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

Rajasthan Board RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4 Textbook Exercise Questions and Answers.

Rajasthan Board RBSE Solutions for Class 7 Maths in Hindi Medium & English Medium are part of RBSE Solutions for Class 7. Students can also read RBSE Class 7 Maths Important Questions for exam preparation. Students can also go through RBSE Class 7 Maths Notes to understand and remember the concepts easily. Students can access the data handling class 7 extra questions with answers and get deep explanations provided by our experts.

RBSE Class 7 Maths Solutions 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.
Answer:
Length of garden = 90 m
Breadth of garden = 75 m
Area of garden = l × b
= 90 m × 75 m
= 6750 m2
RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4 1

Length of garden including path = (90 + 5 + 5) m = 100 m
Breadth of garden including path = (75 + 5 + 5) m = 85 m
Area of garden including path = l × b = (100 m × 85 m) = 8500 m2
Area of path = Area of garden including path - Area of garden
= 8500 m2 - 6750 m2
= 1750 m2
∵ 1 hectare = 10000 m2
Area of garden
∴ 6750 m2 = \(\frac{6750}{10000}\) hectare
= 0.675 hectare
So, the area of path is 1750 m2 and the area of garden is 0.675 hectare.

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

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 rectangular park = 125 m
Breadth of rectangular park = 65 m
Area of rectangular park = l × b
= 125 m × 65 m
= 8125 m2
RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4 2
Length of park including path
= (125 + 3 + 3) m = 131m
Breadth of park including path = (65 + 3 + 3) m = 71 m
Area of park including path
= l × b = 131 m × 71 m = 9301 m2
Area of path = Area of park including path - Area of park
= 9301 m2 - 8125 m2
= 1176 m2
So, the area of path is 1176 m2.

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 cardboard without margin = 8 cm
Breadth of cardboard with margin = 5 cm
Area of cardboard = l × b
= 8 cm × 5 cm = 40 cm2
Length of cardboard = (8 - 1.5 - 1.5) cm = 5 cm
Breadth of cardboard = (5 - 1.5 - 1.5) cm = 2 cm
Area of cardboard with margin
= l × b = 5cm × 2 cm = 10 cm2
RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4 3
Area of margin = Area of cardboard with margin - Area of cardboard
= 40 cm2 - 10 cm2 = 3 0 cm2
So, the total area of margin is 30 cm2.

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 m2.
Answer:
Length of room= 5.5 m
Breadth of room = 4 m
Area of room = l × b
= 4 m × 5.5 m = 22 m2
RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4 4
Length of room with verandah
= (5.5 + 2.25 + 2.25) = 10 m
Breadth of room with verandah
= (4 + 2.25 + 2.25) m = 8.5 m
Area of room with verandah
= l × b = 10 m × 8.5 m = 85 m2

(i) Area of verandah = Area of room with verandah - Area of room
= 85 m2 - 22 m2
= 63 m2

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

(ii) Cost of cementing 1 m2 verandah floor = ₹ 200
Cost of cementing 63 m2 verandah floor = ₹ 200 × 63 = ₹ 12,600
So, the cost of cementing the verandah floor is ₹ 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 m2.
Answer:
(i) Side of square garden = 30 m
Area of square garden = (Side)2
= (30 m)2 = 900 m2
Side of square garden excluding path = (30 - 1 - 1) m = 28 m
Area = (Side)2 = (28 m)2 = 784 m2
∴ Area of path = 900 m2 - 784 m2 = 116 m2
RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4 5

(ii) Cost of planting grass in 1 m2 = ₹ 40
Cost of planting grass in 784 m2 = ₹ 784 × 40 = ₹ 313,60.

