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Example 2: Find the cost of fencing a rectangular plot, 138 m long and 72 m wide at the Example 5: Find the perimeter of a square in which each side is 4 cm. 4 cm
rate of `48 per metre. Solution: Here, side of the square = 4 cm.
Solution: Length of the plot = 138 m and breadth of the plot = 72 m Perimeter = 4 × side = 4 × 4 cm = 16 cm 4 cm

Perimeter of the plot = 2(Length + Breadth)
Be Aware Thus, the perimeter of the square is 16 cm.
= 2(138 + 72) m While calculating the Example 6: Find the side of a square sheet whose perimeter is 32 cm.

= 2 × 2 1 0 m perimeter, the length Solution: Given, perimeter of a square sheet = 32 cm
and breadth must have
= 4 2 0 m the same units. Perimeter of a square = 4 × side

The cost of fencing the plot is `48 per metre. Therefore, 32 cm = 4 × side.
So, the total cost of fencing = `(420 × 48) = `20,160. 32

Hence, the cost of fencing the plot is `20,160. Thus, the side of the square sheet = 4 = 8 cm
Example 3: Suyash jogs 6 rounds around a park every day in the morning. The park is Example 7: If the cost of framing a picture is `3 per cm, then find the
156 m long and 112 m wide. How much distance does he cover? cost of framing a square picture whose side is 32 cm long.

Solution: Length of the park = 156 m and breadth of the park = 112 m Solution: Here, length of the side of a picture = 32 cm.
Perimeter of the park = 2(Length + Breadth) Perimeter of a square = 4 × side

= 2(156 + 112) m = 4 × 32 cm = 128 cm
= 2 × 268 m Now, the cost of framing a picture = `3 per cm.

= 536 m Therefore, the total cost of framing the picture = 128 × `3 = `384.
Distance covered by Suyash in 1 round = 536 m
Think Tank Critical Thinking
Total distance covered by Suyash in 6 rounds = (6 × 536) m = 3216 m
Thus, Suyash covers 3216 m in 6 rounds. 76 m What is the perimeter of your Maths textbook and notebook? Which has a greater
perimeter and by how much?
Example 4: The perimeter of a rectangular field is 182 m and its length
is 76 m. Find its breadth. ? perimeter of a triangle
Solution: Perimeter of the rectangular field = 2(Length + Breadth) In the given figure, ABC is a triangle whose sides are AB, BC and CA. A

182 = 2(76 + Breadth) ( Perimeter = 182 m given) Perimeter of DABC = AB + BC + CA
182 ÷ 2 = 76 + Breadth Perimeter of a triangle = Sum of its sides

91 = 76 + Breadth B C
So, the breadth of the rectangle = 91 m – 76 m = 15 m Example 8: Find the perimeter of a triangle whose sides are 4 cm, 7 cm and 8 cm,
Thus, the breadth of the rectangular field is 15 m. respectively.
perimeter of a Square Solution: Perimeter of the triangle = Sum of all its sides
= (4 cm + 7 cm + 8 cm) 7 cm 4 cm
We know that a square is a 4-sided polygon in which all four sides are of equal length. = 19 cm
In the given figure, ABCD is a square. 8 cm
A D Example 9: Find the perimeter of an isosceles triangle in which each
So, perimeter of a square = AB + BC + CD + DA of the equal sides is 6 cm and the third side is 5 cm.
= AB + AB + AB + AB (Since AB = BC = CD = DA) Solution: Perimeter of the given triangle = Sum of all its sides 6 cm 6 cm

= 4 × A B = (6 cm + 6 cm + 5 cm)
Hence, perimeter of a square = 4 × Side B C = 1 7 c m . 5 cm

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