## NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4

- Class 7 Maths Perimeter and Area Exercise 11.1Â
- Class 7 Maths Perimeter and Area Exercise 11.2
- Class 7 Maths Perimeter and Area Exercise 11.3
- Class 7 Maths Perimeter and Area Exercise 11.4

**NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.4**

Ex 11.4 Class 7 Maths Question 1.

A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.

Solution:

Given: Length = 90 m

Breadth = 75 m

Area of the garden = l Ã— b

= 90 m Ã— 75 m = 6750 m^{2}

Length of the garden including path

= 90m + 5m + 5m = 100 m

Breadth of the garden including path

= 75m + 5m + 5m = 85m

Area of the garden including path

= l Ã— b

= 100 m Ã— 85 m = 8500 m^{2}

Area of the path = 8500 m^{2} – 6750 m^{2} = 1750 m^{2}

Hence, required area of path = 1750 m^{2} and area of the garden = 6750 m^{2} = 0.675 ha

Ex 11.4 Class 7 MathsÂ Question 2.

A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.

Solution:

Length of the park = 125 m

Breadth of the park = 65 m

Area of the park = l Ã— b

= 125 m Ã— 65 m = 8125 m^{2}

Length of the park including path

= 125 m + 3m + 3m = 131 m

Breadth of the park including path

= 65m + 3m + 3m = 71m

Area of the park including path

= 131 m Ã— 71 m = 9301 m^{2}

âˆ´ Area of the path

= 9301 m^{2} – 8125 m^{2} = 1176 m^{2}

Hence, the required area = 1176 m^{2}.

Ex 11.4 Class 7 MathsÂ Question 3.

A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1. 5 cm along each of its sides. Find the total area of the margin.

Solution:

Length = 8 cm, breadth = 5 cm

Area of the cardboard = l Ã— b

= 8 cm Ã— 5 cm = 40 cm^{2}

Width of the margin = 1.5 cm

Length of the inner cardboard

= 8 cm – 1.5 Ã— 2 cm

= 8 cm – 3 cm = 5 cm

Breadth of the inner cardboard

= 5 cm – 1.5 Ã— 2 cm

= 5 cm – 3 cm = 2 cm

Area of the inner rectangle = l Ã— b

= 5 cm Ã— 2 cm = 10 cm^{2} Area of the margin

= 40 cm^{2} – 10 cm^{2} = 30 cm^{2}

Hence, the required area = 30 cm^{2}.

Ex 11.4 Class 7 MathsÂ Question 4.

A verandah of width 2.25 m is constructed all along outside a room which is 5.5 m long and 4 m wide. Find:

(i) the area of the verandah.

(ii) the cost of cementing the floor of the verandah at the rate of â‚¹ 200 per m^{2}.

Solution:

Length of the room = 5.5 m

Breadth of the room = 4 m

âˆ´ Area of the room = l Ã— b = 5.5 m Ã— 4 m = 22 m^{2}

Width of the verandah = 2.25 m

Length of the room including verandah

= 5.5 m + 2 Ã— 2.25 m = 10 m

Breadth of the room including verandah

= 4 m + 2 Ã— 2.25 m = 8.50 m^{2}

Area of the room including verandah = l Ã— b

= 10 m Ã— 8.50 m = 85 m^{2}

(i) Area of the verandah = 85 m^{2} – 22 m^{2}

= 63 m^{2}

(ii) Cost of cementing the floor of the verandah = â‚¹ 63 Ã— 200 = â‚¹12600

Ex 11.4 Class 7 MathsÂ Question 5.

A path 1 m wide is built along the border and inside a square garden of side 30 m. Find:

(i) the area of the path.

(ii) the cost of planting grass in the remaining portion of the garden at the rate of â‚¹ 40 per m^{2}.

Solution:

Area of the square garden = (Side)^{2}

= 30 m Ã— 30 m = 900 m^{2}

Length of the garden excluding the path = 30 m – 2 Ã— 1 m = 28 m

âˆ´ Area of the garden excluding the path = 28 m Ã— 28 m = 784 m^{2}

(i) Area of the path = 900 m^{2} – 784 m^{2}

= 116 m^{2}

(ii) Cost of the planting the remaining portion at the rate of â‚¹ 40 per m^{2}

= â‚¹ 40 Ã— 784 = â‚¹ 31,360

Ex 11.4 Class 7 MathsÂ Question 6.

