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Think and Answer

Rubeena’s mother wants to fence the kitchen garden in front
of her house on three sides with lengths 20 m, 12 m, and 12 m.
Find the cost of fencing at the rate of `150 per metre.

Example 7: The length and the breadth of a rectangular garden are 48 m and 23 m, respectively.
There is a path, 1.5 m broad, all around outside the garden. Find the area of the path. Also, find
the cost of paving the path at the rate of `45 per square metre.
Solution: Length of the garden with the path = (48 + 1.5 + 1.5) m = 51 m

Breadth of the garden with the path = (23 + 1.5 + 1.5) m = 26 m
2
Area of the garden with the path = (51 × 26) m = 1326 m 2
2
Area of the garden = (48 × 23) m = 1104 m 2

Thus, the area of the path

= Area of the garden with the path – Area of the garden (23 + 1.5 + 1.5) m = 26 m
= (1326 − 1104) m = 222 m 2
2
Now, the total cost of paving the path = `(222 × 45) = `9990

(48 + 1.5 + 1.5) m = 51 m
Example 8: Two cross roads, each of width 5 m, run at right angles through the centre of a
rectangular park of length 70 m and breadth 45 m and parallel to its sides. Find the area of the
roads. Also find the cost of constructing the roads at the rate of `105 per m .
2
Solution: Area of the cross roads is the area of shaded A P Q B
portion, i.e., the area of the rectangle PQRS and the area
of the rectangle EFGH. But while doing this, the area of
the square KLMN is taken twice, which is to be subtracted. K L
E
Now, PQ = 5 m and PS = 45 m 45 m H N M F
G
EH = 5 m and EF = 70 m

KL = 5 m and KN = 5 m
Area of the roads D S R C
70 m
= Area of the rectangle PQRS + area of the
rectangle EFGH – Area of the square KLMN

= PS × PQ + EF × EH – KL × KN

= (45 × 5 + 70 × 5 − 5 × 5) m 2

2
= (225 + 350 − 25) m = 550 m 2
Cost of constructing the path = `105 × 550 = `57,750

Mathematics-7 284