These NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4 Questions and Answers are prepared by our highly skilled subject experts.

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

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.

Answer:

Length of the garden (1) = 90 m

Breadth of the garden (b) = 75 m

Area of the garden = 1 x b sq units = 90 m x 75 m = 6750 m^{2}

Length of the outer rectangle (L)

= 90 + 5 + 5 m

= 100 m

Breadth of the outer rectangle (B)

= 75 + 5 + 5 m

= 85 m

Area of the outer rectangle

= L x B = 100 x 85 m^{2}

= 8500 m^{2}

Area of the pathway = Area of the outer rectangle- Area of the inner rectangle

= (8500 – 6750)m^{2}

= 1750m^{2}

(i) Area of the garden = 6750 m^{2}

= \(\frac{6750}{10000}\) ha

= 0.675 ha (1 m^{2} = \(\frac{1}{10000}\) ha)

(ii) Area of the path = 1750 m^{2}

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.

Answer:

Length of the park (1) = 125 m

breadth of the park (b) = 65 m

Area of the park = l x b

= 125 x 65 m^{2}

= 8125m^{2}

Length of the outer region (L)

= (125 + 3 + 3) m

= 131 m

Breadth of the outer region (B)

= (65 + 3+ 3) = 71 m

Area of the outer region

= L x B m^{2}

= (131 x 71) m^{2}

= 9301 m^{2}

Area of the path = Area of the outer region – Area of the park

= 9301 m^{2} – 8125 m^{2}

= 1176 m^{2}

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.

Answer:

Length of the cardboard (l) = 8 cm

Width of the cardboard (b) = 5 cm

Area of the card board = 1 x b sq.

= 8 x 5 = 40 cm^{2}

Width of the margin = 1.5 cm

Length of the inner rectangle

= 8 – (1.5 + 1.5) cm

= 5 cm

Breadth of the inner rectangle

= 5 – (1.5 + 1.5)cm

= 2 cm

Area of the inner rectangle

= 5 x 2 = 10 cm^{2}

Area of the margin = Area of the cardboard – Area of the inner rectangle

= 40 cm^{2} – 10 cm^{2}

= 30 cm^{2}

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 ₹ 200 per m^{2} .

Answer:

Length of the room= 5.5 m

Breadth of the room = 4 m

Area of the room = 5.5 m x 4 m = 22 m^{2}

Width of the verandah = 2.25 m

Length of the verandah

= 5.5 + (2.25 +2.25) m

= 10 m

Breadth of the Verandah

= 4 + (2.25 + 2.25) m

= 8.5 m

Area of the outer rectangle

= 10 m x 8.5 m ‘

= 85 m^{2}

Area of the verandah = Area of the outer rectangle – Area of the inner rectangle

= 85 m^{2} – 22 m^{2}

= 63 m^{2}

Cost of cementing the verandah

= ₹ 200 x 63

= ₹ 12600

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? 40 per m^{2}.

Answer:

(i) Length of the outer square = 30 m

Area of the outer square = side x side

= 30 x 30 = 900 m^{2}

Let width of the path = 1 m

Side of the inner square

= [30 – (1 + 1)3 m

= 30m-2m = 28m

Area of the inner square

= (28 x 28) m^{2}

= 784 m^{2}

Area of the path = Area of the outer square – Area of the inner square

= (900 – 784) m^{2}

Area of the path = 116 m^{2}

(ii) Rate of planting grass

= ₹ 40 per m^{2}

Cost of planting grass = ₹ 40 x 784

= ₹ 31,360

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.

Answer:

Length of the rectangular park (l) = 700 m

Breadth of the rectangular Park (b) = 300 m

Area of the park = l x b sqm

= 700 x 300 m^{2}

= 210000 m^{2}

Area of the road HEFG (length wise)

= 700 m x 10 m

= 7000 m^{2}

Area of the road PQRS (breadth wise)

= 300 x 10 = 3000 m^{2}

Area of KLMN = 10 x 10 = 100 m^{2}

Area of the roads = Area of the road HEFG + Area of the Road PQRS – Area of KLMN (which is repeated two times.)

= (7000 +3000-100) m^{2}

= 9900 m^{2}

Area of the park excluding cross roads = Area of the park – Area of the cross roads

= 21,0000 m^{2} – 9900 m^{2} = 200100 m^{2}

= \(\frac{200100}{10000}\) ha

∴ Area of the park = 20.01ha

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 the rate of? 110 per m2.

