Page 225 - ICSE Math 6

Example 12: The perimeter of a triangular field is 250 m. If one of its sides is 60 m and the other
side is twice of it, find the length of the third side. Also, find the cost of fencing the
field at the rate of ` 45 per.
Solution: Length of one side of the triangular field = 60 m

\ Length of second side of the triangular field = 2 × 60 m = 120 m

Let the length of third side be ‘x’ m.
Perimeter of triangular field = 250 m
\ 250 = 60 + 120 + x

⇒ x = 250 – 180 = 170

Thus, the length of third side is 170 m.
The cost of fencing 1 m is ` 45.
\ Cost of fencing 250 m = ` 45 × 250 = ` 11,250

Different Shapes with the Same Perimeter

Consider a rectangle with length 10 cm and breadth 5 cm.
Perimeter of the rectangle = 2(l + b) 5 cm

= 2(10 cm + 5 cm)
10 cm
= 2(15 cm)

= 30 cm
3 cm 2 cm
Now, consider the shape given alongside.

Perimeter of the given shape = 2 cm + 10 cm + 8 cm + 7 cm + 3 cm 7 cm
= 30 cm 10 cm

Observe that even when the dimensions are not equal, perimeters of 8 cm
different shapes are equal.

Example 13: The length and breadth of a rectangle are 13 cm and 7 cm respectively. Find the
length of the side of a square whose perimeter is equal to the perimeter of the given
rectangle.
Solution: Perimeter of rectangle = 2(l + b)

= 2(13 cm + 7 cm)
= 2(20 cm) = 40 cm

Let, the length of the side of the square be ‘x’ cm.
Given that, perimeter of rectangle = perimeter of square

\ 40 = 4 × x

40
⇒ x = = 10
4
Thus, the length of the side of the square is 10 cm.

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