Class 8

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 11 Mensuration Ex 11.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4

1. Given a cylindrical tank, in which situation will you find surface area and in which situation volume:

Question (a)
To find how much it can hold.
Solution:
To find how much a cylindrical tank can hold, we will find its volume.

Question (b)
Number of cement bags required to plaster it.
Solution:
To find number of cement bags required to plaster a cylindrical tank, we will find its surface area.

Question (c)
To find the number of smaller tanks that can be filled with water from it.
Solution:
To find the number of smaller tanks that can be filled with water from it, we will find volume of the tank.

2. Diameter of cylinder A is 7 cm and the height is 14 cm.

Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?
Solution:
Radius of cylinder B is double them that of cylinder A.
∴ Volume of cylinder B should be more than that of cylinder A.

For cylinder A:
radius (r) = \(\frac{diameter}{2}\) = \(\frac{7}{2}\)
height (h) = 14 cm
Volume of cylinder A = πr²h
= \(\frac{22}{7}\) × (\(\frac{22}{7}\))² × 14
= \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 14
= 11 × 7 × 7
= 593 cm³

For cylinder B:

radius (r) = \(\frac{diameter}{2}\) = \(\frac{14}{2}\) = 7cm and
height (h) = 7 cm
Volume of cylinder B = πr²h
= \(\frac{22}{7}\) × 7² × 7
= \(\frac{22}{7}\) × 7 × 7 × 7
= 22 × 7 × 7
= 1078 cm³

Total surface area:
For cylinder A:

radius (r) = \(\frac{7}{2}\) cm
height (h) = 14 cm
Total surface area of cylinder A
= 2πr (r + h)
= 2 × \(\frac{22}{7}\) × \(\frac{7}{2}\)(\(\frac{7}{2}\) +14)
= 22(\(\frac{35}{2}\))
= 11 × 35
= 385 cm²

For cylinder B :
radius (r) = 7 cm, height (h) = 7 cm
Total surface area of cylinder B
= 2πr (r + h)
= 2 × \(\frac {22}{7}\) × 7(7 + 7)
= 44(14) = 616 cm²
So the surface area of cylinder B is greater than that of cylinder A. Hence, the cylinder with greater volume also has greater surface area.

3. Find the height of a cuboid whose base area is 180 cm² and volume is 900 cm³?
Solution:
Let the height of cuboid be h cm.
Now, = Area of base × Height
∴ 900 = 180 × h
∴ h = \(\frac {900}{180}\) = 5cm
Hence, height of cuboid is 5 cm.

4. A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?
Solution:
Volume of a cuboid = 60 × 54 × 30 cm³
Volume of a cube = 6³ = 6 × 6 × 6 cm³
∴ Number of small cubes
= \(\frac{\text { Volume of cuboid }}{\text { Volume of one cube}}\)
= \(\frac{60 \times 54 \times 30}{6 \times 6 \times 6}\)
= 10 × 9 × 5 = 450
Hence, 450 cubes can be placed in the given cuboid.

5. Find the height of the cylinder whose volume is 1.54 m³ and diameter of the base is 140 cm?
Solution:
For given cylinder:
Volume = 1.54 m³
Radius(r) = \(\frac {diameter}{2}\) = \(\frac {140}{2}\) = 70cm = 0.7 m
Volume of cylinder = πr²h
∴ 1.54 = \(\frac {22}{7}\) × 0.7 × 0.7 × h
∴ h = \(\frac{1.54 \times 7}{22 \times 0.7 \times 0.7}\) = 1m
Hence, height of the cylinder is 1 m.

6. A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in litres that can be stored in the tank?
Solution:
For given cylindrical milk tank:
Radius (r) = 1.5 m = \(\frac {15}{10}\) m
Height (h) = 7m
Volume of cylindrical milk tank
= πr²h
= \(\frac {22}{7}\) × (\(\frac {15}{10}\))² × 7
= \(\frac {22}{7}\) × \(\frac {15}{10}\) × \(\frac {15}{10}\) × 7
= \(\frac{11 \times 3 \times 3}{2}=\frac{99}{2}\) = 49.5m³
= 49.5 m³
1 m³ = 1000 litres
∴ 49.5 m³ = 49.5 × 1000 = 49500 litres
Hence, 49,500 litres of milk can be stored in the tank.

7. If each edge of a cube is doubled:

Question (i)
How many times will its surface area increase?
Solution:
Let the edge of the original cube be x.
Then, its new edge (by doubling) = 2x
Original surface area of the cube
= 6 (side)²
= 6 (x)²
= 6x²
New surface area of the cube
= 6 (side)²
= 6 (2x)²
= 6 (2x × 2x)
= 6 (4x²) = 24X²
= \(\frac{\text { New surface area of the cube }}{\text { Original surface area of the cube }}\) = \(\frac{24 x^{2}}{6 x^{2}}\) = 4
Hence, surface area of a cube will increase 4 times.

Question (ii)
How many times will its volume increase ?
Solution:
Original volume of the cube
= (side)³
= (x)³
= x³
New volume of the cube = (side)³
= (2x)³
= (2x × 2x × 2x)
= 8x³
Now,
\(\frac{\text { New volume of the cube }}{\text { Original volume of the cube }}\) = \(\frac{8 x^{3}}{x^{3}}\) = 8
Hence, volume of the cube will increase 8 times.

8. Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m³, find the number of hours it will take to fill the reservoir.
Solution:
Volume of the cuboidal reservoir =108 m³ 1 m³ = 1000 litres
∴ 108 m³ = 108 × 1000 litres
= 1,08,000 litres
Water poured in a minute = 60 litres
∴ Water poured in an hour
(∵ 1 hour = 60 minutes) = 60 × 60
= 3600 litres
Time taken to fill reservoir = \(\frac{\text { volume of reservoir }}{\text { water poured in an hour }}\) = \(\frac {108000}{3600}\)
= 30 hours
Hence, 30 hours it will take to fill the reservoir.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 6 Squares and Square Roots Ex 6.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.2

1. Find the square of the following numbers:

Question (i).
32
Solution:
Let us use: (a + b)² = a² + 2ab + b²
(32)² = (30 + 2)²
= (30)² + 2 (30) (2) + (2)²
= 900 + 120 + 4
= 1024

Question (ii).
35
Solution:
(35)² = (30 + 5)²
= (30)² + 2 (30) (5) + (5)²
= 900 + 300 + 25
= 1200 + 25
= 1225
Second method :
[Note : The unit digit of 35 is 5.]
(35)² = 3 × (3 + 1) × 100 + 25
= 3 × 4 × 100 + 25
= 1200 + 25
= 1225

Question (iii).
86
Solution:
(86)² = (80 + 6)²
= (80)² + 2 (80) (6) + (6)²
= 6400 + 960 + 36
= 7396

Question (iv).
93
Solution:
(93)² = (90 + 3)²
= (90)² + 2 (90) (3) + (3)²
= 8100 + 540 + 9
= 8649

Question (v).
71
Solution:
(71 )² = (70 + 1)²
= (70)² + 2 (70) (1) + (1)²
= 4900 + 140 + 1
= 5041

Question (vi).
46
Solution:
(46)² = (40 + 6)²
= (40)² + 2 (40) (6) + (6)²
= 1600 + 480 + 36
= 2116

2. Write a Pythagorean triplet whose one member is:

Question (i).
6
Solution:
Here, 2n = 6
∴ n = 3

Now, n² – 1 = 3² – 1
= 9 – 1
= 8
and n² + 1 = 3² + 1
= 9 + 1 = 10
Thus, the required Pythagorean triplet is 6, 8, 10.

