
See this isosceles triangle, the left and right sides are the same length.

Look at the sides of the small triangles ABM and ACM on the left and right.

• AB = AC as mentioned above.
• BM = CM since M is the midpoint of BC.
• AM is a side of both these triangles.

Thus, the lengths of the sides of the triangle ABM and ACM are the same.
So, the angles opposite equal sides must also be the same.

If two sides of a triangle are equal, then their opposite angles are also equal.

A triangle with all three sides equal is called an equilateral triangle.
In any equilateral triangle, each angle is 60°.

If two angles of a triangle are equal, then their opposite sides are also equal.

In any isosceles triangle, the perpendicular from the vertex joining the equal sides to its opposite side splits this side and the angle at this vertex into equal parts.

Reviewing SCERT Class 8 Basic Science Solutions and Kerala Syllabus Class 8 Basic Science Chapter 9 Cells and Cell Clusters Question Answer Notes Pdf can uncover gaps in understanding.

Class 8 Basic Science Chapter 9 Cells and Cell Clusters Question Answer Notes

Class 8 Basic Science Chapter 9 Notes Kerala Syllabus Cells and Cell Clusters Question Answer

Cells and Cell Clusters Class 8 Questions and Answers Notes

Let’s Assess

Question 1.
Identify the word pair relationship and fill the blanks.
a) compound microscope : visible light
……………….. : electron beam
b) Plant cells : eukaryote
………………… : prokaryote
c) Endoplasmic reticulum : Material transport in the cytoplasm
………………….. : Centre for production and distribution of energy
Answer:
a) Electron microscope
b) Bacteria
c) Mitochondria

Question 2.
Choose the statements related to plant cell from the following.
(a) Centrosome is present
(b) Plasma membrane is absent
(c) Plastids are present
(d) Presence of comparatively large vacuoles
Answer:
(c) Plastids are present and (d) Presence of comparatively large vacuoles

Question 3.
Information related to certain cell organelles are given in the table. Complete it by selecting suitable terms from the box.

Answer:
A – f) Golgi apparatus
B – e) Chloroplast
C – d) Ribosome
D – b) Leucoplast

Basic Science Class 8 Chapter 9 Question Answer Kerala Syllabus

Answers to the indicators on page no. 136
Question 1.
Where should the slide be placed in a microscope?
Answer:
The Slide should be placed on the stage of the microscope, and the stage clip is used to keep the slide in place.

Question 2.
Which is the part that regulates the light on the slide?
Answer:
The diaphragm in the condenser controls the amount of light. The condenser lens above this focuses the light onto the object being observed. Mirror- Reflects light from the light source onto the diaphragm.

Question 3.
Which are the lenses used in microscopes?
Answer:
The eyepiece, objective lens, and condenser are the main parts of a microscope that are attached to the lens.

Question 4.
Completed illustration 9.2

Answer:

Question 5.
Completed table 9.1

Answer:

Part of a microsope	Function
Eye piece lens	The image formed by the objective lens is again magnified
Knobes	Used to focus the lenses on the object being observed. The larger knob is called coarse adjustment and the smaller knob is called fine adjustment
Objective lens	The image of the object being observed is magnified and projected into the inner surface of the eyepiece. The part where the objective lenses are mounted is called the nosepiece.
Stage	Used to place slide
Clip	Used to keep the slide in place
Diaphragm in condenser	Control the amount of light
Condenser lens	Helps focus light onto the object.
Mirror	Light from the light source is reflected onto the diaphragm

Question 6.
Completed table 9.2 comparing the peculiarities of different types of microscopes.

Answer:

Peculiarity	Compound microcope	Electron microscope
Need of light	Light is needed	Electron beam is used instead of light
Magnifying power	Up to a thousand times	It magnifies objects more than a million times more than a normal microscope
Lens	Using multiple lenses	Electro magnets are used

Question 7.
Note prepared based on the illustration 9.3 page 140

Answer:
A normal microscope can observe mitochondria, bacteria, red blood cells, yeast, animal cells, plant cells, paramecium, and ova. An electron microscope is an instrument that magnifies objects more than a million times larger than a normal microscope. It helps in the detailed observation of living cells, viruses, and molecular structures. An electron microscope helps in the detailed observation of DNA, viruses, and proteins, which cannot be observed under a normal microscope.

Answers to the indicators on page 141
Question 8.
While preparing the slide:
Answer:

• Use a sharp blade for smooth, thin sections.
• Take the thinnest sections possible. Thin sections allow light to pass through easily.
• Cut straight across to get a proper cross-section.
• Place the thin section on a clean glass slide.
• Add a few drops of water or glycerin to prevent drying.
• Cover with a cover glass using a needle to prevent air bubbles from entering.

Question 9.
Staining:
Answer:

• Use a suitable stain such as safranin or methylene blue.
• Stain for only a few seconds
• Rinse gently with water to remove excess stain before covering with a coverslip.

Question 10.
Observing in high power and low power:
Answer:
Low Power: Observe with a low-power objective lens. Helps to see the epidermis, cortex, pith, etc. Adjust the mirror for a clear view.
High Power: Switch to high power to see things like xylem and phloem. Use the fine adjustment knob to adjust the clarity. Do not move the slide when switching from low power to high power.

Question 11.
Observing and illustrating:
Answer:

• Observe carefully and patiently.
• Draw neatly with a pencil.
• Draw only what you see.
• Use arrows for labelling and write names neatly.

Question 12.
Practical record on page 142 of the textbook:
Answer:
Experiment Name: Plant Stem Observation
Objective: Observe the cells in a plant stem through a microscope.
Materials Required: Microscope, plant stem, Petri dish, glass slide, cover glass, glycerin, blade, tissue paper, needle, brush, water, safranin stain, watch glass.

Procedure:
To prevent dehydration of the object to be observed, place it in Water in a Petri dish. Take thin cross-sections of the object to be observed. Transfer the sections to the water in a watch glass using a brush. Place the thinnest and most complete sections in a diluted stain taken in another watch glass. Add one or two drops of glycerin to prevent the object from drying out. Use a brush to pick up the stained section gently, then wash it in clean water and place it in glycerin on the slide. Cover it with a cover glass using a needle to prevent air bubbles from entering. Wipe off excess glycerin with tissue paper. Observe under a microscope.

Observation:

Conclusion:
Various types of cells are found in plant stems.

Question 13.
Completed illustration 9.5 page 143

Answer:

Question 14.
Completed illustration 9.7 page 147

Answer:

Question 15.
Completed table 9.3 page 149

Answer:

Plant tissues	Characteristics	Functions
Parenchyma	Contain living cells. Thin cell wall. Intercellular spaces are present. Cells in some parts have the ability to divide.	Perform functions such as photosynthesis and food storage.
Collenehyma	A tissue made up of living cells. The cell wall is made of substances such as cellulose and pectin. The cell wall is thicker only in some parts.	Helps in maintaining the shape of plant parts.
Selerenehyma	A tissue consists of dead cells. Thick-cell-wall. The cell wall is of uniform thickness throughout.	Providing strength and support to plant parts.
Xylem	Made up of cells of different shapes and sizes.	Conduction of water and salt to the leaves take place through xylem.
Phloem	Made up of cells of different shapes and sizes.	Carries food synthesised by the leaves to different parts of the plant body.

Question 16.
Completed illustration 9.9 page 151

Answer:

Question 17.
Completed illustration 9.10 page 152

Answer:

Question 18.
Completed table 9.4 page 153

Answer:

Part	Plant cell	Animal cell	Function
Cell wall	present	Absent	Provides protection and shape to the cell
Centrioles	Absent	Present	Helps in cell division
Plastids	Present	Not present	Helps in the production and storage of food.
Lysosome	Absent	Present	Destroying foreign objects

Class 8 Basic Science Chapter 9 Question Answer Extended Activities

Question 1.
Prepare models of Plant cell and Animal cell using materials available in your surroundings.

