# NCERT Solutions for Class 8 Maths Chapter 11 | NCERT Maths Class 8

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration. class 8th maths. ncert class 8 Maths. Here We learn what is in ncert maths class 8 chapter 11 and how to solve questions with easiest method. In this chapter we solve the question of NCERT 8th class maths chapter 11. NCERT class 8 maths solutions Mensuration are part of NCERT Solutions for class 8 maths chapter 11 solution PDF. Ncert solutions for class 8 chapter 11 with formula and solution. ncert class 8th maths

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# NCERT Solutions for Class 8 Maths chapter 11 Mensuration ncert 8th class maths

## Exercise 11.1

1. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?
Solutions:
Side of square = 60 m
Perimeter of square = 240 m
4 × side = 240
side = $$\displaystyle \frac{{600}}{6}$$
side = 60 m
Area of square = (side)2
Area of square = (60)2
= 3600 m2
According to the question both square and rectangular field have the same perimeter.
So, 2 × (l + b) = 240
Length of rectangular field = 80 m
Let’s rectangular field breadth = b
2 × (80 + b) = 240
160 + 2b = 240
2b = 240 − 160
2b = 80
b = 80/2
b = 40 m
Area of rectangular field = l × b
= 80 × 40
= 3200 m2
So that the area of square is larger than rectangular field.

2. Mrs. Kaushik has a square plot with the measurement as shown in the 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 ₹ 55 per m2.
Solution:
Side of square plot = 25 m
Area of plot = (side)2
= (25)2
= 625 m2
Length of house = 20 m
Breadth of house = 15 m
Area of house = length × breadth
= 20 ⨯ 15
= 300 m²
Area of garden = Area of square plot − Area of house
= 625 − 300
= 325 m²
∵ 1 m² developing cost is = ₹ 55
∴ 325 m² developing cost are = 55 ⨯ 325
= ₹ 17875

3. The shape of a garden is rectangular in the middle and semicircular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 – (3.5 + 3.5) metres].
Solution:

Diameter of semi-circle = 7 m
Radius of semi-circle = 3.5 m
Perimeter of semi-circle = $$\displaystyle \frac{{2\pi r}}{2}$$
$$\displaystyle =\frac{2}{2}\times \frac{{22}}{7}\times 3.5$$
= 22 × 0.5
= 11 m
Semicircle II will also have the same circumference as I because they have the same radius = 11 m
Middle rectangular garden length = 20 m
Total perimeter of garden = first semi-circle perpendicular + meddle rectangular garden both length + II semi-circle perpendicular
= 11 + 13 + 13 + 11
= 48 m
Area of I semi-circle = $$\displaystyle \frac{{\pi {{r}^{2}}}}{2}$$
$$\displaystyle =\frac{{22}}{7}\times \frac{{3.5\times 3.5}}{2}$$
= 11 × 0.5 × 3.5
= 19.25 m2
Semi-circle II area is equal to semi-circle I because the radius are same = 19.25 m²
middle rectangle length = 13 m
middle rectangle breadth = 7 m
Area of middle rectangle = length ⨯ breadth
= 13 ⨯ 7
= 91 m2
Total area of garden = Area of semi-circle I + Area of middle rectangle + Area of semi-circle II
= 19.25 + 91 + 19.25
= 129.5 m2

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).
Solution:
Height of parallelogram = 10 cm
Base of parallelogram = 24 cm
Area of parallelogram = base ⨯ height
= 240 cm²
Area of floor = 1080 m² or 1080 ⨯ 10000 = 10800000 cm²
numbers of tiles = $$\displaystyle \frac{{\text{Area of floor}}}{{\text{Area of one tile}}}$$
$$\displaystyle =\frac{{10800000}}{{240}}$$
= 45000 tiles

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.
Solution:

because we know that the ant will make a circle equal to the perimeter of the shape.
So that perimeter of shape I = 2.8 + perimeter of semi-circle
diameter of semi-circle = 2.8 m
perimeter of semi-circle = $$\displaystyle =2.8+\frac{2}{2}\times \frac{{22}}{7}\times 1.4$$
= 2.8 + 22 × 0.2
= 2.8 + 4.4
= 7.2 m
perimeter of shape II = 1.5 + 2.8 + 1.5 + perimeter of semi-circle
$$\displaystyle =1.5+2.8+1.5+\frac{2}{2}\times \frac{{22}}{7}\times 1.4$$
= 1.5 + 2.8 + 1.5 + 4.4
= 10.2 m
perimeter of shape III = 2 + 2 + perimeter of semi-circle
$$\displaystyle =2+2+\frac{2}{2}\times \frac{{22}}{7}\times 1.4$$
= 2 + 2 + 4.4
= 8.4 m
So the ant will have to take the longest round for piece b because piece b has the largest perimeter.