Question (ii).
14
Solution:
Here, 2n = 14
∴ n = 7
Now, n² – 1 = 7² – 1 = 49 – 1 = 48
and n² + 1 = 7² + 1 = 49 + 1 = 50
Thus, the required Pythagorean triplet is 14, 48, 50.

Question (iii).
16
Solution:
Here, 2n = 16
∴ n = 8
Now, n² – 1 = 8² – 1 = 64 – 1 = 63
and n² + 1 = 8² + 1 = 64 + 1 = 65
Thus, the required Pythagorean triplet is 16, 63, 65.

Question (iv).
18
Solution:
Here, 2n = 18
∴ n = 9
Now n² – 1 = 9² – 1 = 81 – 1 = 80
and n² + 1 = 92 + 1 = 81 + 1 = 82
Thus, the required Pythagorean triplet is 18, 80, 82.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 6 Squares and Square Roots Ex 6.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.1

1. What will be the unit digit of the squares of the following numbers?

Question (i).
81
Solution:
81
Here, the ending digit is 1.
1 × 1 = 1
∴ The unit digit of (81)² will be 1.

Question (ii).
272
Solution:
272
Here, the ending digit is 2.
2 × 2 = 4
∴ The unit digit of (272)² will be 4.

Question (iii).
799
Solution:
799
Here, the ending digit is 9.
9 × 9 = 81
∴ The unit digit of (799)² will be 1.

Question (iv).
3853
Solution:
3853
Here, the ending digit is 3.
3 × 3 = 9
∴ The unit digit of (3853)² will be 9.

Question (v).
1234
Solution:
1234
Here, the ending digit is 4.
4 × 4 = 16
∴ The unit digit of (1234)² will be 6.

Question (vi).
26387
Solution:
26387
Here, the ending digit is 7.
7 × 7 = 49
∴ The unit digit of (26387)² will be 9.

Question (vii).
52698
Solution:
52698
Here, the ending digit is 8.
8 × 8 = 64
∴ The unit digit of (52698)² will be 4.

Question (viii).
99880
Solution:
99880
Here, the ending digit is 0.
0 × 0 = 0
∴ The unit digit of (99880)² will be 0.

Question (ix).
12796
Solution:
12796
Here, the ending digit is 6.
6 × 6 = 36
∴ The unit digit of (12796)² will be 6.

Question (x).
55555
Solution:
55555
Here, the ending digit is 5.
5 × 5 = 25
∴ The unit digit of (55555)² will be 5.

2. The following numbers are obviously not perfect squares. Give reason:

Question (i).
1057
Solution:
1057
Here, the ending digit is 7, which is not one of the digits 0, 1, 4, 5, 6 or 9.
∴ 1057 is not a perfect square.

Question (ii).
23453
Solution:
23453
Here, the ending digit is 3, which is not one of the digits 0, 1, 4, 5, 6 or 9.
∴ 23453 is not a perfect square.

Question (iii).
7928
Solution:
7928
Here, the ending digit is 8, which is not one of the digits 0, 1, 4, 5, 6 or 9.
∴ 7928 is not a perfect square.

Question (iv).
222222
Solution:
222222
Here, the ending digit is 2, which is not one of the digits 0, 1, 4, 5, 6 or 9.
∴ 222222 is not a perfect square.

Question (v).
64000
Solution:
64000
Here, the number of zeros at the end is odd.
∴ 64000 is not a perfect square.

Question (vi).
89722
Solution:
89722
Here, the ending digit is 2, which is not one of the digits 0, 1, 4, 5, 6 or 9.
∴ 89722 is not a perfect square.

Question (vii).
222000
Solution:
222000
Here, the number of zeros at the end is odd.
∴ 222000 is not a perfect square.

Question (viii).
505050
Solution:
505050
Here, the ending digit is 0. (Number of odd zero.)
∴ 505050 is not a perfect square.

3. The squares of which of the following would be odd numbers ?

(i) 431
(ii) 2826
(iii) 7779
(iv) 82004
Solution:
[Note: The square of an odd natural number is odd and that of an even number is an even number.]
(i) 431
This is an odd number.
∴ (431)² is an odd number.

(iii) 7779
This is an odd number.
∴ (7779)² is an odd number.

4. Observe the following pattern and find the missing digits:

Solution :
Observing the above pattern, we can find the missing digits as follows :

5. Observe the following pattern and supply the missing numbers :

Solution:
Observing the above pattern, we can find the missing numbers as follows :

6. Using the given pattern, find the missing numbers:
1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 12² = 13²
4² + 5² + …….² = 21²
5² + ……² + 30² = 31²
6² + 7² + ……² = ……²
[To find pattern:
Third number is related to first and second number. How?
Fourth number is related to third number. How?]
Solution:
(a)² + (a + 1)² + {a (a + 1 )}²
= {a (a + 1) + 1}²
The missing numbers are :
1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 12² = 13²
4² + 5² + 20² = 21²
5² + 6² + 30² = 31²
6² + 7² + 42² = 43²

7. Without adding, find the sum:

Question (i).
1 + 3 + 5 + 7 + 9
Solution:
The sum of first five odd numbers = 5²
= 5 × 5
= 25

Question (ii).
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
Solution:
The sum of first ten odd numbers = 10²
= 10 × 10
= 100

Question (iii).
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Solution:
The sum of first twelve odd numbers = 12²
= 12 × 12
= 144

8.

Question (i).
Express 49 as the sum of 7 odd numbers.
Solution:
49 = 7² = Sum of first seven odd numbers
∴ 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13

Question (ii).
Express 121 as the sum of 11 odd numbers.
Solution :
121 = 11² = Sum of first eleven odd numbers
∴ 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

9. How many numbers lie between squares of the following numbers?
[Note : Between n² and (n + 1)², there are 2n non-square numbers.]

Question (i).
12 and 13
Solution:
Natural numbers between 12² and 13²
= 2 × 12
= 24. (2n, n = 12)

Question (ii).
25 and 26
Solution:
Natural numbers between 25² and 26²
= 2 × 25
= 50. (2n, n = 25)

Question (iii).
99 and 100
Solution:
Natural numbers between 99² and 100²
= 2 × 99
= 198. (2n, n = 99)

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 5 Data Handling Ex 5.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.3

1. List the outcomes you can see in these experiments:

Question (a).
Spinning a wheel

Solution :
Outcomes of spinning the given wheel are :
A, B, C or D.

Question (b).
Tossing two coins together
Solution:
Outcomes in tossing two coins together are : HH, HT, TT, TH
(H = Head, T = Tail)

2. When a die is thrown, list the outcomes of an event of getting

Question (i).
(a) a prime number.
(b) not a prime number.
Solution:
Possible outcomes are: 1, 2, 3, 4, 5 or 6
Prime numbers = 2, 3, 5
(a) Outcomes of getting a prime number are: 2, 3 or 5.
(b) Outcomes of getting not a prime number are : 1, 4 or 6.

(ii) (a) a number greater than 5.
(b) a number not greater than 5.
Solution:
(a) Outcomes of getting a number greater than 5 is 6.
(b) Outcomes of getting a number not greater than 5 are: 1, 2, 3, 4 or 5.

