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NCERT Solutions for Class 8 Maths Chapter 9 Mensuration

NCERT Solutions for Class 8 Maths Chapter 9 Mensuration

Edited By Komal Miglani | Updated on Apr 21, 2025 01:53 PM IST

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.

NCERT Solutions for Class 8 Maths Chapter 9 Mensuration
NCERT Solutions for Class 8 Maths Chapter 9 Mensuration

Excelling in these Class 8 Maths Chapter 9 solutions will help us 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 visualization is often considered the best way to learn. Experienced Careers 360 experts have made these solutions following the latest CBSE guidelines

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Mensuration Class 8 Questions And Answers PDF Free Download

Mensuration Class 8 Solutions - Important Formulae

Perimeter: The length of the outline of any simple closed figure is known as the perimeter.


  • Perimeter of a Rectangle: 2 × (Length + Breadth)

  • Perimeter of a Square: 4 × Side
  • Perimeter of a Circle (Circumference): 2πr (where r is the radius of the circle)
  • Perimeter of a Parallelogram: 2(Base + Height)
  • Perimeter of a Triangle: a + b + c (where a, b, and c are the side lengths)
  • Perimeter of a Trapezium: a + b + c + d (where a, b, c, and d are the sides of a trapezoid)
  • Perimeter of a Kite: 2a + 2b (where a is the length of the first pair of sides and b is the length of the second pair)
  • Perimeter of a Rhombus: 4 × Side
  • Perimeter of a Hexagon: 6 × Side

Curved Surface Area of a Cone: πrl, where 'r' is the base radius and 'l' is the slant height, l = r2+h2

  • Volume of a Cuboid: Base Area × Height = Length × Breadth × Height
  • Volume of a Cone: 13πr2h
  • Volume of a Sphere: 43πr3
  • Volume of a Hemisphere: 23πr3

Mensuration Class 8 NCERT Solutions (Exercise)

NCERT Solutions for Class 8 Maths Chapter 0 Mensuration: Topic 9.2 Area Of A Polygon

Question:(i) Divide the following polygons (Fig 11.17) into parts (triangles and trapezium) to find out its area.

1643871408929174521768792616438714362091745217687961

FI is a diagonal of polygon EFGHI NQ is a diagonal of polygon MNOPQR

Answer:

(i)

Area of polygon EFGHI = area of △ EFI + area of quadrilateral FGHI

Draw a diagonal FH.

Area of polygon EFGHI = area of △ EFI + area of △ FGH + area of △ FHI

ii)

Area of polygon MNOPQR = area of quadrilateral NMRQ+area of quadrilateral NOPQ

Draw diagonals NP and NR.

Area of polygon MNOPQR = area of △ NOP + area of △ NPQ + area of △ NMR + area of △ NRQ

Question:(ii) Polygon ABCDE is divided into parts as shown below (Fig 11.18). Find its area if AD=8cm, AH=6cm, AG=4cm, AF=3cm and perpendiculars BF=2cm, CH=3cm, EG=2.5cm. Area of Polygon ABCDE = area of ΔAFB+...

Area of ΔAFB=12×AF×BF=12×3×2=....

Area of trapezium FBCH
=FH×(BF+CH)2

=3×(2+3)2[FH=AH−AF]

Area of ΔCHD=12×HD×CH=.....
Area of ΔADE=12×AD×GE=...

So, the area of polygon ABCDE = ....

25252331745217687997

Answer:

25252331745217688039

Area of Polygon ABCDE= area of △AFB+ area of trapezium BCHF + area of △CDH +area of △AED=(12×AF×BF)+(FH×(BF+CH)÷2)+(12×HD×CH)+(12×AD×EG)=(12×3×2)+(3×(2+3)÷2)+(12×2×3)+(12×8×2.5)=3+7.5+3+10=23.5 cm2

Question:(iii) Find the area of polygon MNOPQR (Fig 11.19) if MP=9cm, MD=7cm, MC=6cm, MB=4cm, MA=2cm NA, OC, QD and RB are perpendiculars to diagonal MP.

