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.

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

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.

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

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

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

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

(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

(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:

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.

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:

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)

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.

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.
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:

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.

(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.

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.