3. Find the:

Question (a).
Probability of the pointer stopping on D in (Question 1 – (a)) ?
Solution:
There are 5 sectors containing A, B, C and D.
Since, there is only 1 sector containing D.
∴ Favourable outcomes = 1
Number of total outcomes = 5
∴ Probability of the pointer stopping on D = \(\frac {1}{5}\)

Question (b).
Probability of getting an ace from a well shuffled deck of 52 playing cards ?
Solution:
Number of possible outcomes = 52
(Total number of cards in a deck = 52)
Since, there are 4 aces in a pack of 52 cards.
∴ Favourable outcomes = 4
∴ Probability of getting an ace = \(\frac{4}{52}=\frac{1}{13}\)

Question (c).
Probability of getting a red apple. (See figure below)

Solution:
Total number of apples = 7
∴ Possible number of ways = 7
Since, there are 4 red apples.
∴ Favourable outcomes = 4
∴ Probability of getting a red apple = \(\frac {4}{7}\)

4. Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box and mixed well. One slip is chosen from the box without looking into it. What is the probability of.

Question (i).
getting a number 6 ?
Solution :
We can get a slip written the number ‘6’ only once.
∴ Number of favourable outcome = 1
Probability of getting the number 6 = \(\frac {1}{10}\)

Question (ii).
getting a number less than 6 ?
Solution:
Numbers less than 6 are 1, 2, 3, 4 and 5. These are five numbers.
∴ Favourable outcomes = 5
Probability of getting the number less than 6 = \(\frac {5}{10}\) = \(\frac {1}{2}\)

Question (iii).
getting a number greater than 6 ?
Solution :
Numbers greater than 6 are 7, 8, 9 and 10. These are four numbers.
∴ Favourable outcomes = 4
Probability of getting the number greater than 6 = \(\frac {4}{10}\) = \(\frac {2}{5}\)

Question (iv).
getting a 1-digit number ?
Solution :
1-digit numbers are 1, 2, 3, 4, 5, 6, 7, 8 and 9. These are nine numbers.
∴ Favourable outcomes = 9
Probability of getting a 1-digit number = \(\frac {9}{10}\)

5. If you have a spinning wheel with 3 green sectors, 1 blue sector and 1 red sector, what is the probability of getting a green sector? What is the probability of getting a non-blue sector?
Solution:
There are 5 sectors in all.
(3 + 1 + 1 = 5)
∴ Number of total possible outcomes = 3
Probability of getting a green colour sector = \(\frac {3}{5}\)
There are 4 non-blue sectors. (5 – 1 = 4)
∴ Number of favourable outcomes = 4
Probability of getting a non-blue sector = \(\frac {4}{5}\)

6. Find the probabilities of the events given in Question 2.
Solution:
There are 6 outcomes.
(i) 2, 3 and 5 are prime numbers. These are three numbers.
∴ Favourable outcomes = 3
Probability of getting a prime number = \(\frac {3}{6}\) = \(\frac {1}{2}\)

(ii) 1, 4 and 6 are non-prime numbers. These are three numbers.
∴ Favourable outcomes = 3
Probability of getting a non-prime number = \(\frac {3}{6}\) = \(\frac {1}{2}\)

(iii) 6 is greater than 5. This is only one number.
∴ Favourable outcome = 1
Probability of a number greater than 5 = \(\frac {1}{6}\)

(iv) 1, 2, 3, 4 and 5 are not greater than 5. There are five numbers.
∴ Favourable outcomes = 5
Probability of a number which is not greater than 5 = \(\frac {5}{6}\)

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 11 Mensuration Ex 11.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3

1. There are two cuboidal boxes as shown in the figures. Which box requires the lesser amount of material to make?

Solution:
For 1st cuboidal box:
length (l) = 60 cm, breadth (b) = 40 cm and height (h) = 50 cm
Total surface area of a cuboid
= 2 (lb + bh + lh)
= 2 [(60 × 40) + (40 × 50) + (50 × 60)]
= 2 (2400 + 2000 + 3000)
= 2(7400)
= 14,800 cm²

For 2nd cuboidal box:
l = b = h = 50 cm
Total surface area of a cuboid
= 2 (lb + bh + lh)
= 2 [(50 × 50) + (50 × 50) + (50 × 50)]
= 2(2500 + 2500 + 2500)
= 2 × 7500
= 15,000 cm²
As, total surface area of cuboid (a) is less, so the box (a) requires the less amount of material to make.

2. A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Solution:
Let us find total surface area of a suitcase (cuboidal in shape).
length (l) = 80 cm, breadth (b) = 48 cm and height (h) = 24 cm
∴ Total surface area of a suitcase = 2 (lb + bh + lh)
= 2 [(80 × 48) + (48 × 24) + (24 × 80)]
= 2(3840 + 1152 + 1920)
= 2 (6912)
= 13,824 cm²
Total surface area of such 100 suitcases = 13,824 × 100
= 13,82,400 cm²
Area of 1 metre trapezium
= length × breadth
= 100 × 96 = 9600 cm²
Tarpaulin required to cover 100 suitcases = \(\frac{1382400}{9600}\) = 144 metres
Hence, 144 m tarpaulin is required to cover 100 suitcases.

3. Find the side of a cube whose surface area is 600 cm².
Solution:
Let the side of the cube be x cm
Total surface area of a cube = 6x²
Total surface area of the cube = 600 cm² (Given)
∴ 6x² = 600
∴ x² = \(\frac {600}{6}\)
∴ x² = 100
∴ x² = 10²
∴ x = 10
Hence, the side of the cube is 10 cm.

4. Rukhsar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet.

Solution:
For given cabinet:
length (l) = 2 m, breadth (b) = 1 m and height (h) = 1.5 m
Total surface area of the cabinet
= 2 (lb + bh + lh)
= 2 [(2 × 1) + (1 × 1.5) + (2 × 1.5)]
= 2 (2 + 1.5 + 3)
= 2(6.5) = 13 m²
She has not painted bottom. So subtract area of bottom from total surface area of the cabinet.
Area of bottom = l × b = 2 × 1 = 2m²
∴ Painted surface area = (13 – 2) m² = 11 m²
Hence, she has painted 11 m² area of a cabinet.

5. Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m² of area is painted. How many cans of paint will she need to paint the room?
Solution:
For given wall:
length (l) = 15 m, breadth (b) = 10 m and height (h) = 7 m
∴ Area to be painted
= Area of 4 walls + Area of ceiling
= [2 (l + b) × h] + l × b
= [2(15 + 10) × 7] + 15 × 10
= [2(25) × 7] + 150
= 350 + 150 = 500 m²
Now, one can of paint covers 100 m² area.
∴ Number of paint cans needed
= \(\frac{\text { Area to be painted }}{\text { Area painted by one can }}=\frac{500 \mathrm{~m}^{2}}{100 \mathrm{~m}^{2}}\)
Hence, 5 cans of paint will be needed.

6. Describe how the two figures given are alike and how they are different. Which box has larger lateral surface area?

Solution:
Here, figure (i) is cylinder and figure (ii) is cube.
Similarity: Both have the same height.
Difference: One is the cylinder and the other is a cube.
For cylinder:
Radius = \(\frac{\text { diameter }}{2}=\frac{7}{2}\) = cm
Height = 7 cm
∴ Lateral (curved) surface area of cylinder
= 2πrh
= 2 × \(\frac{22}{7} \times \frac{7}{2}\) × 7
= 22 × 7 = 154 cm²

For given cube:
side = 7 cm
Lateral surface area of the cube
= 4l² = 4 × 7²
= 4 × 49 = 196 cm²
Hence, cubical box has a larger area.

