NCERT Solutions for Class 8 Maths Chapter 11 – Mensuration. Furthermore, here we’ve provided you with the latest solution for Class 8 Maths Chapter 11 – Mensuration. As a result here you’ll find solutions to all the exercises. This NCERT Class 8 solution will help you to score good marks in your exam.
Students can refer to our solution for NCERT Class 8 Maths Chapter 11 – Mensuration. The Chapter 11 Solution of NCERT will help students prepare for the exams and easily crack the exam. Below we’ve provided you with the exercise-wise latest solution.
NCERT Solutions for Class 8 Maths Chapter 11 – Mensuration Exercise Wise Solution
Exercise 11.1 – Page 171 of NCERT
Exercise 11.2 – Page 177 of NCERT
Exercise 11.3 – Page 186 of NCERT
Exercise 11.4 – Page 191 of NCERT
NCERT Solutions for Class 8 Maths Chapter 11 – Mensuration Exercise 11.1 Solution
Here you’ll find NCERT Chapter 11 – Mensuration Exercise 11.1 Solution
Exercise 11.1: Solutions of Questions on Page Number: 171
Q1: A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area
Answer:
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 = 
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.
Q2: 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.
Answer:
Area of the square plot = (25 m)2 = 625 m2 Area of the house = (15 m) x (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 x 325)
= Rs 17,875
Q3: 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]
Answer:
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
Q4: 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).
Answer:
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.
Q5: 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 = 2Àr, where r is the radius of the circle.
Answer:
Radius (r) of semi-circular part =
Perimeter of the given figure = 2.8 cm + πr
- Radius (r) of semi-circular part =
Perimeter of the given figure = 1.5 cm + 2.8 cm + 1.5 cm +π (1.4 cm)
- 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.
NCERT Solutions for Class 8 Maths Chapter 11 – Mensuration Exercise 11.2 Solution
Here you’ll find NCERT Chapter 11 – Mensuration Exercise 11.2 Solution
Exercise 11.2: Solutions of Questions on Page Number: 177
Q1: 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.
Answer:
Area of trapezium = (Sum of parallel sides) × (Distances between parallel sides)
Q2: 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.
Answer:
It is given that,area of trapezium = 34 cm2 and height = 4 cm
Let the length of one parallel side be a. We know that,
Area of trapezium = (Sum of parallel sides) × (Distances between parallel sides)
Thus, the length of the other parallel side is 7 cm.
Q3: 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.
Answer:
Length of the fence of trapezium ABCD = AB + BC + CD + DA 120 m = AB + 48 m + 17 m + 40 m
AB = 120 m – 105 m = 15 m
Area of the field ABCD
Q4: 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.
Answer:
It is given that,
Length of the diagonal, d = 24 m
Length of the perpendiculars, h1 and h2, from the opposite vertices to the diagonal are h1 = 8 m and h2 = 13 m
Area of the quadrilateral
Thus, the area of the field is 252 m2.
Q5: The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Answer: Area of rhombus = 1/2 (Product of its diagonals) Therefore, area of the given rhombus
=
= 45 cm2
Q6: 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.
Answer:
Let the length of the other diagonal of the rhombus be x. A rhombus is a special case of a parallelogram.
The area of a parallelogram is given by the product of its base and height. Thus, area of the given rhombus = Base × Height = 6 cm × 4 cm = 24 cm2
Also, area of rhombus =
Thus, the length of the other diagonal of the rhombus is 6 cm.
Q7: 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 Rs 4.
Answer:
Area of rhombus =
= 675 cm2
Area of 3000 tiles = (675 × 3000) cm2 = 2025000 cm2 = 202.5 m2 The cost of polishing is Rs 4 per m2.
Cost of polishing 202.5 m2 area = Rs (4 × 202.5) = Rs 810 Thus, the cost of polishing the floor is Rs 810.
Q8: Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. It 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.
Answer:
Let the length of the field along the road be l m. Hence, the length of the field along the river will be 2l m. Area of trapezium =
Thus, length of the field along the river = (2 × 70) m = 140 m
Q9: 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.
Answer :
Side of regular octagon = 5 cm
Area of trapezium ABCH = Area of trapezium DEFG
Area of rectangle HGDC = 11 × 5 = 55 m2
Area of octagon = Area of trapezium ABCH + Area of trapezium DEFG
+ Area of rectangle HGDC
= 32 m2 + 32 m2 + 55 m2 = 119 m2
Q10: 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
Answer:
Jyoti’s way of finding area is as follows.