16438715606441745217688090

Answer:
the area of polygon MNOPQR

= area of △MAN+ area of trapezium ACON+area of △CPO+ area of △MBR+ area of trapezium BDQR+area of △DPQ=(12×2×2.5)+(4(2.5+3)÷2)+(12×3×3)+(12×4×2.5)+(3(2.5+2)÷2)+(12×2×2)=2.5+11+4.5+5+6.75+2=31.75 cm2


Class 8 maths chapter 9 ncert solutions: Exercise: 9.1
Total Questions: 11
Page number: 105-106

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.16438715847691745217688130

Answer:

Area of table =

= 12 × sum of parallel sides × distance between parallel sides

= 12 ×(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 10cm 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, 12×(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.

16438716102761745217688165

Answer:

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

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

⇒ 120 = AB+48+17+40

⇒ AB = 15m

Area of trapezium

= 12 × sum of parallel sides × height

= 12×(40+48)×15

= 12×(88)×15

= (44)×15

= 660m2

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.

16438716423651745217688202

Answer:

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

The perpendiculars are 8 m and 13 m.

the area of the field

= 12×d×(h1+h2)

= 12×24×(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 rhombus =

12 × product of diagonals

= 12×7.5×12

= 7.5×6

= 45cm2

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 rhombus =

= 12 × product of diagonals

⇒ 24 = 12×8×x

⇒ x = 6cm

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

= 12× product of diagonals

= 12×45×30

= 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.

26261745217688234

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 = 12h(a+b)

So the equation becomes:

12100(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.


16438716754801745217688266

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 =

12 × 4 (11+5) = 2 × 16 = 32m2

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.

36363331745217688291

Find the area of this park using both ways. Can you suggest some other way of finding its area?
Answer:
Area of pentagonal park according to Jyoti's diagram:

=2( Area of trapezium )=2(12×152×45)=15×452=337.5 m2

Area of pentagonal park according to Kavita's diagram:
= Area of triangle + Area of square .

=12×15×15+15×15=112.5+225=337.5 m2

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.

3563561745217688316

Answer:

Area of opposite sections will be the same.

So the area of horizontal sections,

= 12×4(16+24) = 2(40) =80 cm2

And area of vertical sections,

= 12×8(20+28) = 4(48) = 96 cm2


NCERT Solutions for Class 8 Maths Chapter 9 Mensuration: Topic 9.4.1 Cuboid

Question:1 Find the total surface area of the following cuboids (Fig 11.31):

35631745217688340

Answer:

(i) Total surface area
= 2(lb + bh + hl)
= 2(6×4+4×2+6×2)
= 2(24+8+12)
= 2(24+8+12)
= 88 cm2

(ii) Total surface area

= 2(lb + bh + hl)

= 2(4×4+4×10+10×4)

= 2(16+40+40)

= 2(96)

= 192 cm2


NCERT Solutions for Class 8 Maths Chapter 9 Mensuration - Topic 9.4.2 Cube

Question: Find the surface area of cube A and the lateral surface area of cube B (Fig 11.36).

16438717003241745217688366

Answer:

Surface area of cube A = 6l2

= 6(10)2 = 600 cm2

Lateral surface area of cube B = 4l2

= 4(8)2 = 4×64 = 256 cm2


NCERT Solutions for Class 8 Maths Chapter 9 Mensuration: Topic 9.4.3 Cylinders

Question:1 Find the total surface area of the cylinders following figure

3651745217688392

Answer:

Total surface area of cylinder = 2πr (r + h)

(i) Area =

2π×14(14+8) = 2π×14(22) = 1935.22 cm2

(ii) Area = 2π×1(1+2) = 2π×1(3) =18.84 cm2


Class 8 maths chapter 9 ncert solutions: Exercise: 9.2
Total Questions: 10
Page number: 114

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

16593385803561745217688417

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 .

Area of such 100 suitcases will be 1382400 cm2

So length of tarpaulin cloth = 1382400296 cm=14400 cm or 144m

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

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?

4561745217688443

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 100m2 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

No. of cans of paint required

= 500100 = 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?