Question 2.
Prepare a picture chart showing the diversity of cells and exhibit it in the class.
Answer:

Question 3.
Prepare a timeline showing the development of cytology and present it in the class.
Answer:

Cells and Cell Clusters Class 8 Notes

Class 8 Basic Science Cells and Cell Clusters Notes Kerala Syllabus

• Cell biology is the study of cells.
• Human eye can distinguish two spots having a distance of 0.2 mm in adequate light. This distance is called resolution of the eye.
• A simple microscope in which a single lens is used, can magnify an object upto 10 times than the original size, while in a compound microscope, in which more than one lens is used, can magnify an object up to 1000 times.
• Magnification power of a lens is its ability to magnify objects.
• Electron microscope is an instrument that magnifies objects more than a million times. It helps to observe cells, viruses and molecular structure in detail.
• In electron microscope electron beam is used instead of light.
• In 1934 German scientists Ernest Raska and Max Knoll invented electron microscope.
• The various parts of a plant cell are the cell wall, cell membrane or plasma membrane, cytoplasm, and cell organelles.
• Organelles are parts found in the cytoplasm that perform functions necessary for the survival of the cell.
• The organelles in plant cells are nucleus, Golgi apparatus, endoplasmic reticulum, and mitochondria, chloroplast, ribosome, vacuole, etc.
• Plastids are specialized parts of plant cells that help produce and store food. Chloroplast, chromoplast and leucoplasts are different forms of plastids.
• The group of similar type of cells are known as tissues. Tissues that are composed of similar type of cells are known as simple tissues. Xylem and phloem are composed of cells with different shape and size. So they are called complex tissues.
• Meristermatic tissues are composed of cells that are capable of continuous division. Meristematic cells cause growth of plants. Cell clusters or tissues that have lost their ability to divide are known as permanent tissues, Parenchyma, sclerenchyma, collenchyma, xylem arid phloem are examples for permanent tissues.
• Some organelles, such as lysosomes and centrioles, are found in animal cells that are not found in plant cells.
• Tissues composed of group of similar cells perform various functions in the body of animals are called animal tissues. Epithelial tissue, Connective tissue, Muscle tissue, and Neural tissue are different types of animal tissues.
• Stem cells are specialised cells that can develop into various types of cells such as muscle cells, nerve cells, blood cells, etc. Stem cells help to eliminate damaged cells, promote the growth of new cells and maintain healthy tissues.

INTRODUCTION

Imagine looking at your hand, or a leaf from a tree, or even a tiny ant. Have you ever wondered what these living things are made of? Well, just like a building is made of many bricks, all living organisms, from the smallest bacteria to the largest elephant, are made up of tiny basic units called cells.

Cells are often called the “building blocks of life” because they are the smallest units that can perform all the functions necessary for life, like taking in nutrients, growing, and reproducing. They are so tiny that you can’t see them with your naked eye; you need a special instrument called a microscope to observe them.

In this chapter, we’re going to explore the amazing world of cells. We’ll learn about what a cell looks like and its different parts, how different cells come together to form cell clusters (Tissues), Cell theory, and contributions of different scientists in this field.

HISTORY OF CELL BIOLOGY

• Cell biology is the study of cells.
• Matthias Jakob Schleiden discovered that all plants are composed of cells.
• Theodor Schwann discovered that all animals are composed of cells.
• Rudolf Virchow introduced the concept that new cells are formed from pre-existing cells.

Cell Theory

• • All organisms are made up of one or more cells.
  • The cell is the basic unit of life.
  • New cells originate from pre-existing cells.
• Different types of microscopes are used to observe very small or tiny cells. In simple microscopes, lenses are used to magnify objects.
• The human eye can distinguish two spots having a distance of 0.2 mm in adequate light. This distance is called the resolution of the eye.
• A lens is required to distinguish spots having a distance of less than 0.2 mm.
• A simple microscope in which a single lens is used can magnify an object upto 10 times than its original size.
• A compound microscope, in which more than one lens is used, can magnify an object up to 1000 times.
• The magnification power of a lens is its ability to magnify objects. If eye piece lens magnifies the object to 10 x (10 times) and objective lens magnifies it into 40 x (40 times), the magnification power of the lens is 400 x.

Electron microscope

• An instrument that magnifies objects more than a million times.
• It helps to observe cells, viruses and molecular structure in detail.
• In an electron microscope electron beam is used instead of light.
• In 1934, German scientists Ernest Raska and Max Knoll invented the electron microscope.
• In an electron microscope, electromagnetic lenses are used to focus a beam of electrons onto the specimen, allowing for much higher resolution imaging compared to light microscopes.

LET’S OBSERVE CELLS

• Living organisms have cells of different shapes and sizes in their bodies.
• The basic structure of all the different cells in a leaf and a stem is the same.

• Cell Wall – The cell wall is the rigid outer layer of the cell membrane.
• The cell membrane, or plasma membrane – Thin, flexible layer that surrounds the cell. Substances enter and leave the cell through the cell membrane.
• The plasma membrane does not allow all substances to pass through. Therefore, the plasma membrane is known as the semi-permeable membrane.
• Protoplasm – It consists of all the components inside the cell membrane, including the nucleus and cytoplasm.
• Cytoplasm – A jelly-like fluid that fills the cell. It maintains all the organelles in their place and serves as the medium for chemical reactions.
• Organelles are the parts found in the cytoplasm and perform the functions necessary for the survival of the cell.
• The organelles of a plant cell are the cytoplasm, Golgi apparatus, endoplasmic reticulum, and mitochondria.
• Nucleus – The centre that controls the cell. The chromatin network is the structure that appears like a network of threads within the nucleoplasm. During cell division, these chromatin networks condense into chromosomes.
• Golgi Apparatus – Cell organelles that appear as stacked membrane layers. These organelles transport proteins and lipids to various parts of the cell and outside the cell, wrapped in membranous sacs.
• Endoplasmic Reticulum – Organelles appear as a network of tubules within the cell, and serve as pathways for conducting materials. It helps in the synthesis and removal of materials required by the cell.
• Mitochondria – The energy production centre of the cell.
• Plastids – Specific parts in the plant cells that help in the synthesis and storage of food materials. They have a two-layered membrane. Different types of plastids perform different functions.

• Vacuole – Stores water, nutrients and wastes. Usually, large vacuoles are seen in plant cells.
• Ribosome – Found in the cytoplasm either freely or attached to the endoplasmic reticulum. They function as the centre for protein synthesis.

TISSUES
PLANT TISSUES

• The group of similar type of cells are known as tissues.

• Tissues that are composed of similar types of cells are known as simple tissues.
• Xylem and phloem are composed of cells with different shapes and sizes. So they are called complex tissues.
• The transport of water and salts to the leaves takes place through the xylem.
• The food produced by the leaves is transported to various parts by the phloem tissue.
• Meristematic tissues are composed of cells that are capable of continuous division. Meristematic cells are responsible for the growth of plants.

• Cell clusters or tissues that have lost their ability to divide are known as permanent tissues. Parenchyma, sclerenchyma. Collenchyma, xylem and phloem are examples of permanent tissues.

ANIMAL TISSUES

• The animal body is also made up of different types of cells (Animal tissues).
• Some organelles, such as lysosomes and centrioles, are found in animal cells that are not found in plant cells.

Stem cells are specialised cells that can develop into various types of cells, such as muscle cells, nerve cells, blood cells, etc.

• Stem cells help to eliminate damaged cells, promote the growth of new cells and maintain healthy tissues.
• In modern medicine, stem cells have great significance.
• Despite these differences, there are many similarities among living organisms.
• These common characteristics of living things lead us to the reality that the basis of life is one and the same.

Reviewing SCERT Class 8 Basic Science Solutions and Kerala Syllabus Class 8 Basic Science Chapter 5 Elements and Compounds Question Answer Notes Pdf can uncover gaps in understanding.

