NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 11 Mensuration.

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Maths
Chapter	Chapter 11
Chapter Name	Mensuration
Exercise	Ex 11.1, Ex 11.2, Ex 11.3, Ex 11.4
Number of Questions Solved	34
Category	NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Chapter 11 Mensuration Exercise 11.1

Question 1.
A square and a rectangular field with measurements as given in the following figure have the same perimeter. Which field has a larger area?
Solution:
Let x be the breadth of the rectangle. It is given that the perimeter of a rectangle = perimeter of the square.
∴ 2 (80 + x) = 4 x 60
⇒ 80 + x = 120
⇒ x = 120 – 80 =40
i.e., breadth of the rectangle = 40 m
Now area of the square = (60 x 60) m = 3600 m²
and the area of the rectangle = (40 x 80) m² = 3200 m²
Hence, the square field has a larger area.

Question 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:
Area of the garden = Area of the outer square – Area of the inner rectangle
= 25 x 25 m² – 20 x 15 m²
= (625 – 300) m² = 325 m²
Cost of developing a garden at the rate of ₹ 55 per sq. metre
= ₹ (55 x 325) = ₹ 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 this garden [Length of rectangle is 20 – (3.5 + 3.5) metres].
Solution:
Total area of the garden = Area of the rectangular portion + The sum of the areas of the pair of semi-circles

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?
Solution:
Area of one tile = base x height
= (24 x 10) cm² = 240 cm²

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.
Solution:
A, B, C and D in the given figures as shown. Let A be the point in each figure from where the ant start moving on the food pieces. She is to reach the initial point after moving around the boundary of each food piece.

Chapter 11 Mensuration Exercise 11.2

Question 1.
The shape of the top surface of a table is 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:
Area of top surface of a table
= Area of the trapezium

Question 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:
Let the required side be x cm.

Question 3.
Length of the fence of a trapezium A 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 B and BC.
Solution:
Let ABCD be the given trapezium in which BC = 48 m, CD = 17 m and
AD = 40 m.

Question 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:
Let ABCD be the given quadrilateral in which BE ⊥ AC and DF ⊥ AC.
It is given that
AC =24m, BE = 8m and DF = 13 m.
Now, area of quad. ABCD
= area of ∆ABC + area of ∆ACD

Question 5.
The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution:
Area of a rhombus = x (product of diagonals)

Question 6.
Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Solution:
Let ABCD be a rhombus of side 6 cm and whose altitude DF = 4 cm. Also, one of its diagonals, BD =8 cm.
Area of the rhombus ABCD

Question 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:
Area of the floor = 3000 x Area of one tile

Question 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 100m, find the length of the side along the river.
Solution:
Let the parallel sides of the trapezium shaped field berm and 2x m. Then, its area

∴ The length of the side along the river is 2 x 70, i.e., 140 metres.

Question 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:
Area of the octagonal surface ABCDEFGH = Area (trap. ABCH) + Area (rect. HCDG) +Area (trap. GDEF)

Question 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:
Taking Jyoti’s diagram :

Question 11.
Diagram of the adjacent picture frame has outer dimensions = 24 cm x 28 cm and inner dimensions 16 cm x 20 cm. Find the area of each section of the frame, if the width of each section is same.
Solution:

Chapter 11 Mensuration Exercise 11.3

Question 1.
There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?
Solution:
Total surface area of first box
= 2(lb + bh + lh)
= 2 (60 x 40 + 40 x 50 + 60 x 50) cm²
= 200 (24+20+30) cm²
= 200 x 74 cm²
=14800 cm²
Total surface area of second box
= 6 (Edge)² = 6 x 50 x 50 cm²
= 15000 cm²
Since the total surface area of first box is less than that of the second, therefore the first box i.e., (a) requires the least amount of material to make.

Question 2.
A suitcase of measures 80 cm x 48 cm x 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 the suitcase
= 2 (80 x 48 + 48 x 24 + 80 x 24) cm²
= 2(3840 + 1152 + 1920) cm²
= 2 x 6912 cm² =13824 cm²
Area of 1 metre of trapaulin = (100 x 96) cm²
= 9600 cm²
Surface area of 100 suitcases =100 x 13824 cm²
Trapaulin required to cover 100 suitcases

= 144 metres

Question 3.
Find the side of a cube whose surface area is 600 cm2.
Solution:
Let a be the side of the cube having surface area 600 cm².
∴ 6a² =600 ⇒ a² =100 ⇒ a =10
Hence, the side of the cube = 10 cm.

