Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 8 Perimeter and Area Ex 8.1 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 5 Maths Chapter 8 Perimeter and Area Ex 8.1

1. Find the perimeter :

Question 1.

Solution:

Perimeter of a rectangle = Length + Breadth + Length + Breadth

= 2 (Length + Breadth)

= 2 (8 m + 3 m)

= 2 × 11 m

= 22 m.

Question 2.

Solution:

Perimeter of a square = side + side + side + side

= 4 × side

= 4 × 5 cm

= 20 cm.

2. Find the perimeter of the rectangle whose length and breadth are as follows :

Question 1.

3 cm, 2 cm

Solution:

Length of the rectangle = 3 cm

Breadth of the rectangle = 2 cm

Perimeter of the rectangle = 2 (Length + Breadth)

= 2 (3 cm + 2 cm)

= 2 × 5 cm

= 10 cm.

Question 2.

12 m, 10 m

Solution:

Length of the rectangle = 12 m

Breadth of the rectangle = 10 m

Perimeter of the rectangle = 2 (Length + Breadth)

= 2 (12 m + 10 m)

= 2 × 22 m

= 44 m.

Question 3.

15 cm, 8 cm.

Solution:

Length of the rectangle = 15 cm

Breadth of the rectangle = 8 cm

Perimeter of the rectangle = 2 (Length + Breadth)

= 2 (15 cm +8 cm)

= 2 × 23 cm

= 46 cm.

3. Find the perimeter of the square, whose side is :

Question 1.

4 cm

Solution:

Side of the square = 4 cm

Perimeter of the square = 4 × side

= 4 × 4 cm

= 16 cm.

Question 2.

8 cm

Solution:

Side of the square = 8 cm

Perimeter of the square = 4 × side

= 4 × 8 cm

= 32 cm.

Question 3.

10 cm

Solution:

Side of the square = 10 cm

Perimeter of the square = 4 × side

= 4 × 10 cm

= 40 cm.

Question 4.

72 mm

Solution:

Side of the square = 72 mm

Perimeter of the square = 4 × side

= 4 × 72 mm

= 288 mm.

4. Find the side of the square whose perimeter is :

Question 1.

48 cm

Solution:

Perimeter of the square = 48 cm

Side of the square = \(\frac{\text { Perimeter }}{4}\)

= \(\frac{48}{4}\) cm

= 12 cm

Question 2.

80 m

Solution:

Side of the square = \(\frac{\text { Perimeter }}{4}\)

= \(\frac{80}{4}\) cm

= 20 cm

Question 3.

24 m

Solution:

Side of the square = \(\frac{\text { Perimeter }}{4}\)

= \(\frac{24}{4}\) cm

= 6 cm

Question 5.

The length and breadth of a rectangular park is 96 m and 64 m respectively. Find the length of wire which can fence it all around.

Solution:

Length of rectangular park = 96 m

Breadth of rectangular park = 64 m

Perimeter of rectangular park = 2 (Length + Breadth)

= 2 (96 m + 64 m)

= 2 × 160 m

= 320 m.

∴ Length of wire which can fence it all round = 320 m.

Question 6.

The perimeter of the rectangular park is 84 m. Find its breadth if length is 24 m.

Solution:

Perimeter of the rectangular park = 84 m

Length of the rectangular park = 24 m

Breadth of the rectangular park = \(\frac{\text { Perimeter }}{2}\) – length

= \(\frac{84 m}{2}\) – 24 m

= 42 m – 24 m

= 18 m

Question 7.

A player runs around a square track of side 50 m. How many rounds will he take to complete the race of 2000 m?

Solution:

Side of the square track = 50 m

Perimeter of the square track = 4 × side

= 4 × 50 m

= 200 m

Total distance of race = 2000 m

Number of rounds = \(\frac{2000}{200}\)

= 10

8. Fill in the blanks :

Question 1.

Perimeter of rectangle = 2 × (length + ………………)

Solution:

Breadth

Question 2.

Perimeter of square = …………… × side

Solution:

Question 3.

The perimeter of a closed figure, made of line segments, is ……………… of lengths of its all sides.

Solution:

sum.