Class 8 Basic Science Chapter 5 Elements and Compounds Question Answer Notes

Class 8 Basic Science Chapter 5 Notes Kerala Syllabus Elements and Compounds Question Answer

Elements and Compounds Class 8 Questions and Answers Notes

Let’s Assess

Question 1.
Match the following

A Element	B Symbol	C Base
Francium	Rb	continent
Curium	Eu	satellite
Rubidium	Ti	colour
Neptunium	Cm	country
Europium	Fr	planet
Titanium	Np	scientist

Answer:

A Element	B Symbol	C Base
Francium	Fr	Country
Curium	Cm	Scientist
Rubidium	Rb	Colour
Neptunium	Np	Planet
Europium	Eu	Continent
Titanium	Ti	Satellite

Question 2.
Cl is the symbol of the element chlorine. How can we represent two chlorine atoms and one chlorine molecule using this symbol?
Answer:

• Two chlorine atoms – 2 Cl
• One chlorine molecule – Cl₂

Question 3.
Find the number of atoms in each of the following and write which of them contains the most number of atoms.
5NH₃, 2H₂O, 5NO₂, 4CO₂

Element	Number of Atoms
5NH3	5 N + (5 × 3) H = 5 N + 15 H = 20
2H2O	(2 × 2)H + 2 O = 4H + 2 O = 6
5NO2	5 N + (5 × 2) O = 5 N + 10 O = 15
4CO2	4C + (4 × 2) O = 4C + 8 O = 12

Ammonia contains the most number of atoms.

Question 4.
Complete the table.

Answer:

Element	Basis of nomenclature	Symbol
Indium	Indigo – colour	In
Rutherfordium	Rutherford – Scientist	Rf
Germanium	Germany – Country	Ge
Silver	Argentum – Latin name	Ag

Question 5.
The names of some scientists are given.
(Dobereiner, Lavoisier, Newlands, Mendeleev)
Match the statements given below with the names given in brackets
i. The chemical and physical properties of the elements are functions of their atomic masses.
ii. Elements were divided into groups containing three elements (triads) with similar properties.
iii. The elements were classified into metals and non-metals.
iv. When the elements were arranged in ascending order of atomic masses, it was found that every eighth element was a repetition of the first, in terms of its properties.
Answer:
i. Mendeleev
ii. Dobereiner
iii. Lavoisier
iv. Newlands

Basic Science Class 8 Chapter 5 Question Answer Kerala Syllabus

Question 1.
What do you know about molecules? Try to write it.
Answer:
Molecules are tiny particles made of two or more atoms stuck together by invisible bonds, forming the basic building blocks of all matter.

Question 2.
Can you formulate a definition for molecules?
Answer:
Molecules are the smallest, independent and stable particles that have all the fundamental properties of a pure substance.
New molecules are formed through chemical reactions.

Question 3.
Let’s do an experiment.
a) Burn a magnesium ribbon which is scrubbed clean. Record your observations.
Answer:
Observation: Magnesium burns brightly in air. A white powder is obtained.

b) What is the white powder obtained as a result of the reaction?
Answer:
It is magnesium oxide, produced when magnesium reacts with oxygen.
This is an example of the formation of new molecules through chemical reactions between different types of molecules.
We can dissociate molecules to make new molecules.

Question 4.
Take some sugar in a dry boiling tube. Cover the mouth of the boiling tube with a piece of cotton. Heat the boiling tube strongly for a few minutes. Write your observations.
Observation – A black substance remains in the boiling tube.

a) What is the black substance left in the boiling tube?
Answer:
Carbon

b) What is the substance seen on the sides of the boiling tube?
Answer:
Water

We can observe that when sugar is heated, carbon and water are formed. Sugar molecules can be broken down through chemical reactions.

Question 5.
Let’s do another experiment.
Take some dilute hydrochloric acid in a test tube and add a small piece of magnesium to it. Introduce a burning incense stick to the mouth of the test tube. Record your observations.
Observation – A vigorous chemical reaction takes place. A gas is released. When the incense stick is introduced to the mouth of the test tube, the gas burns with a pop sound.

a) What is the gas produced as a result of the reaction?
Answer:
Hydrogen (H₂)

b) How can we represent this reaction?
Answer:
Magnesium + Hydrochloric Acid → Magnesium Chloride + Hydrogen

Question 6.
Collect the hydrogen gas obtained in this experiment into another moisture-free test tube and introduce a burning incense stick to it. Record your observations.
Answer:
The gas burns with a popping sound. Water droplets are visible on the sides of the test tube.

а) What do you understand from the presence of water droplets on the sides of the test tube?
Answer:
Hydrogen in the test tube reacts with oxygen to form water.
Hydrogen + Oxygen → Water

Henry Cavendish proved that water is a substance made up of more than one element. His experiment of burning hydrogen gas in the presence of oxygen helped to know about the components of water.

In 1806, Sir Humphry Davy discovered that water could be split into hydrogen and oxygen by passing electricity through it.

Question 7.
Which are the substances obtained when water is dissociated?
Answer:
Hydrogen and Oxygen

Question 8.
Can hydrogen and oxygen be further split in to components?
Answer:
No

Question 9.
Look at the table below

Molecules	Atoms in a molecule
Ammonia	One atom of nitrogen Three atoms of hydrogen
Oxygen	Two atoms of oxygen
Chlorine	Two atoms of chlorine
Methane	One atom of carbon Four atoms of hydrogen
Hydrogen	Two atoms of hydrogen
Carbon dioxide	One atom of carbon Two atoms of oxygen

a) Which molecules in the table consist of the same type of atoms?
Answer:
Oxygen, Chlorine, Hydrogen

b) Which of these molecules consist of different types of atoms?
Answer:
Ammonia, Methane, Carbon dioxide

• Elements are pure substances made up of the same type of atoms. They cannot be further divided into components through chemical reactions.
• Compounds are pure substances formed by the combination of atoms of two or more elements in a fixed ratio. These can be dissociated into constituent elements through chemical reactions.

Question 10.
Categorise the following into elements and compounds.
Oxygen, water, salt, sugar, carbon, nitrogen, carbon dioxide, ammonia, magnesium oxide

Answer:

Elements	Compounds
• Oxygen • Carbon • Nitrogen	• Water • Salt • Sugar • Carbon dioxide • Ammonia • Magnesium oxide

Question 11.
Note the list of other elements represented by the first letter of their name. Expand the list by adding more elements.

Answer:

Element	Symbol
Carbon	C
Nitrogen	N
Sulphur	S
Oxygen	O
Boron	B
Phosphorus	P
Fluorine	F
Iodine	I
Vanadium	V
Yttrium	Y
Uranium	U

Question 12.
The element Helium is represented by the symbol ‘He’. The first and second letters of the English name are used here. What could be the reason?
Answer:
H is the symbol for hydrogen. More than one element may have the same first letter.

• Some elements are represented by another, letter along with the first letter of the English name in upper case. See more examples of using two letters as symbols.

Element	Symbol
Aluminium	Al
Chlorine	Cl
Curium	Cm
Calcium	Ca

• The first letter of the English name is used upper ease along with the second letter or any other prominent letter in lower case.

Question 13.
Find more examples and expand the list.
Answer:

Element	Symbol
Germanium	Ge
Neon	Ne
Manganese	Mn
Cobalt	Co
Zinc	Zn
Beryllium	Be
Magnesium	Mg
Barium	Ba
Radium	Ra
Silicon	Si

Some elements have symbols based on their Latin names.

Question 14.
Find more examples and expand the list.

Answer:

Element	Latin name	Symbol
Sodium	Natrium	Na
Potassium	Kalium	K
Iron	Ferrum	Fe
Gold	Aurum	Au
Copper	Cuprum	Cu
Silver	Argentum	Ag
Tin	Stannum	Sn
Lead	Plumbum	Pb
Antimony	Stibium	Sb
Mercury	Hydrargyrum	Hg
Element	Latin name	Symbol

• The symbol “W” is derived from the German name ‘Wolfram’ for the metal tungsten.