7. A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?
Solution:
For given cylindrical tank:
radius (r) = 7 m, height (h) = 3 m
Total surface area of tank
= 2πr (r + h)
= 2 × \(\frac {22}{7}\) × 7(7 + 3)
= 44(10) = 440 m²
Hence, 440 m² sheet of metal is required.

8. The lateral surface area of a hollow cylinder is 4224 cm². It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?
Solution:
Lateral surface area of a hollow cylinder = 4224 cm² (Given)
Let length of a rectangular sheet made from hollow cylinder be l cm.
∴ l × b = 4224
∴ 1 × 33 = 4224
∴ l = \(\frac {4224}{33}\) = 128 cm
∴ Length of rectangular sheet =128 cm

Perimeter of rectangular sheet
= 2 (l + b)
= 2 (128 + 33)
= 2 (161) = 322 cm
Hence, perimeter of rectangular sheet is 322 cm.

9. A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road levelled if the diameter of a road roller is 84 cm and length is 1 m.
Solution:
Shape of road roller is cylindrical.

For given cylinder:
Radius (r) = \(\frac{\text { diameter }}{2}\) = \(\frac {84}{2}\) = 42cm
length (height) (h) = 1 m = 100 cm
∴ Curved surface area of a roller
= 2 πrh
= 2 × \(\frac {22}{7}\) × 42 × 100
= 26,400 cm²
Roller covers area of 26,400 cm² in 1 revolution.
∴ Area of covered by roller in 750 revolutions
= 26400 × 750 cm²
= \(\frac{26400 \times 750}{100 \times 100}\) m²
= 1980 m²
Hence, area of road levelled is 1980 m².

10. A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm.

Company places a lable around the surface of the container (as shown in the figure). If the lable is placed 2 cm from top and bottom, what is the area of the label.
Solution:
For given cylindrical container:
radius (r) = \(\frac {diameter}{2}\) = \(\frac {14}{2}\) = 7 cm
height (h) = 20 cm
Now, label is placed 2 cm away from top and bottom.
∴ Height of label = (20 – 2 – 2) cm
= (20 – 4) cm = 16 cm
For label:
radius (r) = 7 cm, height (h) = 16 cm
∴ Area of cylindrical label = 2πrh
= 2 × \(\frac {22}{7}\) × 7 × 16
= 44 × 16
= 704 cm²
Hence, area of the label is 704 cm².

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 5 Data Handling Ex 5.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2

1. A survey was made to find the type of music that a certain group of young people liked in a city. Given pie chart shows the findings of this survey.

From this pie chart answer the following:

Question (i).
If 20 people liked classical music, how many young people were surveyed ?
Solution:
Let the required number of total young people surveyed = x
∴ 10% of x = 20
∴ \(\frac {10}{100}\) × x = 20
∴ x = \(\frac{20 \times 100}{10}\)
∴ x = 200
200 young people were surveyed.

Question (ii).
Which type of music is liked by the maximum number of people ?
Solution:
Maximum number of people like the light music.

Question (iii).
If a cassette company were to make 1000 CD’s, how many of each type would they make ?
Solution:
Total number of CD’s = 1000
(a) Number of CD’s for semi classical music = 20 % of 1000
= \(\frac {20}{100}\) × 1000 = 200
(b) Number of CD’s for classical music = 10 % of 1000
= \(\frac {10}{100}\) × 1000 = 100
(c) Number of CD’s for folk music = 30 % of 1000
= \(\frac {30}{100}\) × 1000 = 300
(d) Number of CD’s for light music = 40 % of 1000
= \(\frac {40}{100}\) × 1000 = 400

2. A group of 360 people were asked to vote for their favourite season from the ? three seasons rainy, winter and summer:

Question (i).
Which season got the most votes ?
Solution:
Winter season got the most votes (150).

Question (ii).
Find the central angle of each sector.
Solution:
Total Votes = 90 + 120 + 150 = 360
∴ Central angle of the sector corresponding to :
Summer season = \(\frac {90}{360}\) × 360° = 90°
Rainy season = \(\frac {120}{360}\) × 360° = 120°
Winter season = \(\frac {150}{360}\) × 360° = 150°

Question (iii).
Draw a pie chart to show this information.
Solution:
Draw three radi in such a way that it makes angles of 90°, 120° and 150° at the centre.

3. Draw a pie chart showing the following information. The table shows the colours preferred by a group of people.

Colours	Number of people
Blue	18
Green	9
Red	6
Yellow	3
Total	36

Solution:
Central angle of the sector corresponding to :
(i) The blue colour = \(\frac {18}{36}\) × 360°
= 18 × 10° = 180°
(ii) The green colour = \(\frac {9}{36}\) × 360° = 90°
(iii) The red colour = \(\frac {6}{36}\) × 360° = 60°
(iv) The yellow colour = \(\frac {3}{36}\) × 360° = 30°
Thus, the required pie chart is given below.

4. The given pie chart gives the marks scored in an examination by a student in Hindi, English, Mathematics, Social Science and Science. If the total marks obtained by the students were 540, answer the following questions.

Question (i).
In which subject did the student score 105 marks ?
[Hint: For 540 marks, the central angle = 360°. So for 105 marks, what is the central angle ?]
Solution:
Toted marks = 540
∴ Central angle corresponding to 540 marks = 360°
∴ Central angle corresponding to 105 marks = \(\frac {360}{540}\) × 105° = 70°
The sector having central angle 70° is corresponding to Hindi.
Thus, the student scored 105 marks in Hindi.

Question (ii).
How many more marks were obtained by the student in Mathematics than in Hindi?
Solution:
The central angle corresponding to the sector of Mathematics = 90°
∴ Marks scored in Mathematics = \(\frac{90^{\circ}}{360^{\circ}}\) × 540 = 135
30 (135 – 105) more marks were obtained by the student in Mathematics than in Hindi.

Question (iii).
Examine whether the sum of the marks obtained in Social Science and Mathematics is more than that in Science and Hindi.
[Hint:Just study the central angles.]
Solution:
The sum of the central angles for Social Science and Mathematics = 65° + 90° = 155°
The sum of the central angles for Science and Hindi = 80° + 70° = 150°
∴ Marks obtained in Social Science and Mathematics is more than the marks scored in Science and Hindi.

5. The number of students in a hostel, speaking different languages is given below. Display the data in a pie chart:

Solution:
Central angle of the sector representing:
(i) Gujarati language = \(\frac {40}{72}\) × 360°
= 40 × 5° = 200°
(ii) English language = \(\frac {12}{72}\) × 360°
= 12 × 5°= 80°
(iii) Urdu language = \(\frac {9}{72}\) × 360°
= 9 × 5° = 45°
(iv) Hindi language = \(\frac {7}{72}\) × 360°
= 7 × 5° = 35°
(v) Sindhi language = \(\frac {4}{72}\) × 360°
= 4 × 5° = 20°
The required pie chart is as follows.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 5 Data Handling InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 5 Data Handling InText Questions

Textual (Textbook Page No. 69 – 70)

1. A Pictograph : Pictorial representation of data using symbols.

Question (i).
How many cars were produced in the month of July ?
Solution:
250 cars were produced in the month of July.

Question (ii).
In which month were maximum number of cars produced?
Solution:
Maximum number of cars were produced in the month of September.

2. A Bar Graph : A display of information using bars of uniform width, their heights being proportional to the respective values.

Question (i).
What is the information given by the bar graph ?
Solution:
Here, the bar graph gives information about number of students of class VIII in different academic years.

Question (ii).
In which year is the increase in the number of students maximum ?
Solution:
In year 2004-05, the increase in the number of students is maximum.

