NCERT Solutions for Class 8 Maths Chapter 9 Mensuration

In the realm of mathematics, mensuration governs shapes, ensuring every inch is accounted for. Children often wonder about the geometry around them—"What shape is my book?", "How were the pyramids measured?", or even "How much wax is needed to make a candle for my school project?" and "What shape is a Christmas tree?" All these curious questions relate directly to the concept of Mensuration. They will clearly understand these shapes in detail after completing the Mensuration chapter, which is Class 8 Maths NCERT Chapter 9. In this section of Mathematics, we discuss the length, volume, and surface areas of various types of quadrilaterals. These NCERT solutions for Class 8 on Mensuration are designed to guide students step-by-step whenever they feel confused while solving exercise problems.

Mensuration turns geometry into a tool, letting us calculate the space we occupy in this vast universe. Solving the NCERT Solutions for Class 8 Maths of Chapter 9 will help us excel in board exams and lay a smooth path to learning about these concepts in higher classes. Other competitive exams frequently include many questions from this chapter. These class 8 Mensuration solutions contain many images and diagrams clearly illustrating the shapes because visualisation is often considered the best way to learn. Experienced Careers 360 experts have made these solutions following the latest CBSE guidelines. Download your syllabus, revision notes, and all chapter PDFs here: NCERT.

Mensuration Class 8 Questions And Answers PDF Free Download

Students who wish to access the NCERT solutions for class 8, chapter 9, Mensuration, can click on the link below to download the entire solution in PDF.

NCERT Solutions for Class 8 Maths Chapter 9: Exercise Questions

Question 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 the perpendicular distance between them is 0.8 m.

Answer:

Area of the table

= $\frac{1}{2}$ × sum of parallel sides × distance between parallel sides

= $\frac{1}{2}$ × (1+1.2) × 0.8

= 0.4×2.2

= 0.88 m²

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.

Answer:

Let the length of the other parallel side be x

area of a trapezium, $\frac{1}{2}$ ×(10+x)×4=34

⇒ (10+x)=17

⇒ x = 17−10

⇒ x = 7 cm

Hence, the length of the other parallel side is 7 cm.

Question 3: Length of the fence of a trapezium-shaped field ABCD is 120m. 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:

BC=48 m, CD=17 m and AD=40 m ,

Length of the fence of a trapezium-shaped field ABCD
= 120m = AB+BC+CD+DA

⇒ 120 = AB+48+17+40

⇒ AB = 15 m

Area of the trapezium

= $\frac{1}{2}$ × sum of parallel sides × height

= $\frac{1}{2}$×(40+48)×15

= $\frac{1}{2}$×(88)×15

= (44)×15

= 660 m²

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.

Answer:

The diagonal of a quadrilateral-shaped field is 24 m.

The perpendiculars are 8 m and 13 m.

The area of the field

= $\frac{1}{2}$ ×d×(h1+h2)

= $\frac{1}{2}$ ×24×(13+8)

= 12×21

= 252 m²

Question 5: The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

Answer:

Area of the rhombus

= $\frac{1}{2}$ × product of diagonals

= $\frac{1}{2}$×7.5×12

= 7.5×6

= 45 cm²

Question 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.

Answer:

A rhombus is a type of parallelogram, and the area of the parallelogram is the product of base and height.

So, Area of rhombus = base × height = 5×4.8 = 24

Let the other diagonal be x.

Area of the rhombus = $\frac{1}{2}$ × product of diagonals

⇒ 24 = 12×8×x

⇒ x = 6 cm

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.

Answer:

Area of 1 tile

= $\frac{1}{2}$ × product of diagonals

= $\frac{1}{2}$ ×45×30

= 675 cm²

Area of 3000 tiles

= 675 ×3000

= 202.5 m²

Total cost of polishing

= 202.5 × 4

= Rs. 810

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 100 m, find the length of the side along the river.

Answer:

Let the length of the side along the road be x m.

Then, according to the question, the length of the side along the river will be 2x m.

Area of trapezium = $\frac{1}{2}$h(a+b)

So the equation becomes:

$\frac{1}{2}\times$100(x+2x)=10500

⇒ 3x = 210

⇒ x = 70

So the length of the side along the river is 2x = 140m.

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.

Answer:

Area of octagonal surface = Area of rectangular surface + 2(Area of trapezium surface)

Area of rectangular surface = 11×5 = 55 m²

Area of trapezium surface =

$\frac{1}{2}$ × 4 (11+5) = 2 × 16 = 32 m²

So, the total area of the octagonal surface = 55 + 2(32) = 55 + 64 = 119 m²

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?

