Exercise 11.1
Introduction - Mensuration - Chapter 11 - NCERT Class 8th Maths
Q 1 - Ex 11.1 - Mensuration - NCERT Maths Class 8th - Chapter 11
Question 1
A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?
Sol :
Perimeter of square = 4 (Side of the square) = 4 (60 m) = 240 m
Perimeter of rectangle = 2 (Length + Breadth)
= 2 (80 m + Breadth)
= 160 m + 2 × Breadth
It is given that the perimeter of the square and the rectangle are the same.
160 m + 2 × Breadth = 240 m
Breadth of the rectangle =
= 40 m
Area of square = (Side)2 = (60 m)2 = 3600 m2
Area of rectangle = Length × Breadth = (80 × 40) m2 = 3200 m2
Thus, the area of the square field is larger than the area of the rectangular field.
Question 2
Mrs. Kaushik has a square plot with the measurement as shown in the following figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of Rs 55 per m2.
Sol :
Area of the square plot = (25 m)2 = 625 m2
Area of the house = (15 m) × (20 m) =300 m2
Area of the remaining portion = Area of square plot − Area of the house
= 625 m2 − 300 m2 = 325 m2
The cost of developing the garden around the house is Rs 55 per m2.
Total cost of developing the garden of area 325 m2 = Rs (55 × 325)
= Rs 17,875
Question 3
The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of the garden [Length of rectangle is 20 − (3.5 + 3.5) metres]
Sol :
Length of the rectangle = [20 − (3.5 + 3.5)] metres = 13 m
Circumference of 1 semi-circular part = πr
Circumference of both semi-circular parts = (2 × 11) m = 22 m
Perimeter of the garden = AB + Length of both semi-circular regions BC and
DA + CD
= 13 m + 22 m + 13 m = 48 m
Area of the garden = Area of rectangle + 2 × Area of two semi-circular regions
Question 4
A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m2? (If required you can split the tiles in whatever way you want to fill up the corners).
Sol :
Area of parallelogram = Base × Height
Hence, area of one tile = 24 cm × 10 cm = 240 cm2
Required number of tiles =
= 45000 tiles
Thus, 45000 tiles are required to cover a floor of area 1080 m2.
Question 5
An ant is moving around a few food pieces of different shapes scattered on the floor. For which food − piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression c = 2πr, where r is the radius of the circle.
Sol :
(a)Radius (r) of semi-circular part =
Perimeter of the given figure = 2.8 cm + πr
(b)Radius (r) of semi-circular part =
Perimeter of the given figure = 1.5 cm + 2.8 cm + 1.5 cm +π (1.4 cm)
(c)Radius (r) of semi-circular part =
Perimeter of the figure(c) = 2 cm + πr + 2 cm
Thus, the ant will have to take a longer round for the food-piece (b), because the perimeter of the figure given in alternative (b) is the greatest among all.
A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?
Sol :
Perimeter of square = 4 (Side of the square) = 4 (60 m) = 240 m
Perimeter of rectangle = 2 (Length + Breadth)
= 2 (80 m + Breadth)
= 160 m + 2 × Breadth
It is given that the perimeter of the square and the rectangle are the same.
160 m + 2 × Breadth = 240 m
Breadth of the rectangle =
= 40 mArea of square = (Side)2 = (60 m)2 = 3600 m2
Area of rectangle = Length × Breadth = (80 × 40) m2 = 3200 m2
Thus, the area of the square field is larger than the area of the rectangular field.
Q 2 - Ex 11.1 - Mensuration - NCERT Maths Class 8th - Chapter 11
Mrs. Kaushik has a square plot with the measurement as shown in the following figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of Rs 55 per m2.
Sol :
Area of the square plot = (25 m)2 = 625 m2
Area of the house = (15 m) × (20 m) =300 m2
Area of the remaining portion = Area of square plot − Area of the house
= 625 m2 − 300 m2 = 325 m2
The cost of developing the garden around the house is Rs 55 per m2.
Total cost of developing the garden of area 325 m2 = Rs (55 × 325)
= Rs 17,875
Q 3 - Ex 11.1 - Mensuration - NCERT Maths Class 8th - Chapter 11
Question 3
The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of the garden [Length of rectangle is 20 − (3.5 + 3.5) metres]
Sol :
Length of the rectangle = [20 − (3.5 + 3.5)] metres = 13 m
Circumference of 1 semi-circular part = πr
Circumference of both semi-circular parts = (2 × 11) m = 22 m
Perimeter of the garden = AB + Length of both semi-circular regions BC and
DA + CD
= 13 m + 22 m + 13 m = 48 m
Area of the garden = Area of rectangle + 2 × Area of two semi-circular regions
Q 4 - Ex 11.1 - Mensuration - NCERT Maths Class 8th - Chapter 11
Question 4
A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m2? (If required you can split the tiles in whatever way you want to fill up the corners).
Sol :
Area of parallelogram = Base × Height
Hence, area of one tile = 24 cm × 10 cm = 240 cm2
Required number of tiles =
= 45000 tilesThus, 45000 tiles are required to cover a floor of area 1080 m2.
Q 5 - Ex 11.1 - Mensuration - NCERT Maths Class 8th - Chapter 11
An ant is moving around a few food pieces of different shapes scattered on the floor. For which food − piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression c = 2πr, where r is the radius of the circle.
Sol :
(a)Radius (r) of semi-circular part =
Perimeter of the given figure = 2.8 cm + πr
(b)Radius (r) of semi-circular part =
Perimeter of the given figure = 1.5 cm + 2.8 cm + 1.5 cm +π (1.4 cm)
(c)Radius (r) of semi-circular part =
Perimeter of the figure(c) = 2 cm + πr + 2 cm
Thus, the ant will have to take a longer round for the food-piece (b), because the perimeter of the figure given in alternative (b) is the greatest among all.
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