Gujarat BoardĀ GSEB Textbook Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 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.

Solution:

Length of the garden (l) = 90 m

Breadth of the garden (b) = 75 m

Since, the garden along-with the path makes the outer rectangle.

ā“ Length of the outer rectangle

= (90 + 5 + 5) m = 100 m

Breadth of the outer rectangle

= (75 + 5 + 5) m = 85 m

ā“ Area of the outer path = 100 m x 85 m = 8500 mĀ²

Now, area of the path = [Area of outer rectangle] – [Area of the inner rectangle]

= 8500 mĀ² – 6750 mĀ² = 1750 mĀ².

Question 2.

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

Solution:

Area of the inner region = 125 m x 65 m = 8125 mĀ²

Length of outer region = (125 + 3 + 3)m = 131m

Breadth of the outer region = (65 + 3 + 3) m = 71 m

Area of the outer region

= (131 x 71) mĀ² = 9301 mĀ²

Now, Area of the path

= Area of the shaded region

= [Area of the outer region] – [Area of the inner region]

= 9301 mĀ² – 8125 mĀ² = 1176 mĀ²

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:

Here,

Length of the outer rectangle = 8 cm

Breadth of the outer rectangle = 5 cm

ā“ Area of the outer rectangle

= 8 x 5 cmĀ²

= 40 cmĀ²

Width of the margin = 1.5 cm

ā“ For the inner rectangle,

Length = (8 cm – 1.5 cm – 1.5 cm) = 5 cm

Breadth = (5 cm – 1.5 cm – 1.5 cm) = 2 cm

ā“ Area of inner rectangle = 5 x 2 cmĀ² = 10 cmĀ²

ā“ Area of the margin = [Area of the outer rectangle] – [Area of the inner rectangle]

= 40 cmĀ² – 10 cmĀ² = 30 cmĀ²

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 ā¹ 200 per mĀ².

Solution:

(i) Length of the room = 5.5 m

Breadth of the room = 4 m

Area of the room = 5.5 m x 4 m = 22 mĀ²

Now, width of the verandah = 2.25 m

For the outer rectangle:

Length = [5.5 + 2.25 + 2.25] m = 10 m

Breadth = [4 + 2.25 + 2.25] m = 8.5 m

ā“ Area of the outer rectangle = 10 x 8.5 mĀ² = 85 mĀ²

So, area of the verandah = [Area of the outer rectangle] – [Area of the inner rectangle]

= 85 mĀ²- 22 mĀ² = 63 mĀ²

(ii) Rate of cementing = ā¹ 200/mĀ²

ā“ Cost of cementing the floor of the verandah = ā¹ 200 x 63 = ā¹ 12,600.

Thus, the required cost of cementing the floor of the verandah is ā¹ 12,600.

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 ā¹ 40 per mĀ²

Solution:

(i)

Length of the outer square = 30 m

ā“ Area of the outer square = 30 m x 30 m = 900 mĀ²

āµ Width of the path = 1 m

ā“ Side of the inner square = [30 – 1 – 1] m = 28 m

ā“ Area of the inner square = (28 x 28) mĀ² = 784 mĀ²

ā“ Area of the path = [Area of the outer square] – [Area of the inner square]

= 900 mĀ² – 784 mĀ² = 116 mĀ²

(ii) āµ Rate of planting grass = ā¹ 40/mĀ².

ā“ Cost of planting grass in the remaining portion = ā¹ 40 x 784 = ā¹ 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.

Solution:

Length of the rectangular park = 700 m

Breadth of the rectangular park = 300 m

Width of the road = 10 m

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 ā¹ 110 per mĀ².

Solution:

Length of the rectangular field = 90 m

Breadth of the rectangular field = 60 m

Width of each road = 3 m

ā“ Area of ABCD = 90 x 3 mĀ² = 270 mĀ²

Area of EFGH = 60 x 3 mĀ² = 180 mĀ²

Area of PQRS = 3 x 3 mĀ² = 9 mĀ²

(i) ā“ Area of roads = [(Area of ABCD) + (Area of EFGH)] – [Area of PQRS]

= [270 mĀ² + 180 mĀ²] – 9 mĀ²

= 450 mĀ² – 9 mĀ² = 441 mĀ²

(ii) Rate of construction of roads = ā¹ 110/mĀ²

ā“ Cost of construction of roads = ā¹ 110 x 441 = ā¹ 48510

Question 8.