16593386954971745217688468

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 cylinder = 2×π×r×h

= 2×π×3.5×7

= 154 cm2

lateral surface area of 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 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

=422433=128 cm

So perimeter of 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

16438717841041745217688493

Answer:

Area in one complete revolution of roller = 2πrh

= 2π×0.42×1 = 2.638 m2

So, area of road = 2.638 m2 × 750 = 1979.20 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? 16438718066941745217688518
Answer:

Area of label = 2πrh

=2π×(7)×(16)=703.71 cm2

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


NCERT class 8 maths ch 11 question answer: Topic 9.5.1 Cuboid

Question:1 Find the volume of the following cuboids 16438718351321745217688544

Answer:

(i) Volume of a cuboid is given as:

Volume of cuboid = length×breadth×height,

So, Given that Length = 8cm, Breadth = 3cm, and height = 2 cm so,

Its volume will be = 8cm×3cm×2cm = 48 cm3

Also, for the Given Surface area of a cuboid 24m2 and height = 3 cm,
We can easily calculate the volume:

Volume = Surface area×height

So, Volume = 24×3100 = 0.72m3


NCERT Solutions for Class 8 Maths Chapter 9 Mensuration: Topic 9.5.2 Cube

Question: Find the volume of the following cubes

(a) with a side 4 cm (b) with a side 1.5 m

Answer:

(a) The volume of a cube with a side equal to 4 cm will be

Volume = Side × side × side
⇒ Volume = 4cm × 4cm × 4cm = 64 cm3 .

(b) When having side length equal to 1.5m
then,

Volume = 1.5m × 1.5m × 1.5m = 3.375 m3


NCERT Solutions for Class 8 Maths Chapter 9 Mensuration: Topic 9.5.2 Cube

Question:1 Find the volume of the following cylinders. 16438718640481745217688569

Answer:
(i) The volume of a cylinder given as = π × r2 × length

⇒ Given radius of cylinder = 7 cm and length of cylinder = 10 cm.

So, we can calculate the volume of the cylinder = π × (7cm)2 × 10cm
(Take the value of π=227 )

The volume of the cylinder = 1,540 cm3.

(ii) Given for the Surface area = 250 m2 and height = 2m

we have,

Volume of cylinder = 250 m2 × 2m = 500 m3


Class 8 maths chapter 9 ncert solutions: Exercise: 9.3
Total Questions: 8
Page number: 119-120

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, 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 the cylinders. Check whether the cylinder with greater volume also has greater surface area.

16556179528661745217688594

Answer:

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

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

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 × (72cm)2×14cm = 539 cm3

and Volume of B : π×r2×height=π×(142cm)2×7cm=1,078cm3 .

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×14cm2(142+7)=616cm2 .

and the surface area of cylinder A = 2×π×r(r+h)=2×π×7cm2×(72+14)=385cm2 .

The cylinder with greater volume also has greater surface area.

Question:3 Find the height of a cuboid whose base area is 180 cm 2 and volume is 900 cm 3.

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 is 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 6cm, which occupies the volume
= 6cm × 6cm × 6cm = 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
=97200216=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 cubiodal 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, 1m3 = 1000L

Then 108 m3 = 108,000 Litres

Time taken to fill the tank will be:
10800060=1800 minutes

1800 minutes = 180060=30 hours

NCERT Solutions for Class 8 Maths: Chapter Wise

Importance of NCERT Class 8 Maths Chapter 9 Mensuration Solutions

NCERT Class 8 Maths chapter 9 solutions discuss some important 2D and 3D figures like cubes, cuboids, and cylinders. Here are the important features of these solutions.

  • These solutions are well-explained and easy to understand. Students can analyze these solutions and try to solve the same types of problems on their own.
  • Important formulas and alternative solving methods are also been provided in this article.
  • The consistent practice of these solutions will reduce errors and increase the speed of problem-solving.
  • After analyzing these solutions, students will get the confidence to solve other problems from this chapter.
  • This solution article contains important links, like NCERT exemplar solutions, NCERT notes, Reference books, and the 2025-26 Maths syllabus.

NCERT Solutions for Class 8: Subject Wise

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

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.

Frequently Asked Questions (FAQs)

1. What are the important topics of NCERT syllabus chapter Algebraic Expressions and Identities ?

Addition and subtraction of the algebraic expression, multiplication of the algebraic expression, standard identities, and application of identities are important topics in this chapter.

2. Does CBSE class maths is tough ?

CBSE class 8 maths is not tough at all. It teaches a very basic and simple maths.

3. How many chapters are there in the CBSE class 8 maths ?

There are 16 chapters starting from rational number to playing with numbers in the CBSE class 8 maths.

4. Where can I find the complete solutions of NCERT for class 8 ?

Here you will get the detailed NCERT solutions for class 8 by clicking on the link.

5. Where can I find the complete solutions of NCERT for class 8 maths ?

Here you will get the detailed NCERT solutions for class 8 maths by clicking on the link.

6. Which is the official website of NCERT ?

ncert.nic.in is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.

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