Math Revision ch-Perimeter and Area
Date: 4/3/21
Extra Questions for Practice
1. Perimeter of a square is 48 cm. Find the length of its side?
2. The length of a rectangular garden is 20 m and breadth is 13 m. How much barbed wire will be needed to fence the garden?
3. Find the area of a rectangle whose length is 12 cm and breadth is 10 cm.
4. A square garden has a perimeter 40m. What is its area?
5. A rectangular field is 24 m long and 20 m wide. Find the cost of fencing the field at the rate of 25 rupees per metre.
6. The fencing of a square garden is 32 m in length. How long is one side of the garden?
7. A thin wire 40 cm long is formed into a rectangle. If the width of this rectangle is 8 cm, what is its length?
8. Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm?
9. Find the side of a square whose perimeter is 20m?
10. The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden?
11. Find the length of a rectangle given that its perimeter is 880 m and breadth is 88 m.
12. A rectangular park is 30 metres long and 20 metres broad. A steel wire fence is put up all around it. Find the cost of putting the fence at the rate of ₹15 per metre.
13. Fencing the compound of a house costs ₹5452. If the rate is ₹94 per metre, find the perimeter of the compound. If the breadth is 10 m, find its length.
Solution :
1. Perimeter = 48 cm
Length of a side = perimeter/4 = 48/4 =12cm
2. Given l = 20 m, b = 13 m
Perimeter = 2 x (l + b) = 2 x (20 + 13) = 2 x 33 = 66 m
3. Given l=12 cm, b= 10 cm
Area = l x b = 12 x 10 = 120 sq cm
4. Given perimeter = 40 m
Length of a side = perimeter/ 4 = 40 / 4 = 10 m
Area = side x side = 10 x 10 = 100 sq m
5. Given l= 24 m, b = 20 m
Perimeter = 2 x (l + b) = 2 x (24 + 20) = 2 x 44 = 88 m
Cost of fencing = 88 x 25 = 2200 rupees.
6. Given perimeter = 32 m
Length of one side = 32/ 4 = 8m
7. Perimeter =2 x (l + b) = 40 cm and b = 8 cm
l + b = 20
l + 8 = 20
l = 20 – 8 = 12cm
8. Perimeter = 10 + 14 + 15 = 39 cm
9. Length of the side of a square = perimeter / 4 = 20/4 = 5 m
10. Given l = 50 m
Area = 300 sq m
Width = Area/ length = 300/ 50 = 6 m
11.Perimeter of the rectangle = 2 [length + breadth]
∴ 2 [length + breadth] = 880
length + breadth = 880 ÷ 2 = 440
∵ Breadth = 88 m
∴ Length = 440 m – 88 m = 352 m
Hence, the required length = 352 m.
12. Length of the rectangular park = 30 m
Breadth = 20 m
∴ Perimeter of the rectangular park = 2(length + breadth)
= 2 [30 + 20] = 2 x 50 m = 100 m
∴ Cost of fencing all around the park = ₹15 x 100 = ₹1500
13. Cost of fencing the compound = ₹5452
and the rate of fencing = ₹94 per metre
∴ Perimeter of the compound = 5452 ÷ 94 = 58 metres
Now breadth of the compound = 10 m.
2 [length + breadth] = 58 m
∴ length + breadth = 58 + 2 m = 29 m
∴ Length of the compound = 29 m – 10 m = 19 m.
