Class 7 Maths

Area & Perimeter NCERT Exercise 11.4

Part 2

Question 5: A path of 1m width is built along the border and inside a square garden of side 30m. Find:
(i) the area of the path
(ii) the cost of planting grass in the remaining portion of the garden at the rate of Rs 40 per m2.


Solution: Given, side of the garden = 30m, Width of the path inside the garden = 1m, Cost of the planting of grass = Rs 40 per m2.

Therefore, side of the garden without path = 30m — 1m — 1m = 28m

Area of square = Side2

Area of square garden `= 30^2 = 900` m2

Area of garden without path`= 28^2 = 784` m2

Area of path = Area of garden — Area of garden without path

= 900m2 —784m2 = 116 m2

Cost of planting grass in the remaining portion of garden = Rate × Area

`= Rs. 40 xx 784 = Rs. 31360`

Question 6: Two cross roads, each of width 10 m, cut at right angles through the centre of a rectangular park of length 700 m and breadth 300 m and parallel to its sides. Find the area of the roads. Also find the area of the park excluding cross roads. Give the answer in hectares.


Solution: Let ABCD is a rectangular park. Two roads EFGH and PSRQ is crossing the park at the middle parallel to its side.

Given, Length of the park = 700 m = DC, Width of the park = 300 m = BC, Width of each road = 10m

Therefore, Length of Road PSRQ = Width of the park = 300m, Length of the road EFGH = Length of the park = 700m

Area of the rectangular park = Length `xx` Width

`= 700 xx 300 = 210000` m2 = 21 hectare

Area of Road EFGH = Length `xx` width

`= 70xx10 = 7000` m2

Area of Road PSRQ = Length `xx` width

`= 300xx10= 3000` m2

At the crossing of roads, a square KLMN is formed.

The sides of KLMN = Width of the road = 10m

Hence area of KLMN = Side × Side = 10m × 10m = 100m2

As both roads are overlapping each other, so to calculate the area of shaded portion area of one square should be deducted from the area of one of the road.

Hence, area of cross roads

= Area of one road (PSRQ) + Area of another road(EFGH) — Area of KLMN

= 3000m2 + 7000m2—100m2

= 10000m2 — 100m2 = 9900m2

= 9900 ÷ 10000 = 0.99 hectare

Now, area of park excluding cross roads = Area of park — Areas of cross roads

= 21 hectare —0.99 hectare = 20.01 hectare

Hence, Area of road = 0.99 hectare, and area of park excluding the roads = 20.01 hectare

Quesiton 7: Through a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 m, find
(i) the area covered by the roads.
(ii) the cost of constructing the roads at the rate of Rs 110 per m2.


Solution: Given, Length of the rectangular field = 90m, Width of the rectangular field = 60m, Width of the road = 3m, Cost of construction of road = Rs 110 per m2

Hence; Length of the field without road = 90m —3m = 87m

Width of the field without road = 60m —3m = 57m

Area of rectangular field = Length `xx` Width

`= 90xx60 = 5400` m

Area of rectangular field without road = Length without road `xx` width without road

`= 87xx57 = 4959` m2

Therefore, Area covered by roads

= Area of rectangular field — Area of rectangular field without roads

= 5400m2 — 4959m2 = 441m2

Cost of construction = Rate `xx` Area

`=Rs. 110 xx441 = Rs. 48510`

Question 8: Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any cord left? (π= 3.14)

circle and square

Solution: Given, Radius of pipe = 4cm, Side of square box = 4cm

Here, cord is wrapped around the circular pipe.

Therefore, length of cord = circumference of pipe

Circumference of circle `= 2πr`

`= 2 xx 3.14 xx 4= 6.28 xx 4 = 25.12` cm

Hence, 25.12 cm long wire is required to wrap around the circular pipe.

The same cord is wrapped around the square box.

The length of wire required = The perimeter of square

`= 4 xx \text(side) = 4 xx 4 = 16` cm

Hence, the length of cord left after wrapping over the square box

= Length of cord required to wrap around circular pipe — length required to wrap around the square box

= 25.12cm — 16cm = 9.12 cm