Question 15.
Find put how the elements given below got their names.
• Chromium
• Iridium
• Neptunium
Answer:
• Chromium – From the world “chroma,” meaning colour.
• Iridium – From the word “Iris,” meaning Rainbow.
• Neptunium – From the name of the planet “Neptune”.

Question 16.
Analyse the Periodic Table and write down the name and symbol of any 5 elements.
Answer:

• Lithium (Li)
• Sodium (Na)
• Beryllium (Be)
• Magnesium (Mg)
• Calcium (Ca)

Question 17.
Find out which elements were added last to the periodic table.
Answer:

• Nihonium (Nh)
• Flerovium (Fl)
• Livermorium (Lv)
• Tennessine (Ts)
• Oganesson (Og)

Question 18.
List the elements existing in solid, liquid and gaseous state with the help of periodic table.

Answer:

Solid	Liquid	Gas
Lithium	Caesium	Hydrogen
Sodium	Francium	Nitrogen
Potassium	Mercury	Oxygen
Rubidium	Gallium	Fluorine
Beryllium	Bromine	Chlorine
Magnesium		Helium
Calcium		Neon
Strontium		Argon
Barium		Krypton
Radium		Xenon
		Radon

Question 19.
Find the synthetic elements with the help of the periodic Table.

• Technetium
• Dubnium
• Bohrium
• Meitnerium
• Roentgenium
• Nihonium
• Moscovium
• Tennessine
• Promethium
• Plutonium
• Curium
• Californium
• Fermium
• Nobelium
• Rutherfordium
• Seaborgium
• Hassium
• Darmstadtium
• Copemicium
• Flerovium
• Livermorium
• Oganesson
• Neptunium
• Americium
• Berkelium
• Einsteinium
• Mendelevium
• Lawrencium

Question 20.
Is there any difference between 2N and N₂?
Answer:
2N refers to two free nitrogen atoms. N₂ refers to a nitrogen molecule formed by two nitrogen atoms bonded together.

• 2N – refers to two free nitrogen atoms.
• N₂ – refers to a nitrogen molecule formed by two atoms.
• 2N₂ – refers to two nitrogen molecules.

Question 21.
What is the total number of atoms in 2N₂?
Answer:
2N₂ refers to two nitrogen molecules (N₂ + N₂)
2 atoms in a nitrogen molecule.
Total number of atoms in 2N₂ = 2 + 2 = 4

Question 22.
Check the abbreviation of the molecules of some elements.

Answer:

Molecule	Number of atoms	Category
He	1	Monoatomic elements.
O2	2	Diatomic elements.
S8	8	Polyatomic elements.
H2	2	Diatomic elements
P4	4	Polyatomic elements
Ne	1	Monoatomic element
Cl2	2	Diatomic elements
Ar	1	Monoatomic element
O3	3	Triatomic element

Question 23.
Is it clear that, monoatomic elements are elements that have only one atom in a molecule? Write the difference between diatomic elements and polyatomic elements.
Answer:
Diatomic elements are elements that have two atoms in a molecule. Elements that have more than three atoms in a molecule are called polyatomic elements.

Chemical formula of the molecule represent one molecule of a substance.

Question 24.
Find the number of molecules and the total number of atoms in the following.

Answer:

Substance	No. of molecules	No. of atoms
H2	1	1 × 2 = 2
5O2	5	5 × 2 = 10
6N2	6	6 × 2 = 12
4Cl2	4	4 × 2 = 8
8	1	1 × 8 = 8

Question 25.
Write the symbols for hydrogen and chlorine.
Answer:
Hydrogen – H
Chlorine – Cl
A molecule of hydrogen chloride contains one hydrogen atom and one chlorine atom.
We represent a molecule of hydrogen chloride as HCl.

Question 26.
How can we represent a molecule of water in the same way?
Answer:
H₂O

Question 27.
Complete the table.

Answer:

A compound has a certain ratio between the number of atoms of its constituent elements.
Example: HCl =1 : 1
H₂O = 2 : 1
NH₃ = 1 : 3
CO₂ = 1 : 2

Question 28.
Find the number of molecules and the total number of atoms present in 3CO₂.
Answer:
Number of molecules = 3
Atoms C = 3 × 1 = 3
0 = 3 × 2 = 6
Total number of atoms = 9

Question 29.
Complete the table given below.

Answer:

Compound	No. of molecules	No. of atoms
2HCl	2	H – 2 × 1 = 2 Cl – 2 × 1 = 2 4
4SO2	4	S – 4 × 1 = 4 0 – 4 × 2 = 8 12
3C2H5OH	3	C – 3 × 2 = 6 H – 3 × 6 = 18 0 – 3 × 1 = 3 27
2C6H12O6	2	C – 2 × 6 = 12 H – 2 × 12 = 24 O – 2 × 6 = 12 48

Question 30.
What are the reactants in this reaction?
Answer:
Carbon, Oxygen

Question 31.
What are the products?
Answer:
Carbon Dioxide

Question 32.
How do you write the reactants using chemical formula?
Answer:
C, O₂

Question 33.
How about products?
Answer:
CO₂
This chemical reaction can be represented as C + O₂ → CO₂

A chemical equation is the representation of a chemical reaction which helps to understand it at a glance. It uses symbols/chemical formulae to represent the reactants and products.

Question 34.
Calcium carbonate (CaCO₃) decomposes to form calcium oxide (CaO) and carbon dioxide (CO₂). How can this chemical reaction be represented?
Answer:
CaCO₃ → CaO + CO₂

Question 35.
Write the chemical equations of some chemical reactions that you are familiar with.
Answer:

• 2H₂ + O₂ → 2H₂O
• H₂ + Cl₂ → 2HCl
• 2Mg + O₂ → 2MgO
• CaO + H₂O → Ca(OH)₂
• Mg + 2HCl → MgCl₂ + H₂

Class 8 Basic Science Chapter 5 Question Answer Extended Activities

Question 1.
Make models of the molecules using materials available in your surroundings.
• Carbon tetrachloride (CCl₄)
• Ammonia (NH₃)
• Water (H₂O)
Answer:
Here are short instructions for making the molecule models:

General Materials:

• Atoms: Different coloured clay/Play-Dough/thermocol balls (e.g., Black/Grey for C, Green for Cl, Blue
  for N, White for H, Red for O).
• Bonds: Toothpicks or short stick pieces.

1. Carbon Tetrachloride (CCl₄)

• Atoms: 1 Carbon (large, black/grey), 4 Chlorine (smaller, green).
• Instructions: Connect the central Carbon ball to four Chlorine balls using toothpicks, spreading them out to form the shape of carbon tetrachloride.

2. Ammonia (NH₃)

• Atoms: 1 Nitrogen (medium, blue), 3 Hydrogen (small, white).
• Instructions: Connect the central Nitrogen ball to three Hydrogen balls using toothpicks. Arrange the Hydrogen atoms to form the base of a pyramid with Nitrogen at the top.

3. Water (H₂O)

• Atoms: 1 Oxygen (medium, red), 2 Hydrogen (small, white).
• Instructions: Connect the central Oxygen ball to two Hydrogen balls using toothpicks. Angle the Hydrogen atoms to create a V-shaped molecule.

Question 2.
Prepare a slide presentation/chart on the history of development of Periodic Table and exhibit it in the classroom.
Answer:
The Periodic Table: A Journey of Discovery
(Slide 1 / Chart Title)
Slide 2 / Section 1: Early Groupings (1800s)

• Dobereiner’s Triads (1829): Grouped elements in threes with similar properties.
• Newlands Law of Octaves (1865): Noted repeating properties every eighth element when arranged by atomic weight.