Question 6.
Two crossroads, 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 crossroads. Give the answer in hectare.
Answer:
RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4 6

For I, Rectangle b = 10 m
l = 700
Area = l × b
= 10 m × 700 m
= 7000 m2

For II l = 10 m
b = 300 m
A = l × b = 10 m × 300 m
= 3000 m2

For III
Side of square = 10 m
Area = (Side)2
= (10 m)2 = 100 m2
Area of crossroads = Area of I + Area of II - Area of III
= 7000 m2 + 3000 m2 - 100 m2
= 9900 m2
∵ Area of park = l × b = 700 m × 300 m
= 210000 m2
∴ Area of park excluding crossroads
= 210000 - 9900
= 200100 m2
= \(\frac{2000100}{10000}\) hectare
= 20.01 hectare.

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

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:
RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4 7

For I l = 90 m
b = 3 m
A = l × b = 90 × 3 = 270 m2

For II l = 3 m
b = 60 m
A = l × b = 180 m2

For III
Side of square = 3 m
Area = (Side)2 = (3 m)2 = 9 m2

(i) Area covered by roads = Area of I + Area of II - Area of III
= (270 + 180 - 9) m2
= 441 m2

(ii) Cost of constructing 1 m2 = ₹ 110
Cost of constructing 441 m2 = ₹ 441 × 110
= ₹ 48,510.

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 ? (Take π = 3.14)
RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4 8
Answer:
Radius of circular pipe = 4 cm
Circumference of circular pipe = 2πr
= 2 × 3.14 × 4 = 25.12 cm
Length of each side of square box = 4 cm
Perimeter = 4 × side = 4 × 4 cm = 16 cm
∵ Circumference of circular pipe > Perimeter of square box
She will have (25.12 - 16) cm = 9.12 cm cord left with her.

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

Question 9.
The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find :
RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4 9
(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.
Answer:
(i) Length of lawn = 10 m
Breadth of lawn = 5 m
Area of lawn = l × b
= 10 m × 5 m = 50 m2

(ii) Radius of flower bed = 2 m
Area of flower bed = πr2
= 3.14 × (2)2
= 3.14 × 4
= 12.56 m2

(iii) Area of lawn excluding the area of flower bed = (50 - 12.56) m2 = 37.44 m2

(iv) crcumference of flower bed
= 2πr = 2 × 3.14 × 2 = 12.56 m

Question 10.
In the following figures, find the area of the shaded portions:
RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4 10
Answer:
(i) Length of rectangle = 18 cm
Breadth of rectangle = 10 cm
Area of AB CD = l × b
= 180 cm2
For triangle AEF Area = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 10 × 6
= 30 cm2
For triangle CBE Area = \(\frac{1}{2}\) × 8 × 10
= 40 cm2
Area of shaded portion = (Area of rectangle) - (Area of triangle AEF + Area of triangle CBE)
= 180 cm2 - (30 cm2 + 40 cm2)
= 110 cm2.

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

(ii) Area of square = (Side)2
= (20 cm)2 = 400 cm2
For ∆QPT Area = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 10 × 20 = 100 cm2
For ∆TSU Area = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 10 × 10 = 50 cm2
For ∆QRU Area = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 10 × 20 = 100 cm2
Area of shaded portion = Area of square - (Area of ∆QPT + Area of ∆TSU + Area of ∆QRU)
= 400 cm2 - (100 + 50 + 100) cm2
= 400 - 250 cm2
= 150 cm2

Question 11.
Find the area of the quadrilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN - 3 cm, and BM ⊥ AC, DN ⊥ AC.
RBSE Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4 11
Answer:
Area of ∆ABC = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × AC × BM
= \(\frac{1}{2}\) × 22 × 3 cm2
= 33 cm2
Area of ∆ADC = frac{1}{2} × base × height
= \(\frac{1}{2}\) × AC × DN
= \(\frac{1}{2}\) × 22× 3 cm2 = 33 cm2
Area of ∆BCD = Area ∆ABC + Area ∆ADC
= 33 cm2 + 33 cm2 = 66 cm2.

Bhagya
Last Updated on June 15, 2022, 4:03 p.m.
Published June 15, 2022