Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 8 Perimeter and Area Ex 8.2 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 5 Maths Chapter 8 Perimeter and Area Ex 8.2

1. Find the area of following rectangles whose length and breadth are as follows :

Question 1.

9 m and 7 m

Solution:

Length of rectangle = 9 m

Breadth of rectangle = 7 m

Area of rectangle = Length × Breadth

= 9 m × 7 m

= 63 m^{2}

Question 2.

85 cm and 76 cm

Solution:

Length of rectangle = 85 cm

Breadth of rectangle = 76 cm

Area of rectangle = Length × Breadth

= 85 cm × 76 cm

= 6460 cm^{2}

Question 3.

23 mm and 18 mm

Solution:

Length of rectangle = 23 mm

Breadth of rectangle = 18 mm

Area of rectangle = Length × Breadth

= 23 mm × 18 mm

= 414 mm^{2}

Question 4.

5 m and 85 cm

Solution:

Length of rectangle

= 5 m

= 5 × 100 cm

= 500 cm

Breadth of rectangle = 85 cm

Area of rectangle = Length × Breadth

= 500 cm × 85 cm

= 42500 cm^{2}

Question 5.

840 cm and 7 m

Solution:

Length of rectangle = 840 cm

Breadth of rectangle = 7 m

= 7 × 100 cm

= 700 cm

Area of rectangle = Length × Breadth

= 840 cm × 700 cm

= 588000 cm^{2}

2. Find the area of a square whose side is :

Question 1.

25 cm

Sol.

Side of square = 25 cm

Area of square = side × side

= 25 cm × 25 cm

= 625 cm^{2}

Question 2.

48 cm

Solution:

Side of the square = 48 cm

Area of the square = side × side

= 48 cm × 48 cm

= 2304 cm^{2}

Question 3.

27 mm

Solution:

Side of the square = 27 mm

Area of the square = side × side

= 27 mm × 27 mm

= 729 mm^{2}

Question 4.

87 m

Solution:

Side of the square = 87 m

Area of the square = side × side

= 87 m × 87 m

= 7569 m^{2}

Question 3.

Find the area of rectangular park whose length is 62 m and breadth is 38 m.

Solution:

Length of the rectangular park = 62 m

Breadth of the rectangular park = 38 m

Area of the rectangular park = Length × Breadth

= 62 m × 38 m

= 2356 m^{2}

Question 4.

The side of a carrom-board is 60 cm. Find its area.

Solution:

Side of the carrom-board = 60 cm

Area of the carrom-board = side × side

= 60 cm × 60 cm

= 3600 cm^{2}

Question 5.

The length and breadth of a rectangular field is 100 m and 45 m. What is the cost of levelling its floor at the rate of ₹ 8 per sq. m?

Solution:

Length of the rectangular field = 100 m

Breadth of the rectangular field = 45 m

Area of the rectangular field = Length × Breadth

= 100 m × 45 m

= 4500 m^{2}

The rate of levelling = ₹ 8 per m^{2}

The cost of levelling = ₹ 8 × 4500

= ₹ 36000

Question 6.

A carpet has a length 8 m and breadth 5 m. In an auditorium, 125 such carpets are being set on the floor. Find the area of the floor of the auditorium.

Solution:

Length of the carpet = 8 m

Breadth of the carpet = 5 m

Area of each carpet = Length × Breadth

= 8 m × 5 m

= 40 m^{2}

Area of 125 carpets = 125 × 40 m^{2}

= 5000 m^{2}

Therefore, area of the floor of the auditorium = 5000 m^{2}.

Question 7.

The verandah of Gurpreet’s home is 52 m long and 32 m wide and the verandah of Pankaj’s home is of square shape with side 41 m. which person’s home has a roof of verandah bigger and by how much ?

Solution:

Gurpreet :

Length of verandah = 52 m

Breadth of verandah = 32 m

Area of verandah = Length × Breadth

= 52 m × 32 m

=1664 m^{2}

Pankaj :

Side of square verandah = 41 m

Area of square verandah = Side × side

= 41 m × 41 m

= 1681 m^{2}

Area of verandah of Pankaj’s home is bigger by

= 1681 m^{2} – 1664 m^{2}

= 17 m^{2}

Question 8.

Roof of Amarjeet’s home is of length 9 m and breadth 6 m. There is a leakage of water from the roof. He wants to fix tiles of size 30 cm long and 20 cm wide for plugging the leakage. How many tiles does he need ?

Solution:

Length of the roof = 9 m

= 9 ^{2} 100 cm

= 900 cm

Breadth of the roof = 6 m

= 6 ^{2} 100 cm

= 600 cm

Area of the roof = Length × Breadth

= 900 cm × 600 cm

= 540000 cm^{2}

Length of each tile = 30 cm

Breadth of each tile = 20 cm

Area of each tile = Length × Breadth

= 30 cm × 20 cm

= 600 cm^{2}

Area of the roof

Number of tiles = \(\frac{\text { Area of the roof }}{\text { Area of each tile }}\)

= \(\frac{540000}{600}\) = \(\frac{900 \times 600}{600}\)

= 900.

9. Fill in the blanks :

Question 1.

Area of rectangle = ……………… × ………………

Solution:

Length × Breadth

Question 2.

Area of square = ……………… × ………………

Solution:

side × side

Question 3.

1 sq. m. = ……………… sq. cm.

Solution:

10000

Question 4.

The space covered by a closed figure is called its ………………

Solution:

Area.

Question 10.

Complete the table :

Solution:

(a) 56 m^{2}

(b) 2 cm

(c) 6 mm

(d) 700 cm^{2}