Slide 3 / Section 2: Mendeleev’s Breakthrough (1869) ‘

• Dmitri Mendeleev: Arranged elements by increasing atomic weight.
• Key Insight: Left gaps for undiscovered elements and predicted their properties.
• Significance: His predictions were later proven correct, validating his table.

Slide 4 / Section 3: Moseley’s Refinement (1913)

• Henry Moseley: Discovered that elements are best arranged by increasing atomic number (number of protons).
• Impact: Fixed drawbacks in Mendeleev’s table, leading to the modern version.

Slide 5 / Section 4: The Modem Periodic Table

• Basis: Organised by increasing atomic number.
• Structure: Elements with similar properties fall into the same groups (columns).
• Importance: Essential tool for understanding chemistry and predicting element behaviour.

Elements and Compounds Class 8 Notes

Class 8 Basic Science Elements and Compounds Notes Kerala Syllabus

• When sugar is heated, carbon and water are formed. Sugar molecules can be broken down through chemical reactions.
• Molecules are the smallest particles obtained by dissociating various substances having their fundamental properties.
• Atoms are obtained when molecules are split.
• Dalton’s Atomic theory
  • All matter is made up of minute particles called atoms.
  • An atom is the smallest particle that can participate in a chemical reaction.
  • An atom cannot be divided during a chemical reaction.
  • An atom cannot be created or destroyed.
  • All atoms of an element are of the same size, property and mass.
  • Atoms of different elements differ in their size, property and mass.
• Elements are pure substances made up of the same type of atoms. They cannot be further divided into components through chemical reactions.
• Compounds are pine substances formed by the combination of atoms of two or more elements in a fixed ratio. These can be dissociated into constituent elements through chemical reactions.
• A symbol represents an atom of an element.
• The modern symbol system of representing elements by letters was introduced by a scientist named Berzelius.
• The element hydrogen is represented by the letter ‘H’. That is, the first letter of name of the element is used here as the symbol.
• Some elements are represented by another, letter along with the first letter of the English name in upper case.
• The first letter of the English name is used upper case along with the second letter or any other prominent
  letter in lower case.
• Some elements have symbols based on their Latin names.
• Elements are named on the basis of country, scientists, colour, name of planets, satellites, continents, etc. Example: Francium (Country – France)
  Einsteinum (Scientist – Albert Einstein)
• Lavoisier made the earliest attempts to classify elements. He classified the then-known elements into metals and non-metals.
• Dobereiner classified the elements of similar properties into groups of three. These groups were called triads. He stated that the atomic mass of the element in the middle of the triads was approximately the average of the atomic masses of the other two elements. Only a few elements could be classified in this way.
• Newlands found that when elements were arranged in ascending order of atomic mass, each eighth element was a repetition of the first in its properties. He compared this to the seven notes in music. This classification was known as the Law of Octaves.
• A scientist named Mendeleev arranged the 63 elements known till that day in ascending order of their atomic masses and prepared a table. It was found that elements with common properties were repeated at regular intervals.
• Mendeleev’s periodic law: The physical and chemical properties of elements are the periodic functions of their atomic mass.
• The horizontal rows in the table are called periods and the vertical columns are called groups. Elements in a group show similar chemical properties.
• Advantages of Mendeleev’s periodic table:
  • Elements were arranged in the ascending order of atomic mass.
  • Blank spaces were left for elements to be discovered and their properties were predicted.
  • Elements with similar properties were grouped together to simplify the study of elements and their compounds.
• • Limitations of Mendeleev’s periodic table
  • The order of atomic mass was not strictly followed in the case of all elements. Hydrogen could not be given an exact position.
  • Elements with very different properties had to be placed in the same group.
• Henry Moseley later discovered that the properties of elements depend on their atomic number. The periodic table developed by him is the modern periodic table used today.
• Chemical formula of the molecule represent one molecule of a substance.
• Atomicity is the number of atoms in a molecule.
• In diatomic and polyatomic elements, the atomicity is written as a subscript on the right of the symbol to obtain their chemical formula.
• The total number of molecules is written on the left side of the chemical formula.
• A chemical equation is the representation of a chemical reaction which helps to understand it at a glance. It uses symbols/chemical formulae to represent the reactants and products.

INTRODUCTION

All the objects we see around us are made up of different molecules. Atoms are the smallest particles that are obtained by dividing molecules. Elements are pure substances that contain the same type of atoms. 118 Elements have been discovered so far. Compounds are formed when atoms of elements combine with each other. In the language of chemistry, symbols are used to represent elements, and chemical formulas are used to represent compounds. The periodic table, which has been prepared by classifying the elements comprehensively, makes the study of compounds and elements easier. In this unit, we will learn about the origins of element names, their symbols, chemical formulas of molecules, and the history of element classification.

We know that all objects around us are made up of different molecules. Let us examine how each of these molecules differ with respect to their structural characteristics.

ATOM

• Molecules are the smallest particles obtained by dissociating various substances having their fundamental, properties.
• Atoms are obtained when molecules are split.
• The molecules of every substance are made up of minute atoms. All substances in the universe are made up of atoms. The word “atom” comes from the Greek word “atomos,” which means indivisible.
• Democritus proposed that all matter is made up of indivisible particles called atoms.
  In 1807, John Dalton proposed the atomic theory to explain the nature of atom. Dalton’s Atomic Theory is one of the important theories in modem chemistry. The ideas put forward by Dalton are given below.
  • All matter is made up of minute particles called atoms.
  • An atom is the smallest particle that can participate in a chemical reaction. .
  • An atom cannot be divided during a chemical reaction.
  • An atom cannot be created or destroyed.
  • All atoms of an element are of the same size, property and mass.
  • Atoms of different elements differ in their size, property and mass.

A water molecule is made up of two atoms of hydrogen and one atom of oxygen.

SYMBOLS

• Symbols are used to represent concepts in a clear and simple manner. Symbols make communication and interpretation easier.
• Similarly, signs are used to represent elements. These are known as chemical symbols. A symbol represents an atom of an element.
• The modern symbol system of representing elements by letters was introduced by a scientist named Berzelius.

A symbol represents an atom of an element.

ORIGIN OF SYMBOLS
• The element hydrogen is represented by the letter ‘H’. That is, the first letter of name of the element is used here as the symbol.

HOW ELEMENTS GOT THEIR NAMES
Elements are named on the basis of country, scientists, colour, name of planets, satellites, continents, etc.
Example: Francium (Country – France)
Einsteinum (Scientist – Albert Einstein)

CLASSIFICATION OF ELEMENTS
In a library we can see that books are arranged by categorising them into essays, poems, stories, novels, etc, Elements are also classified in this manner. So far, 118 elements have been discovered which include natural and synthetic elements. It is very difficult to study all these elements and their compounds separately. It would be easier if elements with similar properties are classified into groups.

• The Periodic Table is a table in which all the elements are categorised and arranged based on their properties.

EARLIER ATTEMPTS FOR THE CLASSIFICATION OF ELEMENTS
The modern Periodic Table that we use today is the result of the efforts of many scientists.

• Lavoisier made the earliest attempts to classify elements. He classified the then-known elements into metals and non-metals.
• Dobereiner classified the elements of similar properties into groups of three. These groups were called triads. He stated that the atomic mass of the element in the middle of the triads was approximately the average of the atomic masses of the other two elements. Only a few elements could be classified in this way.

Element	Atomic mass
Li	7
Na	23
K	39
Ca	40
Sr	87.6
Ba	137.3
Cl	35.5
Br	80
I	127

• Newlands found that when elements were arranged in ascending order of atomie mass, each eighth element was a repetition of the first in its properties. He compared this to the seven notes in music. This classification was known as the Law of Octaves. As more elements were discovered, it became clear that all elements could not be arranged in this way.

Li	Be	B	C	N	O	F
Na	Mg	Al	Si	P	S	Cl

• A scientist named Mendeleev arranged the 63 elements known till that day in ascending order of their atomic masses and prepared a table. It was found that elements with common properties were repeated at regular intervals.