Question (iii).
In which year is the number of students maximum ?
Solution:
In year 2007 – 08, the number of students is maximum.

Question (iv).
State whether true or false:
‘The number of students during 2005 – 06 is twice that of 2003 – 04.’
Solution :
False, the number of students during 2005 – 06 is not twice that of 2003 – 04 but more than twice.

3. Double Bar Graph : A bar graph showing two sets of data simultaneously. It is useful for the comparison of the data.

Question (i).
What is the information given by the double bar graph?
Solution :
Here, the double bar graph provides information about marks obtained by a student in different subjects and comparison of his marks in year 2005 – 06 and 2006 – 07.

Question (ii).
In which subject has the performance improved the most?
Solution :
In the subject Maths, the performance has improved the most.

Question (iii).
In which subject has the performance deteriorated?
Solution :
In the subject English, the performance has deteriorated.

Question (iv).
In which subject is the performance at par?
Solution:
In the subject Hindi, the performance is at par.

Think, Discuss and Write (Textbook Page No. 71)

Question 1.
If we change the position of any of the bars of a bar graph, would it change the information being conveyed? Why?
Solution:
If the height of a bar remains unchanged, then changing of its position does not change the information being conveyed.

Try These (Textbook Page No. 71)

Question 1.
Draw an appropriate graph to represent the given information:

Solution:
To represent the given data by a bar graph, draw two axes perpendicular to each other. Let us represent ‘Months’ on \(\overrightarrow{\mathrm{OX}}\) and ‘Number of watches sold’ \(\overrightarrow{\mathrm{OY}}\) on OY. Let us make rectangles of same width. The heights of the rectangles are proportional to the number of watches, using a suitable scale :
Here, scale is 1 cm = 500 watches
Since, 500 watches = 1 cm
1000 watches = 2 cm
1500 watches = 3 cm
2000 watches = 4 cm
2500 watches = 5 cm

Question 2.

Children who prefer	School A	School B	School C
Walking	40	55	15
Cycling	45	25	35

Solution:
Since, a comparison of two activities is to be represented, therefore a double graph is drawn by taking the ‘Schools’ along X-axis and ‘Number of children’ along Y-axis, using a scale of 1 cm = 5 children.

Question 3.
Percentage wins in ODI by 8 top cricket teams.

Teams	From Champions Trophy to World Cup ’06	Last 10 ODI in ’07
South Africa	75%	78%
Australia	61 %	40%
Sri Lanka	54%	38%
New Zealand	47%	50%
England	46%	50%
Pakistan	45%	44%
West Indies	44%	30%
India	43%	56%

Solution:
To compare the percentage win in ODI achieved by various teams, we represent the data by a double bar graph. We represent the ‘Team’s Names’ along the X-axis and their ‘percentage win’ along Y-axis, using the scale 1 cm – 5%.

Try These (Textbook Page No. 72)

1. A group of students were asked to say which animal they would like most to have as a pet. The results are given below:
dog, cat, cat, fish, cat, rabbit, dog, cat, rabbit, dog, cat, dog, dog, dog, cat, cow, fish, rabbit, dog, cat, dog, cat, cat, dog, rabbit, cat, fish, dog.
Make a frequency distribution table for the same.
Solution:
Using tally-marks, we have

Try These (Textbook Page No. 73-74)

1. Study the following frequency distribution table and answer the questions given below.
Frequency Distribution of Daily Income of 550 Workers of a factory:

Class Interval (Daily Income in ₹)	Frequency (Number of workers)
100 – 125	45
125 – 150	25
150 – 175	55
175 – 200	125
200 – 225	140
225 – 250	55
250 – 275	35
275 – 300	50
300 – 325	20
Total	550

Question (i).
What is the size of the class intervals ?
Solution:
Class size = [Upper class limit] – [Lower class limit]
= 125 – 100
= 25

Question (ii).
Which class has the highest frequency ?
Solution:
The class 200 – 225 is having the highest frequency (140).

Question (iii).
Which class has the lowest frequency ?
Solution :
The class 300 – 325 has the lowest frequency (20).

Question (iv).
What is the upper limit of the class interval 250 – 275?
Solution:
The upper limit of the class interval 250 – 275 is 275.

Question (v).
Which two classes have the same frequency ?
Solution :
The classes 150 – 175 and 225 – 250 are having the same frequency (55).

2. Construct a frequency distribution table for the data on weights (in kg) of 20 students of a class using intervals 30 – 35, 35 – 40 and so on.
40, 38, 33, 48, 60, 53, 31, 46, 34, 36, 49, 41, 55, 49, 65, 42, 44, 47, 38, 39
Solution:
Lowest observation = 31
Highest observation = 65
Class intervals :
30-35, 35-40, 40-45, ……..
The frequency distribution table for above data can be prepared as follows :

Try These (Textbook Page No. 75 – 76)

1. Observe the histogram and answer questions given below:

Question (i).
What information is being given by the histogram ?
Solution:
This histogram represents the heights (in cms) of girls of class VII.

Question (ii).
Which group contains maximum girls ?
Solution:
The group 140- 145 contains maximum number of girls (7).

Question (iii).
How many girls have a height of 145 cms and more?
Solution:
7 girls have a height of 145 cm and more (4 + 2 + 1 = 7).

Question (iv).
If we divide the girls into the following three categories, how many would there be in each?
150 cm and more-Group A
140 cm to less
than 150 cm – Group B
Less than 140 cm – Group C
Solution:
Number of girls in
Group A : 150 cm and more = 2 + 1 = 3
Group B : 140 cm and less than 150 cm = 7 + 4 = 11
Group C : Less them 140 cm = 3 + 2 + 1 = 6

Try These (Textbook Page No. 78)

1. Each of the following pie charts gives you a different piece of information !; about your class. Find the fraction of the circle representing each of these information:

Solution:
(i) Fraction of the circle representing the ‘girls’ 50 % = \(\frac {50}{100}\) = \(\frac {1}{2}\)
Fraction of the circle representing the ‘boys’ 50% = \(\frac {50}{100}\) = \(\frac {1}{2}\)

(ii) Fraction of the circle representing ‘walk’ 40 % = \(\frac {40}{100}\) = \(\frac {2}{5}\)
Fraction of the circle representing ‘bus or car’ 40 % = \(\frac {40}{100}\) = \(\frac {2}{5}\)
Fraction of the circle representing ‘cycle’ 20 % = \(\frac {20}{100}\) = \(\frac {1}{5}\)

(iii) Fraction of the circle representing those who love mathematics = (100 – 15)%
= 85 %
= \(\frac {85}{100}\) = \(\frac {17}{20}\)
Fraction of the circle representing those who hate mathematics = 15%
= \(\frac {15}{100}\) = \(\frac {3}{20}\)

2. Answer the following questions based on the pie chart given:

(i) Which type of programmes are viewed the most?
(ii) Which two types of programmes have number of viewers equal to those watching sports channels ?
Solution:
From the given pie chart,

Type of viewers	Percentage
Sports viewers	25 %
News viewers	15 %
Information viewers	10 %
Entertainment viewers	50%

(i) The entertainment programmes are viewed the most (50 %).
(ii) The news and informative programmes have the number of viewers equal to those watching sports channels (15 % + 10 % = 25 %).

Try These (Textbook Page No. 81)

1. Draw a pie chart of the data given below : The time spent by a child during a day.
Sleep – 8 hours
School – 6 hours
Homework – 4 hours
Play – 4 hours
Others – 2 hours
Solution:
First we find the central angle corresponding to each of the given activities.