Area of pentagon = 2 (Area of trapezium ABCF)
= 337.5 m2
Kavita’s way of finding area is as follows.
Area of pentagon = Area of ΔABE + Area of square BCDE
Q11: 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.
Answer:
Given that, the width of each section is same. Therefore, IB = BJ = CK = CL = DM = DN = AO = AP
IL = IB + BC + CL 28 = IB + 20 + CL
IB + CL = 28 cm – 20 cm = 8 cm IB = CL = 4 cm
Hence, IB = BJ = CK = CL = DM = DN = AO = AP = 4 cm
Area of section BEFC = Area of section DGHA
Area of section ABEH = Area of section CDGF
⇒Area of section ABEH = Area of section CDGF
= [12(16+24)(4)]=80
NCERT Solutions for Class 8 Maths Chapter 11 – Mensuration Exercise 11.3 Solution
Here you’ll find NCERT Chapter 11 – Mensuration Exercise 11.3 Solution
Exercise 11.3: Solutions of Questions on Page Number: 186
Q1: There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make
Answer:
We know that,
Total surface area of the cuboid = 2 (lh + bh + lb) Total surface area of the cube = 6 (l)2
Total surface area of cuboid (a) = [2{(60) (40) + (40) (50) + (50) (60)}] cm2
= [2(2400 + 2000 + 3000)] cm2
= (2 x 7400) cm2
= 14800 cm2
Total surface area of cube (b) = 6 (50 cm)2 = 15000 cm2
Thus, the cuboidal box (a) will require lesser amount of material.
Q2: A suitcase with 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
Answer:
Total surface area of suitcase = 2[(80) (48) + (48) (24) + (24) (80)]
= 2[3840 + 1152 + 1920]
= 13824 cm2
Total surface area of 100 suitcases = (13824 × 100) cm2 = 1382400 cm2 Required tarpaulin = Length × Breadth
1382400 cm2 = Length × 96 cm
Length =
Thus, 144 m of tarpaulin is required to cover 100 suitcases.
Q3: Find the side of a cube whose surface area is 600 cm2.
Answer:
Given that, surface area of cube = 600 cm2 Let the length of each side of cube be l.
Surface area of cube = 6 (Side)2 600 cm2 = 6l2
l2= 100 cm2
l = 10 cm
Thus, the side of the cube is 10 cm.
Q4: 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
Answer:
Length (l) of the cabinet = 2 m Breadth (b) of the cabinet = 1 m Height (h) of the cabinet = 1.5 m
Area of the cabinet that was painted = 2h (l + b) + lb
= [2 x 1.5 x (2 + 1) + (2) (1)] m2
= [3(3) + 2] m2
= (9 + 2) m2
= 11 m2
Q5: 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.
Answer:
Given that,
Length (l) = 15 m, breadth (b) = 10 m, height (h) = 7 m
Area of the hall to be painted = Area of the wall + Area of the ceiling
= 2h (l + b) + lb
= [2(7) (15 + 10) + 15 ×10] m2
= [14(25) + 150] m2
= 500 m2
It is given that 100 m2 area can be painted from each can. Number of cans required to paint an area of 500 m2
=
Hence, 5 cans are required to paint the walls and the ceiling of the cuboidal hall.
Q6: Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area
Answer:
Similarity between both the figures is that both have the same heights.
The difference between the two figures is that one is a cylinder and the other is a cube. Lateral surface area of the cube = 4l2 = 4 (7 cm)2 = 196 cm2
Lateral surface area of the cylinder = 2πrh cm2 = 154 cm2 Hence, the cube has larger lateral surface area.
Q7: 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
Answer:
Total surface area of cylinder = 2πr (r + h)
m2
= 440 m2
Thus, 440 m2 sheet of metal is required.
Q8: 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.
Answer:
A hollow cylinder is cut along its height to form a rectangular sheet. Area of cylinder = Area of rectangular sheet
4224 cm2 = 33 cm × Length
Thus, the length of the rectangular sheet is 128 cm. Perimeter of the rectangular sheet = 2 (Length + Width)
= [2 (128 + 33)] cm
= (2 × 161) cm
= 322 cm
Q9: 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.