Answer:

Length of the rectangular field = 90 m

Breadth of the rectangular field = 60 m

Width of each road = 3m.

Area of the road ABCD = 90 x 3 m^{2}

= 270 m^{2}

Area of the road EFGH = 60 x 3

= 180 m^{2}

Area of PQRS = 3 x 3 = 9 m^{2}

(i) Area covered by the roads = Area of the road ABCD + Area of the road EFGH – Area of PQRS (which is repeated two times)

= (270 + 180 – 9) m^{2}

= 450 m^{2} – 9 m^{2}

= 441 m^{2}

(ii) Rate of construction of roads = ₹ 110/m^{2}

Cost of construction of roads

= ₹ 110 x 441

= ₹ 48510

Question 8.

Pragya wrapped a cord around a circular pipe of radius 4 cm (figure given below) 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)

Answer:

Radius of the circular pipe (r) = 4 cm

Circumference of the pipe = 2πr.

= 2 x 3.14 x 4 cm = 25.12 cm

Side of the square box = 4 cm

Perimeter of the square box

= 4 x 4 cm

= 16 cm

Since, 25.12 cm > 16 cm

Difference in length

= 25.12 cm – 16 cm

= 9.12 cm.

Yes, she has 9.12 cm of length cord left.

Question 9.

The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find: (Take π = 3.14)

(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.

In the following figure, find the area of the shaded portions:

Answer:

Length of the land = 10 m

Breadth of the land = 5 m

(i) Area of the whole land

= (10 x 5) m^{2}

= 50 m^{2}

(ii) Radius of the flower bed (r)

= 2 m.

Area of the flower bed = πr^{2} sq.m

= 3.14 x 2 x 2 m^{2} = 12.56 m^{2}

(iii) Area of the lawn excluding the area of the flower bed = 50 m^{2} – 12.56 m^{2}

= 37.44 m^{2}

(iv) Circumference of the flower bed

= 2πr.

= 2 x 3.14 x 2 m

= 12.56 m

Question 10.

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

Answer:

(i) Area of the whole rectangle ABCD

= 18 x 10 cm^{2} = 180 cm^{2}

Area of the right ΔAEF

= \(\frac { 1 }{ 2 }\) x base x height

= \(\frac { 1 }{ 2 }\) x 6 x 10 cm^{2}

= 30 cm^{2}

Area of the right ΔCBE

= \(\frac { 1 }{ 2 }\) x base x height

= \(\frac { 1 }{ 2 }\) x 8 x 10 cm^{2}

= 40 cm^{2}

Area of the shaded portion = Area of ABCD – (Area of ΔAEF + Area of ΔCBE)

= 180 – (30 + 40)cm^{2} = 180 cm^{2} – 70 cm^{2} = 110 cm^{2}

(ii) Side of the square PQRS = 20 cm

Area of the square PQRS

= Side x Side

= 20 x 20 cm^{2}

= 400 cm^{2}

Area of the right ΔQPT

= \(\frac { 1 }{ 2 }\) x base x height

= \(\frac { 1 }{ 2 }\) x 20 x 10 cm^{2}

= 100 cm^{2}

Area of the rightΔTSU

= \(\frac { 1 }{ 2 }\) x 10 x 10 cm^{2}

= 50 cm^{2}

Area of the right AQRU

= \(\frac { 1 }{ 2 }\) x 10 x 20 cm^{2}

= 100 cm^{2}

Area of the shaded portion = Area of the square PQRS – (Area of ΔQPT + Area of ΔTSU + Area of ΔQRU)

= 400 cm^{2} – (100 + 50 + 100) cm^{2}

= (400 – 250) cm^{2}

= 150 cm^{2}

Question 11.

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

Answer:

Area of ΔABC

= \(\frac { 1 }{ 2 }\) x AC x BM

= \(\frac { 1 }{ 2 }\) x 22 x 3 cm^{2}

= 33 cm^{2}

Area of AACD = \(\frac { 1 }{ 2 }\) x AC x ND

= \(\frac { 1 }{ 2 }\) x 22 x 3 cm^{2}

= 33 cm^{2}

∴ Area of the quadrilateral ABCD = Area of ΔABC + Area of ΔACD

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

= 66 cm^{2}