Answer:

Area of pentagon $\mathrm{ABCDE}=$ Area of trapezium $\mathrm{ABCF}+$ Area of trapezium AEDF

$\begin{array}{rl}& =\frac{1}{2}\times (AF+BC)\times FC+\frac{1}{2}\times (AF+ED)\times DF\\ & =\frac{1}{2}\times (30\mathrm{m}+15\mathrm{m})\times \frac{15}{2}\mathrm{m}+\frac{1}{2}\times (30+15\mathrm{m})\times \frac{15}{2}\mathrm{m}\\ & =2\times \frac{1}{2}(30\mathrm{m}+15\mathrm{m})\times \frac{15}{2}\\ & =45\mathrm{m}\times 7.5\mathrm{m}\\ & =337.5{\mathrm{m}}^2\end{array}$
Thus, the area of the pentagon-shaped park according to Jyoti's way is $337.5{\mathrm{m}}^2$

$\begin{array}{rl}& \text{ Area of pentagon }ABCDE=\text{ Area of triangle }ABE+\text{ Area of square }EBCD\\ & =\frac{1}{2}\times BE\times (AF-OF)+FC\times BC[\text{ Since, }AO=AF-OF]\\ & =\frac{1}{2}\times 15\times (30-15)+(15\times 15)\\ & =(\frac{1}{2}\times 15\times 15){\mathrm{m}}^2+225{\mathrm{m}}^2\\ & =112.5{\mathrm{m}}^2+225{\mathrm{m}}^2\\ & =337.5{\mathrm{m}}^2\end{array}$
Thus, the area of the pentagon-shaped park according to Kavitha's way is $337.5{\mathrm{m}}^2$

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

Answer:

The area of the opposite sections will be the same.

So the area of horizontal sections,

= $\frac{1}{2}$×4(16+24)

= 2(40)

= 80 cm²

And the area of vertical sections,

= $\frac{1}{2}$×4(20+28)

= 2(48)

= 96 cm²

Question 1: There are two cuboidal boxes, as shown in the adjoining figure. Which box requires the lesser amount of material to make?

Answer:

Surface area of cuboid (a) =2(60×40+40×50+50×60)=14800 cm²

Surface area of cube (b) =6(50)²=15000 cm²

So, box (a) requires the lesser amount of material to make.

Question 2: A suitcase with measures 80cm×48cm×24cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96cm is required for over 100 such suitcases?

Answer:

Surface area of suitcase =2(80×48+48×24+24×80) = 13824 cm² .

The area of such 100 suitcases will be 1382400 cm²

So length of tarpaulin cloth $=\frac{1382400}{96}$ cm = 14400 cm or 144 m

Question 3: Find the side of a cube whose surface area is 600 cm².

Answer:

Surface area of cube = 6l²

So, 6l² = 600

⇒ l² = 100

⇒ l = 10 cm

Thus, the side of the cube is 10 cm.

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

Answer:

Required area = Total area - Area of bottom surface

Total area =2(1×2+2×1.5+1.5×1) = 13 m²

Area of bottom surface = 1×2 = 2 m²

So, required area = 13−2 =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?

Answer:

Total area painted by Daniel :

= 2(15×10+10×7+7×15) − (15×10) ( ∵ Bottom surface is excluded.)

So, Area

= 650−150 = 500 m²

Number of cans of paint required

$=\frac{500}{100}$
$=5$

Thus, 5 cans of paint are required.

Question 6: Describe how the two figures at the right are alike and how they are different. Which box has a larger lateral surface area?

Answer:

The two figures have the same height. The difference between them is that one is a cylinder and another is a cube.

Lateral surface area of a cylinder
= 2×π×r×h
= 2×π×3.5×7
= 154 cm²

Lateral surface area of a cube
= 4 × side²
= 4×7²
= 4×49
= 196 cm²

So, the cube has a 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?

Answer:

Total surface area of a cylinder
= 2πr (h + r)
= 2π×7(7+3)
= 14π×10
= 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 the rectangular sheet.

Answer:

Length of rectangular sheet

$=\frac{4224}{33}$
=128 cm

So the perimeter of a rectangular sheet
= 2(l + b)
= 2(128+33)
= 322 cm

Thus perimeter of the rectangular sheet is 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 the length is 1 m

Answer:

Area in one complete revolution of the roller
= 2πrh
= 2π×0.42×1
= 2.638 m²

So, area of road = 2.638 m² × 750 = 1978.5 m²

Question 10: A company packages its milk powder in a 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 the top and bottom, what is the area of the label?

Answer:

Area of label
= 2πrh
=2π×(7)×(16)
=703.71 cm²

Here, the height is found out as = 20-2-2 = 16 cm.

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.

Answer:

(a) To find out how much the cylindrical tank can hold, we will basically find out the volume of the cylinder.

Question 1: Given a cylindrical tank, in which situation will you find surface area and in
which situation volume?

(b) Number of cement bags required to plaster it.

Answer:

(b) If we want to find out the cement bags required to plaster, it means the area to be applied, and we then calculate the surface area of the bags.