Answer:
In one revolution, the roller will cover an area equal to its lateral surface area. Thus, in 1 revolution, area of the road covered = 2πrh
In 750 revolutions, area of the road covered
=
= 1980 m2
Q10: 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 area of the label.
Answer:
Height of the label = 20 cm – 2 cm – 2 cm = 16 cm
Radius of the label
Label is in the form of a cylinder having its radius and height as 7 cm and 16 cm. Area of the label = 2π (Radius) (Height)
Q11: 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 area of the label.
Answer:
Height of the label = 20 cm – 2 cm – 2 cm = 16 cm
Radius of the label
Label is in the form of a cylinder having its radius and height as 7 cm and 16 cm. Area of the label = 2π (Radius) (Height)
NCERT Solutions for Class 8 Maths Chapter 11 – Mensuration Exercise 11.4 Solution
Here you’ll find NCERT Chapter 11 – Mensuration Exercise 11.4 Solution
Exercise 11.4: Solutions of Questions on Page Number: 191
Q1: Given a cylindrical tank, in which situation will you find surface area and in which situation volume.
- To find how much it can hold
- Number of cement bags required to plaster it
- To find the number of smaller tanks that can be filled with water from it.
Answer:
- In this situation, we will find the volume.
- In this situation, we will find the surface area.
- In this situation, we will find the volume.
Q2: 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
Answer:
The heights and diameters of these cylinders A and B are interchanged. We know that,
Volume of cylinder
If measures of r and h are same, then the cylinder with greater radius will have greater area.
Radius of cylinder A = 
Radius of cylinder B =
As the radius of cylinder B is greater, therefore, the volume of cylinder B will be greater.
Let us verify it by calculating the volume of both the cylinders.
Volume of cylinder A
Volume of cylinder B
Volume of cylinder B is greater.
Surface area of cylinder A
Surface area of cylinder B
Thus, the surface area of cylinder B is also greater than the surface area of cylinder A.
Q3: 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.
Answer:
Volume of cuboid = 60 cm × 54 cm × 30 cm = 97200 cm3 Side of the cube = 6 cm
Volume of the cube = (6)3 cm3 = 216 cm3
Required number of cubes =
Thus, 450 cubes can be placed in the given cuboid.
Q4: Find the height of the cylinder whose volume is 1.54 m3 and diameter of the base is 140 m.
Answer:
Diameter of the base = 140 cm
Radius (r) of the base 
Thus, the height of the cylinder is 1 m.
Q5: 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.
Answer:
Radius of cylinder = 1.5 m Length of cylinder = 7 m
Volume of cylinder
1m3 = 1000 L
Required quantity = (49.5 × 1000) L = 49500 L Therefore, 49500 L of milk can be stored in the tank.
Q6: If each edge of a cube is doubled,
- how many times will its surface area increase
- how many times will its volume increase
Answer:
- Let initially the edge of the cube be l.
Initial surface area = 6l2
If each edge of the cube is doubled, then it becomes 2l. New surface area = 6(2l)2 = 24l2 = 4 x 6l2
Clearly, the surface area will be increased by 4 times.
- Initial volume of the cube = l3
When each edge of the cube is doubled, it becomes 2l. New volume = (2l)3 = 8l3 = 8 x l3
Clearly, the volume of the cube will be increased by 8 times.
Q7: Water is pouring into a cubiodal 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.
Answer:
Volume of cuboidal reservoir = 108 m3 = (108 × 1000) L = 108000 L It is given that water is being poured at the rate of 60 L per minute. That is, (60 × 60) L = 3600 L per hour
Required number of hours 
NCERT Class 8 Maths All Chapters Solution
Chapter 1: Rational Numbers
Chapter 2: Linear Equations in One Variable
Chapter 3: Understanding Quadrilaterals
Chapter 4: Practical Geometry
Chapter 5: Data Handling
Chapter 6: Squares and Square root
Chapter 7: Cubes and Cube Roots
Chapter 8: Comparing Quantities
Chapter 9: Arithmetic Expressions
Chapter 10: Visualising Solid Shapes
Chapter 11: Mensuration
Chapter 12: Exponents and Powers
Chapter 13: Direct and Inverse Proportions
Chapter 14: Factorisation
Chapter 15: Introduction to Graphs
Chapter 16: Playing With Numbers