Activity	Duration of the activity in a day out of 24 hours	Central angle corresponding to the activity
Sleep	8 hours	\(\frac {8}{24}\) × 360° = 120°
School	6 hours	\(\frac {6}{24}\) × 360° = 90°
Home­ work	4 hours	\(\frac {4}{24}\) × 360° = 60°
Play	4 hours	\(\frac {4}{24}\) × 360° = 60°
Others	2 hours	\(\frac {2}{24}\) × 360° = 30°

The required pie chart is given below.
[Note: Dividing a circle into sectors with corresponding angle (with protractor) you get the required pie chart.]

Think, Discuss and Write (Textbook Page No. 81)

Which form of graph would be appropriate to display the following data.

Question 1.
Production of food grains of a state.

Solution :
A bar graph will be an appropriate representation to display given data.

Question 2.

Favourite food	Number of people
North Indian	30
South Indian	40
Gujarati	25
Others	25
Total	120

Solution :
A pie chart (circle graph) will be an appropriate representation to display given data.

Question 3.
The daily income of a group of a factory workers:

Daily income (in Rupees)	Number of workers (in a factory)
75 – 100	45
100 – 125	35
125 – 150	55
150 – 175	30

Solution :
A histogram will be an appropriate representation to display given data.

Try These (Textbook Page No. 83 – 84)

Question 1.
If you try to start a scooter, what are the possible outcomes ?
Solution:
It may start.
It may not start.

Question 2.
When a die is thrown, what are the six possible outcomes?
Solution:
When a die is thrown, the possible outcomes are: 1, 2, 3, 4, 5 or 6 on the upper face of the die.

Question 3.
When you spin the wheel shown, what are the possible outcomes ? (Fig 5.9)
List them.
(Outcome here means the sector at which the pointer stops.)

Solution:
When we spin the wheel shown, the possible outcomes are A, B or C.
[Note: The sector at which the pointer stops is outcome.]

Question 4.
You have a pot with five identical balls of different colours and you are to pull out (draw) a ball without looking at it; list the outcomes you would get (Fig 5.10).

Solution:
When we draw a ball from a bag with five identical balls of different colours, the possible outcomes are : W, R, B, G or Y.

Think, Discuss and Write (Textbook Page No. 84)

In throwing a die:

1. Does the first player have a greater chance of getting a six?
Solution:
No.

2. Would the player who played after him have a lesser chance of getting a six?
Solution:
No.

3. Suppose the second player got a six. Does it mean that the third player would not have a chance of getting a six ?
Solution:
No.

Try These (Textbook Page No. 86)

Suppose you spin the wheel:

1. ( i ) List the number of outcomes of getting a green sector and not getting a green sector on this wheel.
Solution:
Number of outcomes of getting a green sector = 5
Number of outcomes of not getting a green sector = 3

(ii) Find the probability of getting a green sector.
Solution :
∴ The total number of outcomes = 8
Number of outcomes of getting a green sector = 5
∴ Probability of getting a green sector = \(\frac {5}{8}\)

(iii) Find the probability of not getting a green sector.
Solution:
∴ The total number of outcomes = 8
Number of outcomes of not getting a green sector = 3
∴ Probability of not getting a green sector = \(\frac {3}{8}\)

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 11 Mensuration Ex 11.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2

1. The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.

Solution:
Area of top surface of a table = \(\frac {1}{2}\) × (sum of parallel sides) × perpendicular distance between the parallel sides
= \(\frac {1}{2}\) × (1.2 + 1) × 0.8
= \(\frac {1}{2}\) × 2.2 × 0.8
= 0.88 m²
Hence, area of top surface of table is 0.88 m².

2. The area of a trapezium is 34 cm² and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.
Solution:

Let the length of the other parallel side be xcm.
Area of trapezium
= \(\frac {1}{2}\) × (sum of parallel sides) × height
= \(\frac {1}{2}\) × (10 + x) × 4
= (10 + x) × 2
= 20 + 2x
Area of trapezium = 34 cm² (given)
∴ 20 + 2x = 34
∴ 2x = 34 – 20
∴ 2x = 14
∴ x = 7
Hence, length of the other parallel side is 7 cm.

3. Length of the fence of a trapezium-shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

Solution:
Perimeter of a field = length of fence of a held
∴ AB + BC + CD + DA = 120
∴ AB + 48 + 17 + 40 = 120
∴ AB + 105 = 120
∴ AB = 120 – 105
∴ AB = 15 m
Now, Area of trapezium ABCD = \(\frac {1}{2}\) × (sum of parallel sides)
× perpendicular distance between the parallel sides
= \(\frac {1}{2}\) × (AD + BC) × AB
= \(\frac {1}{2}\) × (40 + 48) × 15
= \(\frac {1}{2}\) × 88 × 15 = 660 m²
Hence, area of the field is 660 m².

4. The diagonal of a quadrilateral shaped field is 24 m and B the perpendiculars dropped on it from c the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

Solution:
Area of quadrilateral ABCD
Area of Δ ABD + Area of Δ BCD
= \(\frac {1}{2}\) × BD × AM + \(\frac {1}{2}\) × BD x CN
= \(\frac {1}{2}\) × 24 × 13 + \(\frac {1}{2}\) × 24 × 8
= 12 × 13 + 12 × 8
= 156 + 96
= 252 m²
Hence, area of the field is 252 m².

5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution:
d₁ = 7.5 cm, d₂ = 12 cm
Area of a rhombus = \(\frac {1}{2}\) × d₁ × d₂
= \(\frac {1}{2}\) × 7.5 × 12
= 7.5 × 6 = 45 cm².
Hence, area of rhombus is 45 cm².

6. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Solution:
Rhombus is a parallelogram too.
∴ Area of rhombus = base × height
= 5 × 4.8
= 5 × \(\frac {48}{10}\)
= 24 cm²
Now, area of rhombus = \(\frac {1}{2}\) × d₁ × d₂
∴ 24 = \(\frac {1}{2}\) × 8 × d₂
∴ 24 = 4 × d₂
∴ d₂ = \(\frac {24}{4}\) = 6 Cm

Hence, length of the other diagonal of rhombus is 6 cm.

7. The floor of a building consists of 3000 tiles which are rhombus-shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m² is ₹4.
Solution:
The shape of a floor tile is rhombus.
d₁ = 45 cm, d₂ = 30 cm
Area of a tile = \(\frac {1}{2}\) × d₁ × d₂
= \(\frac {1}{2}\) × 45 × 30
= 45 × 15
= 675 cm²
Now, floor of a building consists of total 3000 tiles.
∴ Area of floor = Number of tiles × Area of one tile
= 3000 × 675
= 20,25,000 cm²
1cm = \(\frac {1}{100}\)m
∴ 1cm² = \(\frac{1}{100 \times 100}\)m² = \(\frac {1}{10000}\)m²
∴ 20,25,000 cm² = \(\frac{2025000}{10000}\)m²
= 202.5 m²
∴ Cost of polishing the floor = ₹ 4 per m²
∴ Total cost of polishing the floor
= ₹ 4 × 202.5 = ₹ 810
Hence, total cost of polishing the floor is ₹ 810

8. Mohan wants to buy a trapezium-shaped field. Its side along the river is parallel to and twice the sidealong the road. If the area of this field is 10,500 m² and the perpendicular distance between the two parallel sides is 100 m, And the length of the side along the river.