The physical and chemifcal properties of elements are the periodic functions of their atomic mass.

MENDELEEV’S PERIODIC LAW
Mendeleev’s Periodic Table is shown in the figure. The horizontal rows in the table are called periods and the vertical columns are called groups. Elements in a group show similar chemical properties.

CHARACTERISTICS

• Elements were arranged in the ascending order of atomic mass. Blank spaces were left for elements to be discovered and their properties were predicted.
• Elements with similar properties were grouped together to simplify the study of elements and their compounds.
• However, the order of atomic mass was not strictly followed in the case of all elements. Hydrogen could not be given an exact position.
• Elements with very different properties had to be placed in the same group.

When Mendeleev’s Periodic Table was formed in 1869, there was limited knowledge about the structure of the atom and its elementary particles. Henry Moseley later discovered that the properties of elements depend on their atomic number. The periodic table developed by him is the modern periodic table used today.

CHEMICAL FORMULA
CHEMICAL FORMULA OF MOLECULES
We use symbols to identify elements easily. If we write N, it refers to nitrogen and one atom of nitrogen.

CHEMICAL FORMULA OF ELEMENTS

• In monoatomic elements, the symbol itself is considered as the chemical formula.
• In monoatomic molecules, the number written on the left of the symbol indicates, the number of molecules. This is also the number of atoms.

Example: 2He → 2 Helium atoms / 2 Helium molecules
3Ne → 3 Neon atoms / 3 Neon molecules
Atomicity is the number of atoms in a molecule. In diatomic Mid polyatomic elements, the atomicity is written as a subscript on the right of the symbol to obtain their chemical formula. The total number of molecules is written on the left side of the chemical formula.
We can find the number of molecules and the number of atoms from the chemical formula of the elements.
3O₂ → 3 oxygen molecules .
Total number of atoms = 3 × 2 = 6 oxygen atoms
5Cl₂ → 5 chlorine molecules
Total number of atoms = 5 × 2 = 10 chlorine atoms

CHEMICAL FORMULA OF COMPOUNDS
Hydrogen and Chlorine are the constituent elements in hydrogen chloride.

CHEMICAL EQUATIONS
Carbon reacts with oxygen to form carbon dioxide.

Reviewing SCERT Class 8 Basic Science Solutions and Kerala Syllabus Class 8 Basic Science Chapter 6 Metals and Non Metals Question Answer Notes Pdf can uncover gaps in understanding.

Class 8 Basic Science Chapter 6 Metals and Non Metals Question Answer Notes

Class 8 Basic Science Chapter 1 Notes Kerala Syllabus Metals and Non Metals Question Answer

Metals and Non Metals Class 8 Questions and Answers Notes

Let’s Assess

Question 1.
Match the following.

Answer:

A	B	C
Name of the metal	Characteristic property	Symbol
Silver	High electrical conductivity	Ag
Platinum	High ductility	Pt
Gold	The highest malleability	Au

Question 2.
The names of certain metals are given in the box.

Choose the appropriate metal for the following situations.
a. Reacts with the components of air and forms tarnish on the surface.
b. It is the best conductor of heat.
c. The metal that melts when placed on the palm.
d. It exists in the liquid state in normal temperature.
e. This metal is present in bones and teeth.
Answer:
a. Copper
b. Silver
c. Gallium
d. Mercury
e. Calcium

Question 3.
Certain metals are given in the following table. Find any two uses of each and the characteristic property responsible for it.

Answer:

Metal	Use	Characteristic property
Copper	Manufacture of electric wire	Electrical conductivity, Ductility
	Manufacture of utensils, Idols	Malleability
Gold	Manufacture of ornaments	Malleability, Ductility
	Manufacture of coins	Metallic lustre, Malleability
Aluminium	Manufacture of utensils for cooking	Thermal conductivity, Malleability
	Manufacture of electrical wires	Electrical conductivity, Ductility
Iron	Manufacture of wires, tools etc.	Hardness
	Used for the manufacture of sheets.	Malleability
Silver	Used for the manufacture of ornaments	Malleability, Ductility, Metallic lustre
	Used for the manufacture of coins, idols, etc.	Metallic lustre, Malleability

Question 4.
Oxides of elements A and B are dissolved in water in two separate beakers.
a) How can you identify the acidic solution and the basic solution?
b) Which of these elements is more likely to be a metal? Why?
Answer:
a) Dip red and blue litmus papers in both solutions.

• The solution which turns blue litmus red is the acidic solution.
• The solution which turns red litmus blue is the basic solution.

b) The solution that turns red litmus into blue is formed by a metal oxide, because metal oxides are basic in nature.
Therefore, the element whose oxide formed the basic solution is more likely to be a metal.

Basic Science Class 8 Chapter 6 Question Answer Kerala Syllabus

Question 1.
You must be familiar with different kitchen utensils. What are the common materials used to make utensils for cooking?
Answer:
Aluminium, Copper, Steel, Iron

Question 2.
Let’s do an activity.
Take some water into a flat bowl. Place a wooden ladle and a steel ladle in it and let it boil. Touch the ladles and find out which one is hotter. Why is it so?
Answer:
A steel ladle heats up more because steel conducts heat.

Question 3.
Why are you warned against handling iron-like objects during thunder and lightning?
Answer:
Metals like iron are good conductors of electricity. Therefore, they are susceptible to lightning strikes.

Question 4.
Find out the characteristics of metals as indicated in the above situations.
Answer:
Thermal conductivity
Electrical conductivity

Question 5.
Complete the table below.

Answer:

Question 6.
What are the other characteristics of metals familiar to you?
Answer:

• Very Hard
• High melting point
• High Boiling point
• High Density

Question 7.
Define metals?
Answer:
Metals are substances that exhibit high thermal and electrical conductivity and possess lustre and hardness.

MALLEABILITY

Have a look at the picture.

Question 8.
What is the blacksmith doing?
Answer:
A hammer strikes the piece of metal.

Question 9.
What changes occur when a piece of metal is hammer blown?
Answer:
The piece of metal flattens into a sheet.

Question 10.
Haven’t you seen aluminium foil which is used to wrap tablets and food materials? Which characteristic property of metals is used in this case?

Malleability

Generally, metals can be hammered and flattened into thin sheets. This characteristic property of metals is known as malleability.

• All metals are not malleable to the same extent.
• Gold is the most malleable metal.

Question 11.
Write examples for utilising the malleability of metals.
Answer:

• Metal sheets are used to make roofing of buildings.
• Used for making thin foils. Aluminium foil for packaging food.
• Used for jewellery making.
• Metal sheets are used to make cupboards and furniture.

DUCTILITY

Question 12.
What metal is the filament of the bulb in the picture made of?
Answer:
Tungsten

Question 13.
Can you explain why tungsten is used to make the filament?
Answer:
Tungsten can be drawn into very thin wires.

Question 14.
Define Ductility
Answer:
Metals can be drawn into thin wires. This property of metals is known as ductility.

Metals like gold and copper can be drawn into very fine wires. Gold and Platinum are the most ductile metals.

Question 15.
There are a lot of situations where ductility of metals is utilized. List a few of them.
Answer:

• Electric wires
• Filaments in light bulbs
• Musical instrument strings
• Jewellery
• Construction wires

Question 16.
Which among these objects made the bulb glow?
Answer:
A piece of razor blade, A fine iron wire, aluminium wire, Copper wire,

Question 17.
Are they metals or non-metals?
Answer:
They are metals
The ability of a material to conduct electricity through it is known as electrical conductivity. All metals are electrical conductors.
Silver is the best electrical conductor among the metals. The electrical conductivity of the metals such as silver, copper, gold, and aluminium decreases in the following order.
Silver > Copper > Gold > Aluminium

Question 18.
Which metal is used in household electrical wiring?
Answer:
Copper

Question 19.
Which metal is used to make electric lines for the public distribution of electricity?
Answer:
Aluminium

Question 20.
Why is silver not used for this purpose?
Answer:
High Cost
Low availability

Question 21.
Utensils for cooking are usually made of metal. Which characteristic properties of metals are utilised in this case?
Answer:
Thermal conductivity
Malleability

Question 22.
Define thermal conductivity.
Answer:
The ability of metals to conduct heat is called thermal conductivity. It is one of the fundamental characteristics of metals.

