Page 198 - Start Up Mathematics_6

Example 8: Find the cost of fencing a rectangular park of length 150 m and breadth 100 m at
the rate of ` 10 per metre.

Solution: Length of fencing = Perimeter of rectangular park

= 2(Length + Breadth)
= 2(150 + 100) m = 2(250) m = 500 m

Rate of fencing = ` 10 per metre

∴ Cost of fencing = ` 500 × 10 = ` 5,000

Example 9: Find the cost of fencing a square park of side 12 m on three of its sides as the wall
of the adjoining house acts as a boundary on one side. The rate of fencing is ` 30
per metre.

Solution: Length of fencing = 3 × Side = 3 × 12 m = 36 m

Since, rate of fencing = ` 30 per metre

∴ Cost of fencing = ` 30 × 36 = ` 1,080
Example 10: Aditya runs around a square park of side 50 m three times. Shweta runs twice around
a rectangular park of length 100 m and breadth 60 m. Who covers more distance?

Solution: Perimeter of square park = 4 × Side

= 4 × 50 m = 200 m
Distance covered by Aditya = 3 × 200 m = 600 m

Perimeter of rectangular park = 2(Length + Breadth)

= 2(100 + 60) m = 2(160) m = 320 m
Distance covered by Shweta = 2 × 320 = 640 m

Clearly, Shweta covers more distance.

Example 11: Find the perimeter of each of the following figures. What can be inferred from the
answers?

5 cm 3 cm
4 cm 5 cm
6 cm
3 cm 6 cm
8 cm
(a) (b) (c)

Solution: (i) Perimeter of Fig. (a) = 8 × side = (8 × 3) cm = 24 cm
(ii) Perimeter of Fig. (b) = 2(5 + 6) + 8 = 30 cm

(iii) Perimeter of Fig. (c) = 2(4 + 4 + 3 + 5) + 6 = 38 cm

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