Pragya wrapped a cord 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? (Take Ļ = 3.14)

Solution:

Radius of the circular pipe (r) = 4 cm

ā“ Circumference of the pipe = 2Ļr = 2 x 3.14 x 4 cm = 25.12 cm

ā“ Length of the cord wrapped around the circular pipe = 25.12 cm

Now, the perimeter of the square = 4 x Side = 4 x 4 cm = 16 cm

ā“ Length of the cord wrapped around the square = 16 cm

Since 25.12 cm > 16 cm and (25.12 – 16) cm = 9.12 cm

Thus, we can say, yes, the cord (9.12 cm) is left with Pragya.

Question 9.

The adjoining 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:

Length of the plot = 10 m

Breadth of the plot = 5 m

(i) ā“ Area of the plot (whole land) 5 m = 50 mĀ²

(ii) Radius of the circular flower bed (r) = 2 m

ā“ Area of the flower bed = ĻrĀ²

= 3.14 x 2 x 2 mĀ²

= 12.56 mĀ²

(iii) ā“ Area of the lawn excluding the area of the flower bed = 50 mĀ² – 12.56 mĀ² = 37.44 mĀ²

(iv) Circumference of the circular flower bed = 2Ļr

= 2 x 3.14 x 2 m = 12.56 m

Question 10.

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

Solution:

(i) Area of the whole rectangle ABCD

= 18 x 10 cm = 180 cmĀ²

Area of right āAEF = \(\frac { 1 }{ 2 }\) x Base x Height = \(\frac { 1 }{ 2 }\) x 6 x 10 cm = 30 cmĀ²

Area of right āCBE = \(\frac { 1 }{ 2 }\) x Base x Height = \(\frac { 1 }{ 2 }\) x 8 x 10 cmĀ²

ā“ Area of the shaded portion = [Area of ABCD] – [Area of āAEF + Area of āCBE]

= [180 cmĀ²] – [30 cmĀ² + 40 cmĀ²]

= 180 cmĀ² – 70 cmĀ² = 110 cmĀ²

(ii) Side of the square PQRS = 20 cm

ā“ Area of the square PQRS = Side x Side

= 20 x 20 cmĀ² = 400 cmĀ²

Now, Area of right āQPT = \(\frac { 1 }{ 2 }\) x Base x Height

= \(\frac { 1 }{ 2 }\) x 20 cm x 10 cm = 100 cmĀ²

Area of right āTSU = \(\frac { 1 }{ 2 }\) x Base x Height

= \(\frac { 1 }{ 2 }\) x 10 x 10 cmĀ² = 50 cmĀ²

Area of right āQRU = \(\frac { 1 }{ 2 }\) x Base x Height

= \(\frac { 1 }{ 2 }\) x 10 x 20 cmĀ² = 100 cmĀ²

ā“ Area of the shaded portion = [Area PQRS] – [Area of āQPT + Area of āTSU + Area of āQRU]

= [400 cmĀ²] – [100 cmĀ² + 50 cmĀ² + 100 cmĀ²]

= 400 cmĀ² – 250 cmĀ² = 150 cmĀ²

Question 11.

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

Solution:

Area of āABC = \(\frac { 1 }{ 2 }\) x AC x [Altitude corresponding to AC]

= \(\frac { 1 }{ 2 }\) x 22 x 3 cmĀ² = 33 cmĀ²

Area of āACD =\(\frac { 1 }{ 2 }\) x AC x [Altitude corresponding to AC]

= \(\frac { 1 }{ 2 }\) x 22 x 3 cmĀ²

= 33 cmĀ²

Now, area of quadrilateral ABCD

= Area of āABC + Area of āACD

= 33 cmĀ² + 33 cmĀ² = 66 cmĀ²