Silver is the best conductor of heat among metals. Metals such as aluminium and copper have relatively high thermal conductivity.

Question 23.
Aluminium is extensively used for cooking utensils. Find the reasons.
Answer:

• High Thermal Conductivity
• Malleability
• More availability
• Low cost

The non-metal carbon exists in two main allotro ic forms, diamond and graphite

Diamond	Graphite
High thermal conductivity	High electrical conductivity
It is used to cut glass	It is used as electrode

The strikingly different properties of diamond and graphite are due to the differences in the carbon
carbon bonding in them.

Question 24.
Give examples of situations where sonority is useful.
Answer:

• Bells, church bells
• Gongs of bells
• Musical Instruments

Question 25.
Identify the situations in which metallic lustre is made use of?
Answer:

• For making ornaments.
• Mirrors and Reflectors.
• Utensils and Cookware.

Question 26.
Is it possible to cut metals such as copper, aluminium and gold in the same manner?
Answer:
No. Lithium, sodium and potassium are soft metals. But other metals are harder.

Question 27.
Give an example of a hard metal.
Answer:
Iron, Copper, Aluminium, Gold

Question 28.
Give an example of soft metals.
Answer:
Lithium, Sodium, Potassium

Question 29.
Are there metals that exist in a liquid state?
Answer:
Yes

Question 30.
Write the names of metals that exist in a liquid state.
Answer:
Mercury (Hg)
Gallium (Ga)
Caesium (Cs)

• Metals generally have high melting points
  However, gallium and caesium are metals that have low melting points. They start melting when placed on our palm. In other words, they exist in the liquid state on warmer days.
• Metals generally have high boiling points.
• Metals generally have high density.

Question 31.
Take some articles made up of different metals (iron nail, aluminium wire), a small piece of wood, charcoal, thermocol and cork. Put each of them one by one in water in a beaker. Which among these articles sink into the water? Which of these floats over water?
Answer:
Articles sink into the water: An Iron nail, an Aluminium wire, A small piece of wood
Articles float on water: Charcoal. Thermocol, Cork.

Metals are substances that generally exist in the solid state, exhibit high conductivity of heat and electricity, and possess Metallic lustre and hardness.

Question 32.
What are the characteristics of non-metals in comparison with metals?
Answer:
Generally, non-metals are less hard than metals

• They do not show sonority.
• They generally do not conduct electricity.
• They generally show a very low boiling point and Melting point.

Question 33.
List some non metals in your Science Diary. Classify them into solid, liquid and gas.
Answer:

Solid	Liquid	Gas
Boron	Bromine	Hydrogen
Carbon		Nitrogen
Silicon		Oxygen
Germanium		Fluorine
Phosphorous		Chlorine
Arsenic		Iodine
Sulphur		Helium
		Neon
		Argon
		Krypton
		Radon
		Xenon

Question 34.
Observe the newly formed surface of sodium when it is cut with a knife. Record your observation.
Answer:
Good metallic lustre.

Question 35.
Observe the surface after keeping it exposed to air for some time. What change do you observe?
Answer:
Metallic lustre disappears, and the surface becomes dull or tarnished.

Question 36.
Why did it lose its metallic lustre?
Answer:
It is because sodium reacts with the components of air

Question 37.
Haven’t you noticed the dull appearance of the surface of old objects made of aluminium?
Answer:
Most metals react with the atmosphere to form various compounds, and this is the reason for their dull appearance.

Question 38.
Take an aluminium wire and scrub it with sandpaper. What change do you observe?
Answer:
When you scrub an aluminium wire with sandpaper, you observe that the dull surface becomes shiny or lustrous again.
Most metals, such as copper and magnesium, react with air in a similar manner.

Question 39.
Burn a piece of magnesium in air. What is left behind?
Answer:
Magnesium reacts with oxygen in the air to form the compound magnesium oxide.
Magnesium + Oxygen → Magnesium oxide
2Mg + O₂ → 2MgO

Question 40.
a) Dissolve the magnesium oxide formed, in water taken in a beaker. What is the colour change when a red litmus paper is dipped in it?
b) What is the reason?
Answer:
a) Red litmus paper changes to blue.
b) Magnesium oxide dissolves in water to form magnesium hydroxide, which is a base.

• Most of the metals (except silver, gold and platinum) react with oxygen in air to form the corresponding oxide.
• Almost all the metal oxides show a basic nature in the presence of water.
• Some metals react with the components of air like moisture and carbon dioxide in addition to oxygen.
  Oxides of metals are generally basic in nature.

OXIDES OF NON-METALS
Carbon combines with oxygen to form carbon dioxide.
C + O₂ → CO₂

Question 41.
Do you know how soda water is prepared?
Answer:
Soda water is prepared by dissolving carbon dioxide in water at high pressure.
CO₂ + H₂O → H₂CO₃

Question 42.
Dip a blue litmus paper in soda water. Notice the change in colour.
Answer:
When a blue litmus paper is dipped in soda water, the blue litmus paper will turn red.
It is clear that soda water is acidic in nature. Carbonic acid is present in soda water.

Oxides of nonmetals are generally acidic in nature.

Question 43.
What is Metal Corrosion?
Answer:
Metals react with different components of air and form various compounds. This is known as metal corrosion.
Examples:
Rusting of iron
Copper tarnishes.
Metals lose their lustre.

Question 44.
What are the factors that influence the rusting of iron?
Answer:

• Presence of oxygen and moisture.
• Presence of impurities.
• Temperature.
• Presence of salts like common salt accelerates the rusting process of iron.

EXPERIMENT
Materials required

Clean and dry test tubes (4), cork (4), rust-free and shiny iron nails (4), anhydrous calcium chloride, table salt, ordinary water, distilled water, boiled distilled water, and oil.

Procedure
Take four test tubes. Put a rust-free and shiny iron nail in each of them. Add distilled water in the first test tube, salt solution in the second test tube and distilled water soon after boiling in the third test tube. Iron nails should be fully immersed in these three test tubes. Add oil in the third test tube in such a way that the surface of the water is completely covered by oil. Take anhydrous calcium chloride or quicklime in the fourth test tube. All the four test tubes should be closed immediately with cork after putting the materials.

Question 45.
Keep the test tubes at rest for a few days and watch the iron nails afterwards. Indicate the test tubes in which the iron nails got rusted.
Answer:
The iron nails in the first and second test tubes have rusted.
Air and moisture are present in the first test tube.
The salt solution in the second test tube influences the rusting of iron.

Question 46.
The iron nails in the third and fourth test tubes haven’t undergone any change. What is the reason?
Answer:
Since the third test tube contains distilled water soon after boiling and oil floating on the surface of water, the iron nail cannot come in contact with air. Anhydrous calcium chloride or quicklime absorbs the moisture present in the fourth test tube.

Anhydrous calcium chloride absorbs moisture from atmosphere very quickly. Silica gel and quick lime are also used for this purpose.

Question 47.
Why was boiled distilled water used in this experiment?
Answer:
To completely remove air and achieve a high temperature.

Question 48.
Do you know why iron window bars in houses close to the seashore corrode faster?
Answer:
The presence of salt in the atmospheric air causes faster rusting of iron.

Question 49.
Besides iron, which other metals are used in our daily life that undergo corrosion in the presence of atmospheric air?
Answer:

• Copper
• Aluminium
• Silver
• Zinc

Question 50.
Very reactive metals like sodium and potassium are stored in kerosene. Can you find the reason for it?
Answer:
Highly reactive metals quickly undergo chemical reactions in the presence of air and moisture. They are stored in kerosene to avoid contact with air and moisture.

