MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.1

MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.1

Question 1.
The length and the breadth of a rectangular piece of land are 500 m and 300 m respectively. Find
(i) its area
(ii) the cost of the land, if 1 m2 of the land costs ₹ 10,000.
Solution:
Length (l) = 500 m
Breadth (b) = 300 m
(i) Area = Length × Breadth = 500 × 300 = 150000 m2
(ii) Cost of 1 m2 land = ₹ 10000
∴ Cost of 150000 m2 land
= 150000 × 10000 = ₹ 1500000000

Question 2.
Find the area of a square park whose perimeter is 320 m.
Solution:
Perimeter of the square park = 320 m
∴ 4 × Length of the side of park = 320
Length of the side of park = \(\frac{320}{4}\) = 80 m
Area = (Length of the side of park)2
= (80)2 = 6400 m2

Question 3.
Find the breadth of a rectangular plot of land, if its area is 440 m2 and the length is 22 m. Also find its perimeter.
Solution:
Area of a rectangular plot = 440 m2
Length = 22 m
Area = Length x Breadth = 440 m2
∴ 22 × Breadth = 440
⇒ Breadth = \(\frac{440}{22}\) = 20 m
∴ Perimeter = 2 (Length + Breadth)
= 2 (22 + 20) = 2(42) = 84 m

MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.1

Question 4.
The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.
Solution:
Length = 35 cm
Perimeter = 100 cm
∴ 2 (35 + Breadth) = 100
⇒ 35 + Breadth = 50
⇒ Breadth = 50 – 35 = 15 cm
∴ Area = Length × Breadth
= 35 × 15 = 525 cm2

Question 5.
The area of a square park is the same as of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park.
Solution:
Side of the square park = 60 m
Length of the rectangular park = 90 m
Area of the square park = (side)2 = (60)2 = 3600 m2
Area of rectangular park = Length × Breadth
= 90 × Breadth
It is given that area of square park = area of rectangular park
∴ 3600 = 90 × Breadth
⇒ Breadth = 40 m

Question 6.
A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side. Also find which shape encloses more area?
Solution:
Length of rectangle = 40 cm
Breadth of rectangle = 22 cm
Perimeter of rectangle = Perimeter of square
∴ 2 (Length + Breadth) = 4 × Side of square
⇒ 2 (40 + 22) = 4 × Side of square
⇒ 2 × 62 = 4 × Side of square
∴ Side of square = \(\frac{124}{4}\) = 31 cm
Now, area of rectangle = 40 × 22 = 880 cm2
Area of square = (Side)2 = 31 × 31 = 961 cm2
As 961 > 880.
Therefore, the square-shaped wire encloses more area than rectangle – shaped wire.

MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.1

Question 7.
The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, find its length. Also find the area of the rectangle.
Solution:
Breadth = 30 cm
Perimeter = 130 cm
∴ 2 (Length + 30) = 130
⇒ Length + 30 = 65
⇒ Length = 65 – 30 = 35 cm
Now, area = Length × Breadth
= 35 × 30 = 1050 cm2

Question 8.
A door of length 2 m and breadth 1 m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m (see the given figure). Find the cost of white washing the wall, if the rate of white washing the wall is ₹ 20 per m2.
MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.1 1
Solution:
Length of wall = 4.5 m
Breadth of wall = 3.6 m
Area of wall = Length × Breadth
= 4.5 × 3.6
= 16.2 m2
Area of door = 2 × 1 = 2 m2
Area to be white-washed
= Area of wall – Area of door
= 16.2 – 2 = 14.2 m2
Cost of white-washing 1 m2 area = ₹ 20 2.
∴ Cost of white-washing 14.2 m2 area
= 14.2 × 20 = ₹ 284

MP Board Class 7th Maths Solutions

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