Solution:
The side along the river is parallel to and twice the side along the road.
Let the length of side along the road be x m.
Then, the length of side along the river = 2xm
Area of trapezium
= \(\frac {1}{2}\) × (sum of parallel sides) × height
= \(\frac {1}{2}\) × (x + 2x) × 100
= \(\frac {1}{2}\) × 3x × 100
= 3x × 50 = 150 x
But, area of field = 10500 m² (given)
∴ 150x = 10500
∴ x = \(\frac {10500}{150}\)
∴ x = 70m
∴ 2x = 2 × 70 = 140 m
Hence, the length of the side along the river is 140 m.

9. Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.

Solution:
Let us, divide regular octagon into two congruent trapezium and a rectangle. Then add areas of all these three shapes.
Here, height of trapezium = 4 m Length of two parallel sides = 11 m and 5 m respectively.
∴ Area of a trapezium = \(\frac {1}{2}\) × (11 + 5) × 4
= \(\frac {1}{2}\) × 16 × 4
= 8 × 4
= 32 m²
∴ Area of two trapeziums = 2 × 32
= 64m² …….. (i)
Area of a rectangle = length × breadth
= 11 × 5 = 55m² …….. (ii)
∴ Area of regular octagonal platform
= (64 + 55) m² [From (i) and (ii)]
= 119 m²

10. There is a pentagonal-shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways.

Find the area of this park using both ways. Can you suggest some other way of finding its area?
Solution:
Jyoti has divided given pentagon into two congruent trapeziums.
∴ Area of a trapezium = \(\frac {1}{2}\) × (sum of parallel sides) × perpendicular distance between two parallel sides
= \(\frac {1}{2}\) × (15 + 30) × \(\frac {15}{2}\)
= \(\frac {1}{2}\) × 45 × \(\frac {15}{2}\)
= \(\frac {675}{4}\) m²
∴ Area of two trapeziums = 2 × \(\frac {675}{4}\)
= \(\frac {675}{2}\)
= 337.5 m²
Now, Kavita has divided given pentagon into one triangle and the other square.
∴ Area of a triangle = \(\frac {1}{2}\) × b × h
= \(\frac {1}{2}\) × 15 × 15
= \(\frac {225}{2}\)
= 112.5m² …………….(i)
∴ Area of a square = (side)²
= (15)² = 15 × 15
= 225 m² ……….(ii)
Area of a pentagon
= (112.5 + 225) m² [From (i) and (ii)]
= 337.5 m²
Now, let us use our idea and find another method to find area of a given pentagon. Here, we have divided given pentagon into three triangles.

Area of ∆ a = \(\frac {1}{2}\) × 15 × 15
= \(\frac {225}{2}\)
= 112.5m² … (i)
Area of ∆ b = \(\frac {1}{2}\) × 15 × 15
= \(\frac {225}{2}\)
= 112.5 m² ………… (ii)
Area of ∆ c = \(\frac {1}{2}\) × 15 × 15
= \(\frac {225}{2}\)
= 112.5 m² ……. (iii)
From (i), (ii) and (iii) the area of a given pentagon
= (i) + (ii) + (iii)
= (112.5 + 112.5 + 112.5) m²
= 337.5 m²
Hence, area of the given pentagon is 337.5 m².
[Note : By using such ideas, i.e., dividing any shape into convienient shapes, you can find area of given figure.]

11. Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same.

Solution:
Here, picture frame is divided into four trapeziums, such that a, b, c and d.
The sides of opposite trapeziums have same measurement so they have equal area.
∴ Area of a = area of c and area of b = area of d

For trapezium a and c:
Length of parallel sides = 24 cm, 16 cm
Height = \(\frac{28-20}{2}=\frac{8}{2}\) = 4 cm
∴ Area of trapezium a
= \(\frac {1}{2}\) × (sum of parallel sides) × perpendicular distance between parallel sides
= \(\frac {1}{2}\) × (24 + 16) × 4
= \(\frac {1}{2}\) × 40 × 4
= 80 cm²
So area of trapezium c = 80 cm².

For trapezium b and d:
Length of parallel sides = 28 cm, 20 cm
Height = \(\frac{24-16}{2}=\frac{8}{2}\) = 4 cm
∴ Area of trapezium b
= \(\frac {1}{2}\) × (sum of parallel sides) × perpendicular distance between parallel sides
= \(\frac {1}{2}\) × (28 + 20) × 4
= \(\frac {1}{2}\) × 48 × 4
= 96 cm²
So area of trapezium d = 96 cm².
Hence, area of each section of frame is as follows.
Area of section a = 80 cm²
Area of section b = 96 cm²
Area of section c = 80 cm²
Area of section d = 96 cm²
To find the total surface area, we find the area of each face and then add them.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 11 Mensuration Ex 11.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1

1. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

Question (a)

Solution:
Side of a square field = 60 m
∴ Perimeter of a square field = 4 × side
= 4 × 60 = 240 m
Area of a square field = (side)²
= (60)²
= 60 × 60
= 3600m²

Question (b)

Solution:
Perimeter of a rectangular field = Perimeter of square field
∴ Perimeter of a rectangular held = 240
∴ 2 (length + breadth) = 240
∴ 2 (80 + breadth) = 240
∴ 80 + breadth = \(\frac {240}{2}\)
∴ 80 + breadth =120
∴ breadth = 120 – 80
∴ breadth = 40
Breadth of rectangular field = 40 m
∴ Area of rectangular field = length × breadth
= (80 × 40)
= 3200 m²
Area of square field > Area of rectangular field
Thus, area of square field (a) is larger.

2. Mrs. Kaushik has a square plot with the measurement as shown g in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of ₹ 55 per m².

Solution:
Side of the square plot = 25 m
∴ Area of the square plot = (side)²
= (25 × 25) m²
= 625 m²
In square plot, a rectangular-shaped house is to be constructed.
∴ Area of the constructed house
= length × breadth
= (20 × 15) m²
= 300 m²
∴ Area of the garden = Area of square plot – Area of constructed house
= 625 – 300 = 325 m²
Cost of developing garden of 1 m² is ₹ 55
∴ Cost of developing garden of 325 m²
= ₹ (55 × 325)
= ₹ 17,875
Thus, total cost of developing garden is ₹ 17,875.

3. The shape of a garden is rectangular in the middle and semicircular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 – (3.5+ 3.5) metres].

Solution:
[Note: Here 2 semicircles at the ends of a rectangular garden makes a whole circle. So first find area of a circle and then area of a rectangle. Sum of these two areas is total area. Follow same pattern to find perimeter too. For perimeter of a garden, take only length as rectangle is between two semicircles. Diameter of a circle = Breadth of a rectangle = 7 m]
For semicircle:
∴ Radius = \(\frac{\text { diameter }}{\text { 2 }}\) = \(\frac {7}{2}\)m
Area of circle = πr²
Area of a semicircle = \(\frac {1}{2}\)πr²
∴ Area of 2 semicircles = 2(\(\frac {1}{2}\)πr²)
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)m²
= 38.5 m²
Circumference of two semicircles = 2πr
= 2 × \(\frac {22}{7}\) × \(\frac {7}{2}\)
= 22 m

For rectangle:
length = 20 – (3.5 + 3.5) = 20 – 7 = 13 m
breadth = 7 m
Area of the rectangle = length × breadth
= 13 × 7 = 91 m²
Perimeter of the rectangle
= 2 (length × breadth)
= 2 (13 + 0)
= 2 × 13 = 26 m
∴ Total area of the garden = (38.5 + 91) m²
= 129.5 m²
∴ Perimeter of the garden = (22 + 26) m
= 48 m
Thus, area of the garden is 129.5 m² and the perimeter is 48 m.