Question 1: Given a cylindrical tank, in which situation will you find surface area and in
Which situation volume?

(c) To find the number of smaller tanks that can be filled with water from it.

Answer:

(c) We have to find out the volume.

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

Answer:

Given the diameter of cylinder A = 7 cm and the height = 14 cm.

Also, the diameter of cylinder B = 14 cm and its height = 7 cm.

We can easily suggest whose volume is greater without doing any calculations:

Volume is directly proportional to the square of the radius of the cylinder and directly proportional to the height of the cylinder.
Hence,

B has more Volume as compared to A because B has a larger diameter.

Verifying:

Volume of A : π×r²×height=π × $(\frac{7}{2}{)}^2$ ×14cm = 539 cm³

and Volume of B : π×r²×height=π×($\frac{14}{2}$)²×7cm=1,078 cm³

Hence, clearly we can see that the volume of cylinder B is greater than the volume of cylinder A.

The cylinder B has surface area of = 2×π×r(r+h)=2×π× $\frac{14}{2}$ ($\frac{14}{2}$+7) = 616 cm² .

and the surface area of cylinder A = 2×π×r(r+h)=2×π×$\frac{7}{2}$×($\frac{7}{2}$+14) = 385 cm² .

The cylinder with 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.

Answer:

Given the height of a cuboid whose base area is 180 cm² and volume is 900 cm³

As the Volume of the cuboid = base area × height

So, we have the relation: 900 cm³ = 180cm² × height

⇒ Height = 5 cm

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

Answer:

So given the dimensions of a cuboid are 60 cm × 54 cm × 30 cm
Hence, its volume is equal to = 97,200 cm³

We have to make small cubes with a side of 6 cm, which occupy the volume
= 6 cm × 6 cm × 6 cm = 216 cm³

Hence, we now have one cube having a side length = 6 cm and volume = 216 cm³

So, the total number of small cubes that can be placed in the given cuboid
= $\frac{97200}{216}=450$

Hence, 450 small cubes can be placed in that cuboid.

Question 5: Find the height of the cylinder whose volume is 1.54 m 3 and diameter of the base is 140 cm?

Answer:

Given that the volume of the cylinder is 1.54 m³ and its diameter at the base = 140 cm

So, as the Volume of the cylinder = base area × height

Hence, putting in the relation, we get;

1.54 m³ = (π× (1.4 m²)²) × height

The height of the cylinder would be = 1 metre

Question 6: A milk tank is in the form of a 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:

Volume of the cylinder = V = πr²h = π(1.5)²×7 = 49.48 m³

So, the quantity of milk in litres that can be stored in the tank is 49500 litres.

(∵1 m³ = 1000 litres)

Question 7(i): If each edge of a cube is doubled, how many times will its surface area increase?

Answer:

The surface area of a cube = 6l²

So if we double the edge, l becomes 2l.

New surface area = 6(2l)² = 24l²

Thus surface area becomes 4 times.

Question 7 (ii): If each edge of a cube is doubled, how many times will its volume increase?

Answer:

Volume of cube = l³

Since l becomes 2l, so new volume is : (2l)³ = 8l³

Hence, the volume becomes 8 times.

Question 8: Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of the reservoir is 108 m3, find the number of hours it will take to fill the reservoir.

Answer:

Given that the water is pouring into a cuboidal reservoir at the rate of 60 litres per minute.

The volume of the reservoir is 108 m³, then

The number of hours it will take to fill the reservoir will be:

As we know, 1 m³ = 1000L

Then 108 m³ = 108,000 Litres

Time taken to fill the tank will be:
$\frac{108000}{60}$ =1800 minutes

1800 minutes = $\frac{1800}{60}$ = 30 hours

Mensuration Class 8 Chapter 9: Topics

The topics discussed in the NCERT Solutions for class 8, chapter 9, Mensuration, are:

• Introduction
• Area of a Polygon
• Solid Shapes
• Surface Area of Cube, Cuboid and Cylinder
• Volume of Cube, Cuboid and Cylinder
• Volume and Capacity

Mensuration Class 8 Solutions - Important Formulae

Perimeter Formulae

Surface Area Formula

Volume Formulae

NCERT Solutions for Class 8 Maths: Chapter Wise

The links below allow students to access all the Maths solutions from the NCERT book.

NCERT Solutions for Class 8: Subject Wise

Students can use the following links to find the solutions for other subjects.

• NCERT Solutions for Class 8 Maths
• NCERT Solutions for Class 8 Science

NCERT Books and NCERT Syllabus

During preparation, one of the most important tools is the latest syllabus. The following link will lead you to it. There are also links for some reference books, which are beneficial for students.

• NCERT Books Class 8 Maths
• NCERT Syllabus Class 8 Maths
• NCERT Books Class 8
• NCERT Syllabus Class 8