Two cross roads, each of width 10 m, cut a 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:

Length of the road parallel to the length of the park = 700 m

Width of the road = 10 m

âˆ´ Area of the road = l Ã— b = 700 m Ã— 10 m = 7000 m^{2}

Length of the road parallel to the breadth of the park = 300 m

Width of the road = 10 m Area of this road = l Ã— b = 300 m Ã— 10 = 3000 m^{2}

Area of the both roads

= 7000 m^{2} + 3000 m^{2} – Area of the common portion

= 10,000 m^{2} – 10 m Ã— 10 m

= 10,000 m^{2} – 100 m^{2}

= 9900 m^{2} = 0.99 ha

Area of the park = l Ã— b

= 700 m Ã— 300 m = 210000 m^{2}

Area of the park excluding the roads

= 210000 m^{2} – 9900 m^{2}

= 200100 m^{2} = 20.01 ha

Ex 11.4 Class 7 MathsÂ Question 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 of the rate of â‚¹ 110 per m^{2}.

Solution:

Length of the road along the length of the field = 90 m

Breadth = 3 m

âˆ´ Area of this road = l Ã— b

= 90 m Ã— 3 m = 270 m^{2}

Similarly, the area of the road parallel to the breadth of the field = l Ã— b

= 60 m Ã— 3 m = 180 m^{2} Area of the common portion

= 3m Ã— 3m = 9m^{2}

(i) Area of the two roads

= 270 m^{2} + 180 m^{2} – 9 m^{2}

= 450 m^{2} – 9 m^{2} = 441 m^{2}

(ii) Cost of constructing the roads

= â‚¹ 110 Ã— 441 = â‚¹ 48,510

Ex 11.4 Class 7 MathsÂ Question 8.

Pragya wrapped a card around a circular pipe of radius 4 cm 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? (Ï€ = 3.14)

Solution:

Length of the cord = Circumference of the circular pipe

= 2Ï€r = 2 Ã— 3.14 Ã— 4 = 25.12 cm

Perimeter of the square box

= 4 Ã— side = 4 Ã— 4 cm = 16 cm

Length of the cord left

= 25.12 cm – 16 cm = 9.12 cm

Yes, 9.12 cm cord is left.

Ex 11.4 Class 7 MathsÂ Question 9.

The given figure represents a rectangular lawn with a circular flower bed in the middle. Find:

(i) the area of the whole land.

(ii) the area of the flower bed.

(iii) the area of the lawn excluding the area of the flower bed.

(iv) the circumference of the flower bed.

Solution:

(i) Length of the lawn = 10 m

Breadth of the lawn = 5 m

Area of the lawn = l Ã— b

= 10 m Ã— 5 m = 50 m^{2}

(ii) Area of the circular flower bed = Ï€r^{2}

\(=\frac{22}{7} \times 2 \times 2=\frac{88}{7} \mathrm{m}^{2}=12.57 \mathrm{m}^{2}\)

(iii) Area of the lawn excluding the area of the flower bed

Ex 11.4 Class 7 MathsÂ Question 10.

In the following figures, find the area of the shaded portion.

Solution:

(i) Area of the rectangle = l Ã— b

= 18 cm Ã— (6 cm + 4 cm)

= 18 cm Ã— 10 cm = 180 cm^{2}

Area of right triangle

\(=\frac{1}{2} \times b \times h=\frac{1}{2} \times 6 \times 10=30 \mathrm{cm}^{2}\)

Area of right âˆ†BCE = \(\frac{1}{2}\) Ã— b Ã— h

= \(\frac{1}{2}\) Ã— 8 Ã— 10 =40 cm^{2}

Area of the two right triangles

= 30 cm^{2} + 40 cm^{2} = 70 cm^{2}

Area of the shaded portion

= 180 cm^{2} – 70 cm^{2} = 110 cm^{2}

(ii) Area of the square PQRS = (Side)^{2}

= (20)^{2} = 400 cm^{2}

Area of the three triangles

= 50 cm^{2} + 100 cm^{2} + 100 cm^{2} = 250 cm^{2}

Area of the shaded portion

= 400 cm^{2} – 250 cm^{2} = 150 cm^{2}

Ex 11.4 Class 7 MathsÂ Question 11.

Find the area of the quadrilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm, and BM âŠ¥ AC, DN âŠ¥ AC.

Solution:

Area of the quadrilateral ABCD

= Area of âˆ†ABC + Area of âˆ†ADC

= 33 cm^{2} + 33 cm^{2} = 66 cm^{2}

Hence, the required area = 66 cm^{2}.

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