### EXERCISE 11.2

1. The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m. Solution :
Parallel sides of trapezium = 1 m and 1.2 m
perpendicular distance between them (height) = 0.8 m
Area of trapezium = $$\displaystyle \frac{1}{2}\text{ }\times \text{ sum of parallel sides }\times \text{ height}$$
$$\displaystyle \frac{1}{2}\times (1+1.2)\times 0.8$$
= 2.2 x 0.8
= 1.76 m²

2. The area of a trapezium is 34 cm2 and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.
Solution :
Height of trapezium = 4 cm
one parallel side of trapezium = 10 cm
Let’s the second parallel side of trapezium is = 𝑥
Area of trapezium = $$\displaystyle \frac{1}{2}\text{ }\times \text{ sum of parallel sides }\times \text{ height}$$
$$\displaystyle \frac{1}{2}\times (10+x)\times 4=34$$
(10 + 𝑥) × 2 = 34
20 + 2𝑥 = 34
2𝑥 = 34 – 20
2𝑥 = 14
$$\displaystyle x=\frac{{14}}{2}$$
𝑥 = 7 cm
So, the second parallel side of trapezium is 7 cm.

3. Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC. Solution :
Perimeter of trapezium = 120 m
All three sides sum of trapezium are = 48 + 17 + 40 = 105 m
Fourth side of trapezium (AB) = 120 – 105 = 15 m
Area of trapezium = $$\displaystyle \frac{1}{2}\text{ }\times \text{ sum of parallel sides }\times \text{ height}$$
$$\displaystyle =\frac{1}{2}\times (40+48)\times 15$$
$$\displaystyle =\frac{1}{2}\times 88\times 15$$
= 44 × 15
= 660 m²

4. The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field. Solution :
Area of quadrilateral shaped field = $$\displaystyle \frac{1}{2}\text{ }\times \text{ perpendiculars }\times \text{ sum of opposite vertices}$$
$$\displaystyle =\frac{1}{2}\times (13+8)\times 24$$
$$\displaystyle =\frac{1}{2}\times 21\times 24$$
= 21 × 12
= 252 m²

5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution :
Area of rhombus = $$\displaystyle \frac{1}{2}\text{ }\times \text{ product of diagonals}$$
$$\displaystyle =\frac{1}{2}\times 12\times 7.5$$
= 6 × 7.5
= 45 cm²

6. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Solution :
If we consider the rhombus as a parallelogram then –
Area of parallelogram = Base × Hight
= 6 × 4
= 24 cm²
Let’s Second diagonal is = 𝑥
Area of rhombus = $$\displaystyle \frac{1}{2}\text{ }\times \text{ product of diagonals}$$
$$\displaystyle 24=\frac{1}{2}\times 8\times x$$
24 = 4 × 𝑥
24 = 4𝑥
$$\displaystyle x=\frac{{24}}{4}$$
𝑥 = 6 cm
So the second diagonal is 6 cm.

7. The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m2 is ₹ 4.
Solution :
Rhombus shaped tiles area = $$\displaystyle \frac{1}{2}\text{ }\times \text{ product of diagonals}$$
$$\displaystyle =\frac{1}{2}\times 45\times 30$$
= 45 × 15
= 675 cm²
∵ same 3000 tiles area = 3000 × 675
= 2025000 cm²
or 202.5 m²
∵ Cost of polishing the 1 m2 = ₹ 4
∴ Cost of polishing the 202.5 m² = 202.5 × 4
= ₹ 810

8. Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.
Solution : Let’s length of side along the road = 𝑥 m
length of side along the river = 2𝑥 m
distance the both parallel sides = 100 m
Area of trapezium = 10500 m²
$$\displaystyle \frac{1}{2}\text{ }\times \text{ sum of parallel sides }\times \text{ height}$$ = 10500 m²
$$\displaystyle \frac{1}{2}\times (x+2x)\times 100=10500$$
3𝑥 × 50 = 10500
150𝑥 = 10500
$$\displaystyle =\frac{{10500}}{{150}}$$
𝑥 = 70 m
So, first parallel side of trapezium = 70 m
and second parallel side = 2𝑥 = 2 × 70 = 140 m

9. Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface. Solution :
Shape I is a trapezium.
parallel sides of trapezium = 11 m and  5 m
height of trapezium = 4 m
Area of trapezium = $$\displaystyle \frac{1}{2}\text{ }\times \text{ sum of parallel sides }\times \text{ height}$$
$$\displaystyle =\frac{1}{2}\times (11+5)\times 4$$
$$\displaystyle =\frac{1}{2}\times 16\times 4$$
= 16 × 2
= 32 m²
shape III and shape I are same. so, the area are also same = 32 m²
Area of shape II =  Area of rectangle (because shape is a rectangular shape)
Area shape II = length ⨯ breadth
= 11 × 5
= 55 m²
so, the area of octagonal surface is = area of shape I + area of shape II + Area of shape III
= 32 + 55 + 32
= 119 m²

10. There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways.
Find the area of this park using both ways. Can you suggest some other way of finding its area?
Solution :
To find the area by Jyoti, the pentagon shape was divided into two parts and divided into two trapeziums.
So, the area of pentagonal shaped = 2 × area of trapezium
$$\displaystyle =2\times \frac{1}{2}\times (15+30)\times 7.5$$
= 45 × 7.5
= 337.5 m²

To find the area by poetry, divide the pentagon shape into two parts, first a triangle and second a square.
Area of triangle = $$\displaystyle \frac{1}{2}\text{ }\times \text{ base }\times \text{ height}$$
$$\displaystyle =\frac{1}{2}\times 15\times 15$$
$$\displaystyle =\frac{1}{2}\times 225$$
= 112.5 m²
Area of square = (side)²
= (15)²
= 225 m²
Area of pentagonal shaped = Area of triangle + Area of square
= 112.5 + 225
= 337.5 m²

11. Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same. Solution:
we can divide the picture frame in the V sigment.
Area of segment I = Area of trapezium = $$\displaystyle \frac{1}{2}\text{ }\times \text{ sum of parallel sides }\times \text{ height}$$
$$\displaystyle =\frac{1}{2}\times (24+16)\times 4$$
$$\displaystyle =\frac{1}{2}\times 40\times 4$$
= 40 × 2
= 80 cm²
Shaped III and I are the same. So the area of shaped III is also = 80 cm²
Area of shaped II = Area of trapezium = $$\displaystyle \frac{1}{2}\text{ }\times \text{ sum of parallel sides }\times \text{ height}$$
$$\displaystyle =\frac{1}{2}\times (28+20)\times 4$$
$$\displaystyle =\frac{1}{2}\times 48\times 4$$
= 48 × 2
= 96 cm²
Area of shaped V = Area of rectangular = length ⨯ breadth
= 20 × 16
= 320 cm²

#### EXERCISE 11.3

1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make? Solution:
Shape (a) is a cuboid.
length of cuboid = 60 m
Breadth of the cuboid = 40 m
height of cuboid = 50 m
Area of cuboid = 2 × (l × b + b × h + h × l)
= 2 × (60 × 40 + 40 × 50 + 50 × 60)
= 2 × (2400 + 2000 + 3000)
= 2 × 7400
= 14800 m²
Shape (b) is a cube.
side of cube = 50 cm
Total area of cube = 6 × (side)²
= 6 × (50)²
= 6 × 2500
= 15000 m²
Less material is required to make figure (a).

2. A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Solution:
total surface area of cuboidal suitcase = 2 × (l × b + b × h + h × l)
= 2 × (80 × 48 + 48 × 24 + 24 × 80)
= 2 × (3840 + 1152 + 1920)
= 2 × 6912
= 13824 cm²
so, the area of 100 suitcase are = 13824 × 100
= 1382400 cm²
Area of tarpaulin= length ⨯ breadth
1382400 = 𝑥 × 96
96𝑥 = 1382400
$$\displaystyle =\frac{{1382400}}{{96}}$$
𝑥 = 14400 cm
𝑥 = 144 m
Hence, to cover 100 suitcases, a tarpaulin of length 144 m is required.

3. Find the side of a cube whose surface area is 600 cm2.
Solution:
Side of cube = 𝑥
Total area of cube = 600 cm²
6 × (side)² = 600
6 × 𝑥² = 600
$$\displaystyle =\frac{{600}}{6}$$
𝑥² = 100
$$\displaystyle x=\sqrt{{100}}$$
So, the side of cube is 10 cm.

4. Rukhsar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet. Solution:

Cabinet length = 1 m
Cabinet height = 1.5 m
Total area of cabinet = 2 × (l × b + b × h + h × l)
= 2 × (1 × 2 + 2 × 1.5 + 1.5 × 1)
= 2 × (2 + 3 + 1.5)
= 2 × 6.5
= 13 m²
Area of cabinet bottom surface = length × breadth
= 2 × 1
= 2 m²
Area for paint = Total area of cabinet − Area of cabinet bottom surface
= 13 − 2
= 11 m²