4. A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m²? (If required you can split the tiles in whatever way you want to fill up the corners.)
Solution:
[Note : To find number of tiles, divide the area of the floor by area of a tile. Let us do it in a simple way. Unit of floor area and tile area should be same.] Here, tile is parallelogram shaped.
So it’s area = base × corresponding height
Area of a floor = 1080 m²
Base of a tile = 24 cm = \(\frac {24}{100}\) m
Corresponding height of a tile = 10 cm = \(\frac {10}{100}\) m
Number of tiles = \(\frac{\text { Area of a floor }}{\text { Area of a title }}\)
= \(\frac{1080}{\frac{24}{100} \times \frac{10}{100}}\)
= \(\frac{1080 \times 100 \times 100}{24 \times 10}\)
= 45,000
Thus, 45,000 tiles are required to cover the given floor.

5. An ant is moving around a few food pieces of different shapes scattered on the floor. For which food piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression c = 2 πr, where r is the radius of the circle.

Question (a)

Solution:
Here, the shape is semi-circular.
Diameter = 2.8 cm
Radius = \(\frac{\text { Diameter }}{2}=\frac{2.8}{2}\) = 1.4 cm
Circumference of a semicircle = πr
Perimeter of the given figure
= πr + diameter
= (\(\frac {22}{7}\) × 1.4) + 2.8
= 4.4 + 2.8
= 7.2 cm

Question (b)

Solution:
Here, given shape is semicircular at one side, (radius = \(\frac {2.8}{2}\) = 1.4 cm)
So perimeter of semicircular region (circumference) = πr
= \(\frac {22}{7}\) × 1.4
= \(\frac {22}{7}\) × \(\frac {14}{10}\)
= 4.4 cm … (i)
Perimeter of the other portion
= breadth + length + breadth
= (1.5 + 2.8 + 1.5) cm
= 5.8 cm … (ii)
∴ Perimeter of the given figure
= (4.4 + 5.8) cm [from (i) and (ii)]
= 10.2 cm

Question (c)

Solution:
Perimeter of a given part
(semi circular circumference) = πr
= \(\frac {22}{7}\) × 1.4
= 4.4 cm
∴ Perimeter of the given figure
= (4.4 + 2 + 2) cm
= 8.4 cm
Thus, 7.2 cm < 8.4 cm < 10.2 cm.
Thus, the ant would has to take a longer round for food piece (b), as it has a larger perimeter.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 12 Exponents and Powers InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 12 Exponents and Powers InText Questions

Try These : [Textbook Page No. 194]

1. Find the multiplicative inverse of the following:

Question (i)
2^-4
Solution:
Multiplicative inverse of 2^-4 = 2⁴

Question (ii)
10^-5
Solution:
Multiplicative inverse of 10^-5 = 10⁵

Question (iii)
7^-2
Solution:
Multiplicative inverse of 7^-2 = 7²

Question (iv)
5^-3
Solution:
Multiplicative inverse of 5^-3 = 5³

Question (v)
10^-100
Solution:
Multiplicative inverse of 10^-100 =10¹⁰⁰

Try These : [Textbook Page No. 194]

1. Expand the following numbers using exponents:
(i) 1025.63
(ii) 1256.249
Solution:

Number	Expanded form
(i) 1025.63	(1 × 1000) + (0 × 100) + (2 × 10) + (5 × 1) + (6 × \(\frac {1}{10}\)) + (3 × \(\frac {1}{100}\)) OR (1 × 103) + (2 × 101) + (5 × 100) + (6 × 10-1) + (3 × 10-2)
(ii) 1256.249	(1 × 1000) + (2 × 100) + (5 × 10) + (6 × 1) + (2 × \(\frac {1}{10}\))+ (4 x \(\frac {1}{100}\)) + (9 × \(\frac {1}{1000}\)) OR (1 × 103) + (2 × 102) + (5 × 101) + (6 × 100) + (2 × 10-1) + (4 × 10-2) + (9 × 10-3)

Try These : [Textbook Page No. 195]

1. Simplify and write in exponential form:

Question (i)
(- 2)^-3 × (- 2)^-4
Solution:
= (-2)^-3+(-4)
= (-2)^-3-4
= (-2)^-7 or \(\frac{1}{(-2)^{7}}\)

Question (ii)
p³ × p^-10
Solution:
= p^3+(-10)
= p^3-10
= (p)^-7 or \(\frac{1}{(p)^{7}}\)

Question (iii)
3² × 3^-5 × 3⁶
Solution:
= 3^2+(-5)+6
= 3^2-5+6
= 3^2+6-5
= 3³

Try These : [Textbook Page No. 199]

1. Write the following numbers in standard form:

Question (i)
0.000000564
Solution:
= \(\frac{564}{1000000000}\)
(The decimal point is shifted to nine places to the right.)
= \(\frac{5.64}{10^{9}}\)
= \(\frac{5.64}{10^{7}}\)
= 5.64 × 10^-7
∴ 0.000000564 = 5.64 × 10^-7

Question (ii)
0.0000021
Solution:
\(\frac{21}{10000000}\)
= \(\frac{2.1 \times 10}{10000000}\)
= \(\frac{2.1}{10^{6}}\)
= 2.1 × 10^-6
∴ 0.0000021 = 2.1 × 10^-6

Question (iii)
21600000
Solution:
= 216 × 100000
= 216 × 10⁵
= 2.16 × 10² × 10⁵
= 2.16 × 10⁷
∴ 21600000 = 2.16 × 10⁷

Question (iv)
15240000
Solution:
= 1524 × 10000
= 1.524 × 1000 × 10000
= 1.524 × 10³ × 10⁴
= 1.524 × 10⁷
∴ 15240000 = 1.524 × 10⁷

2. Write all the facts given in the standard form. Observe the following facts: [Textbook Page No. 198 ]

Question 1.
The distance from the Earth to the Sun is 150,600,000,000 m.
Solution:
150,600,000,000 m
= 1.506 × 10¹¹ m

Question 2.
The speed of light is 300,000,000 m/sec.
Solution:
300,000,000 m / sec
= 3 × 10⁸m/sec

Question 3.
Thickness of Class VII Mathematics book is 20 mm.
Solution:
20 = 2 × 10¹ mm

Question 4.
The average diameter of a Red Blood Cell is 0.000007 m.
Solution:
0.000007 = 7 × 10^-6 m

Question 5.
The thickness of human hair is in the range of 0.005 cm to 0.01 cm.
Solution:
0.005 = 5 × 10^-3cm and
0.01 = 1 × 10^-2 cm

Question 6.
The distance of moon from the Earth is 384,467,000 m (approx).
Solution:
384,467,000 = 3.84467 × 10⁸ m

Question 7.
The size of a plant cell is 0.00001275 m.
Solution:
0.00001275 = 1.275 × 10^-5m

Question 8.
Average radius of the Sun is 695000 km.
Solution:
695000 = 6.95 × 10⁵ km

Question 9.
Mass of propellant in a space shuttle solid rocket booster is 503600 kg.
Solution:
503600 = 5.036 × 10⁵ kg

Question 10.
Thickness of a piece of paper is 0.0016 cm.
Solution:
0.0016 = 1.6 × 10^-3 cm

Question 11.
Diameter of a wire on a computer chip is 0.000003 m.
Solution:
0.000003 = 3 × 10^-6 cm

Question 12.
The height of Mount Everest is 8848 m.
Solution:
8848 = 8.848 × 10³ m