Question 4.
Rukhsar painted the outside of the cabinet of measure 1 m x 2 m x 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet.
Solution:
Here l = 2 m, b = 1 m and h = 1.5 m
Area to be painted = 2bh + 2lh + lb
= (2 x 1 x 1.5 + 2 x 2 x 1.5 + 2 x 1) m²
= (3 + 6+2) m² =11 m²

Question 5.
Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m2 of area is painted.
How many cans of paint will she need to paint the room?
Solution:
Here l = 15 m, b = 10 m and h = 7 m
Area to be painted = 2bh + 2lh + lb
= (2 x 10 x 7 + 2 x 15 x 7 + 15 x 10) m²
= (140 + 210 + 150) m² =500 m²
Since each can of paint covers 100 m2, therefore number of cans required .

Question 6.
Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area?
Solution:
The given figure are alike in their heights. They differ as under :
(i) One is named cylinder and the other is a cube.
(ii) Cylinder is a solid obtained by revolving a rectangular lamina about one of its sides whereas cube is solid bounded by six squares.
(iii) Cylinder has two circular faces, whereas cube has six faces.
Given,
r = cm, h = 7 cm = cm² = 154 cm²
Lateral surface area of cylinder = 2πrh and, lateral surface area of cube
= Perimeter of base x height = (4 x 7 x 7) cm² = 196 cm²
Clearly, cube has larger lateral surface area.

Question 7.
A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?
Solution:
Here, r = 7 m and h = 3 m.
Sheet of metal required to make a closed cylinder
= Total surface area of the cylinder
= (2πrh + 2πr²) sq. units
²
=(138 + 308)m² = 440 m²

Question 8.
The lateral surface area of a hollow cylinder is 4224 cm2. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet.
Solution:
When a hollow cylinder is cut along its height and formed into a rectangular sheet. Then, the circumference of the base of cylinder is equal to the length of the sheet and the height of the cylinder is equal to the breadth of the sheet. Therefore, lateral surface area of the cylinder is equal to the area of the rectangular sheet. Let? be the length of the sheet.
∴ Area of the sheet = Lateral surface area of cylinder
⇒ l x 33 = 4224
⇒
Now, perimeter of rectangular sheet
= 2 (l + b) = 2 (128 + 33) cm
= 2 x 161 cm
= 322 cm

Question 9.
A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m.
Solution:
Road roller is a cylinder whose radius,

Question 10.
A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the surface area of the label.
Solution:
Since the company places a label around the surface of the cylindrical container of radius 7 cm and height 20 cm such that it is placed 2 cm from top and bottom.
∴ We have to find the curved surface of a cylinder of radius 7 cm and height (20 – 4) cm i.e., 16 cm. ,
This curved surface area =
= 704 cm²

Chapter 11 Mensuration Exercise 11.4

Question 1.
Given a cylindrical tank, in which situation will you find surface area and in which situation volume.
(a) To find how much it can hold.
(b) Number of cement bags required to plaster it.
(c) To find the number of smaller tanks that can be filled with water from it.
Solution:
(a) Volume
(b) Surface area
(c) Volume

Question 2.
Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?
Solution:
Cylinder B has greater volume.
Surface area of cylinder A
= (2πrh + 2πr² sq. units,

Thus, cylinder having greater volume also has greater surface area.

Question 3.
Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3?
Solution:
Volume of the cuboid =900 cm³
⇒ (Area of the base) x Height = 900 cm³
⇒ 180 x Height = 900
⇒ Height = = 5
Hence, the height of the cuboid is 5 cm.

Question 4.
A cuboid is of dimensions 60 cm x 54 cm x 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?
Solution:

Question 5.
Find the height of the cylinder whose volume is 1.54 m3 and diameter of the base is 140 cm?
Solution:
Let the h be the height of cylinder whose radius, r = cm = 70 cm = m and volume =1.54 m³.

Question 6.
A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in litres that can be stored in the tank?
Solution:
Quantity of milk that can be stored in the tank = Volume of the tank

Question 7.
If each edge of a cube is doubled,
(i) how many times will its surface area increase?
(ii) how many times will its volume increase?
Solution:
Let x units be the edge of the cube. Then, its surface area = 6x² and its volume = x³.
When its edge is doubled,
(i) Its surface area = 6 (2x)² = 6 x 4 x 2 = 24 x 2
⇒ The surface area of the new cube will be 4 times that of the original cube.
(ii) Its volume = (2x)³ = 8x³.
⇒ The volume of the new cube will be 8 times that of the original cube.

Question 8.
Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m3, find the number of hours it will take to fill the reservoir.
Solution:
Volume of the reservoir =108 m³
= 108 x 1000 litres
= 108000 litres
Since water is pouring into reservoir @ 60 litres per minute
∴ Time taken to fill the reservoir = hours
= 30 hours

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