Question 51.
After cutting a lemon, if the iron knife is left without wiping, it starts rusting faster. You are familiar with the acid present in lemon, aren’t you?
Answer:
Iron rusts faster in the presence of acids. Citric acid is present in lemon.

Question 52.
Why are food items with having sour taste, such as pickles and curd, are not stored in vessels made of aluminium and Iron?
Answer:
Sour substances contain acids (pickles – acetic acid, curd – lactic acid). Metal vessels react with these. Therefore, pickles and curd are not stored in metal vessels.

Question 53.
Do all metals similarly undergo corrosion?
Answer:
Metals that react with components in the atmospheric air are susceptible to corrosion. Less reactive metals, such as gold, Platinum, and Silver, do not corrode quickly.

Question 54.
What are the disadvantages of the corrosion of metals

• Causes metals to lose their shine and original appearance, becoming dull or discoloured.
• Environmental Pollution
• Results in economic loss
• Leads to structural failures (bridges, buildings), leaks, and explosions in pipelines
• Mechanical failures in vehicles and aircraft.

Question 55.
What are the methods to prevent corrosion?
Answer:

• Painting
• Greasing/Oiling
• Galvanisation
• Apply Non-Metallic coating
• Convert to alloys

Question 56.
What are alloys?
Answer:
Metals are combined-with other metals or non-metals to make alloys.
It is done to obtain materials with desirable properties and to resist corrosion.
A very small number of non-metals like carbon, phosphorous, etc. are also added to such alloys.
For e.g:-stainless steel, brass, bronze

Question 57.
What are the advantages of alloying metal?
Answer:

• Increased strength and Hardness
• Preventing corrosion
• Improve appearance
• Improved Durability

Question 58.
The presence of certain metals is essential for the normal functioning of the human body. Give examples.
Answer:
Sodium, Potassium, Calcium, Iron, Cobalt and Zinc are examples of such metals.

Question 59.
Name the metal contained in haemoglobin in the blood is?
Answer:
Iron

Question 60.
Name the most abundant metal in the human body?
Answer:
Calcium

• It gives strength to bones and teeth.
• It helps to regulate blood pressure.

As the human body cannot synthesise those metals, they should be made available through food in sufficient quantities.

Class 8 Basic Science Chapter 6 Question Answer Extended Activities

Question 1.
Alloys have a variety of uses. Duralumin, which is used to make the parts of aircrafts and Alnico, which is used to make magnets are examples. Collect information about the alloys and their uses. Present it in your class as slides or charts.
Answer:
Hints

Alloys	Components	Uses
Steel	Fe, C	Used for the manufacture of tools, Building materials, bridges, and railway tracks
Stainless Steel	Fe, Ni, Cr, C	To manufacture household appliances and surgical instruments
Brass	Cu. Zn	To make locks, handles, and musical instruments
Bronze	Cu, Sn	To make medals, statues, musical instruments, etc.
Alnico	Al, Ni, CO, Fe	To make permanent magnets
Duralumin	Al, Cu, Mg, Mn	To make aircraft parts
Magnalium	Mg, Al	To make aircraft parts
Nichrome	Ni, Cr, Fe, C	To make heating coils
Solder	Sn, Pb	To connect electric wires
Ornamental Gold	Au, Zn/Ag	To make jewellery and coins

Question 2.
Metals are present not only in the human body, but in plants as well. Collect information regarding the important metals present in plants and their functions. Prepare a science article based on this.
Answer:
Metals in the plant body

• Magnesium – The main component of chlorophyll Helps in photosynthesis
• Potassium, Calcium – Helps with growth and nutrition
• Cobalt, Iron, Manganese, Copper, Molybdenum, Nickel, Zinc – Plays an important role in growth, enzyme production, and metabolic functions

Metals and Non Metals Class 8 Notes

Class 8 Basic Science Metals and Non Metals Notes Kerala Syllabus

• Metals are substances that generally exist in the solid state. Show high conductivity of heat and electricity, metallic lustre and hardness.
• Generally, metals can be hammered and flattened into thin sheets. This characteristic property of metals is known as malleability.
• Gold is the most malleable metal.
• Metals can be drawn into thin wires. This property of metals is known as Ductility.
• Metals like gold and copper can be drawn into very fine wires. Gold and Platinum are the most ductile metals.
• The ability of a material to conduct electricity through it is known as electrical conductivity. All metals are electrical conductors.
• Silver is the best electrical conductor among the metals.
• The ability of metals to conduct heat, viz., thermal conductivity, is one of the fundamental characteristics of metals.
• Silver is the best conductor of heat among metals.
• The ability of metals to produce characteristic sound when striked with a hard material is known as sonority.
• The shiny appearance of the surface of metals is known as metallic lustre.
• Generally, metals are hard.
• Metals generally have high melting point.
• Metals generally have high boiling points.
• Metals generally have high density.
• Characteristics of Non-Metals
  • Generally, non-metals are less hard.
  • They do not show sonority.
  • They generally do not conduct heat.
• Reactions of metals with atmospheric air
  • Most of the metals (except silver, gold and platinum) react with oxygen in air to form the corresponding oxide. Almost all the metal oxides show basic nature in the presence of water. Some metals react with the components of air like moisture and carbon dioxide in addition to oxygen.
• Oxides of metals are generally basic in nature.
• Oxides of non-metals are generally acidic in nature.
• Metals react with different components of air and form various compounds. This is known as corrosion of metals.
• Metals are combined with other metals or non-metals to make alloys. It is done to obtain materials with
  desirable properties and to resist corrosion. A very small amount of non-metals like carbon, phosphorus, etc. are also added to such alloys.
• Essential metals like sodium, potassium, calcium, iron, cobalt, and zinc are crucial for normal human body functioning, with calcium being the most abundant and vital for bones, teeth, and blood pressure regulation. Since the body cannot synthesise these, they must be obtained in sufficient quantities through diet.

INTRODUCTION

It can be observed that most of the elements discovered so far are metals. The discovery and use of metals greatly accelerated the pace of human progress. Tools, machinery, construction materials, vehicles, bridges, railway tracks, ornaments, and household appliances are just a few of the numerous objects made from metals. Certain characteristics of metals make them beneficial for various uses. This unit will delve into the specific characteristics of metals that make them so versatile and invaluable, covering topics like malleability, ductility, electrical and thermal conductivity, sonority, metallic lustre, and hardness. Metals undergo corrosion due to interaction with the atmosphere. In this unit, we will understand the causes of metal corrosion and methods to prevent it, as well as the properties of non-metals.

ELECTRICAL CONDUCTIVITY
Complete the circuit as shown in the figure

A few materials are given on the table. Put them one by one in the place marked as P, Q. Repeat the experiment using more objects. Record the observation and inference.

Object	Bulb glows/does not glow	Conductor of electricity /not a conductor of electricity
A folded piece of paper	Bulb does not glow	Not a conductor of electricity
A piece of plastic scale /ruler	Bulb does not glow	Not a conductor of electricity
A piece of a razor blade	Bulb glows	Conductor of electricity
A fine iron wire	Bulb glows	Conductor of electricity
Aluminum wire	Bulb glows	Conductor of electricity
Jute	Bulb does not glow	Not a conductor of electricity
Piece of cloth	Bulb does not glow	Not a conductor of electricity
Copper wire	Bulb glows	Conductor of electricity
Rubber band	Bulb does not glow	Not a Conductor of electricity

SONORITY
The ability of metals to produce a characteristic sound when strike with a hard material is known as sonority.

METALLIC LUSTRE
The shiny appearance of the surface of metals is known as metallic lustre.
Certain metals show metallic lustre when they are cut into pieces or polished.

HARDNESS
You know that it is easy to cut sodium metal with a knife.

PROPERTIES OF METAL
• Generally, metals are hard
Metals generally exist in the solid state.