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NCERT Solutions
Class 8
Maths
Chapter 9 Mensuration

NCERT Solutions Class 8 Maths Chapter 9 Mensuration

Mensuration is the ninth chapter of Class 8 Maths. It covers various topics, including an introduction, area of polygon, solid shapes, surface area of cube, cuboid and cylinder, volume of cube, cuboid and cylinder, volume and capacity.

The chapter begins with an introduction to mensuration in section 9.1, followed by significant topics in sections 9.2, 9.3, 9.4, and 9.5.

1.0Download NCERT Class 8 Maths Chapter 9 Mensuration Solutions

ALLEN'S Experts lucidly curated the class 8 mensuration solution to improve the students' problem-solving abilities. For a more precise idea about Mensuration NCERT solutions, students can download the below NCERT solutions class 8 Maths chapter 9 pdf solution.

NCERT Solutions Class 8 Maths Chapter 9 Mensuration

2.0NCERT Solutions Class 8 Maths Chapter 9 : Important Topics

Introduction: Overview of basic mensuration concepts and their applications.

Area of Polygon: Calculating the area of various polygonal shapes.

Solid Shapes: Understanding 3D figures and their properties.

Surface Area of Cube, Cuboid, and Cylinder: Finding the surface area of these 3D shapes.

Volume of Cube, Cuboid, and Cylinder: Calculating the volume of these solids.

Volume and Capacity: Differentiating between volume and capacity and solving related problems.

3.0Class 8 Maths Chapter 9 Mensuration: Exercise Solutions

There are three exercises in chapter 9 (Mensuration) of class 8 Maths. Students can find the split below:

Class 8 Maths Chapter 9 Exercises

No. of Questions

Exercise 9.1

11

Exercise 9.2

10

Exercise 9.3

8

Explore Mensuration and learn how to solve various problems only on NCERT Solutions For Class 8 Maths.

4.0Key Features of Class 8 Maths Chapter 9 NCERT Solutions – Mensuration

  • Practical Application: Helps students understand real-life measurements such as area, volume, and surface area.
  • Foundation for Advanced Topics: Builds a strong base for geometry and advanced math topics in higher classes.
  • Improves Problem-Solving Skills: Enhances logical thinking and analytical abilities through diverse problem-solving exercises.
  • Boosts Confidence: Familiarizes students with various shapes and figures, making complex concepts easier to grasp.

5.0NCERT Questions with Solutions for Class 8 Maths Chapter 9 - Detailed Solutions

Exercise : 9.1

  • 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 .

The shape of the top surface of a table is trapezium.

  • Sol. Area of top surface of a table = Area of the trapezium =21​×( sum of parallel sides )×( distance between them) =[21​×(1+1.2)×0.8]m2 =(21​×2.2×0.8)m2 =(1.1×0.8)m2=0.88 m2
  • 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 other parallel side. Sol. Let the required side be xcm . Then, area of the trapezium =[21​×(10+x)×4]cm2=2(10+x)cm2 But, the area of the trapezium =34 cm2 (given) ∴2(10+x)=34 ⇒10+x=17 ⇒x=17−10=7 Hence, the other side =7 cm
  • 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 .

 A B is perpendicular to the parallel sides AD and BC .

  • Sol. Let ABCD be the given trapezium in which BC=48 m,CD=17 m and AD=40 m

the given trapezium

  • Through D, draw DL ⊥ BC. Now, BL = AD = 40 m and LC=BC−BL=(48−40)m=8 m Applying Pythagoras theorem in right △ DLC, we have DL2=DC2−LC2=172−82=289−64=225 ⇒DL=225​=15 m Now, area of the trapezium ABCD =21​×(BC+AD)×DL =21​×(48+40)×15 m2 =(44×15)m2=660 m2.
  • 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.

Find the area of the field.

  • Sol. Let ABCD be the given quadrilateral in which BE⊥AC and DF⊥AC. It is given that AC=24 m,BE=8 m and DF=13 m.

area of quad

  • Now, area of quad. ABCD = area of △ABC+ area of △ACD =21​×AC×BE+21​×AC×DF =(21​×24×8+21​×24×13)m2 =(12×8+12×13)m2 =(96+156)m2=252 m2
  • The diagonals of a rhombus are 7.5 cm and 12 cm . Find its area. Sol. Area of a rhombus =21​× (product of diagonals) =(21​×7.5×12)cm2=45 cm2
  • 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. Sol. Let ABCD be a rhombus of side 5 cm and whose altitude DE=4.8 cm. Also, one of its diagonals, BD=8 cm. Area of the rhombus ABCD =2× Area (△ABD)=2×21​×AB×DE =(5×4.8)cm2=24 cm2 Also, area of the rhombus ABCD =21​×AC×BD ⇒21​×AC×8=24 ⇒AC=6 cm

area of the floor

  • Hence, the other diagonal is 6 cm .
  • 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 . Sol. Area of the floor =3000× Area of one tile =3000×21​×45×30 cm2 =1500×45×30 cm2 =100×1001500×45×30​ m2=202.5 m2 Cost of polishing the floor ₹ 4 per m 2 =₹(4×202.5)=₹810
  • Mohan wants to buy a trapezium shaped field. It's 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.

find the length of the side along the river.

  • Sol. Let the parallel sides of the trapezium shaped field be x m and 2 x m . Then, its area. =21​(x+2x)×100 m2 =50×3xm2=150xm2 But, it is given that the area of the field is 10500 m2. ∴150x=10500 ⇒x=15010500​=70 ∴ The length of the side along the river is 2×70, i.e., 140 metres.
  • 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.

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.

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

Area of the octagonal surface ABCDEFGH

  • =[21​(5+11)×4+11×5+21​(11+5)×4]m2 =(32+55+32)m2=119 m2
  • 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? Sol. Taking Jyoti's diagram : Area of the pentagonal shaped park

There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways.


  • =2× Area of trapezium ABEF =2×21​×(15+30)×215​ m2 =(45×215​)m2=337.5 m2 Taking Kavita's diagram : Area of the pentagonal shaped park = Area (square ACDF) + Area ( △DEF ) =15×15+21​×15×15 =225+2225​=225+112.5=337.5 m2


Area of the pentagonal shaped park


  • 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.

Diagram of the adjacent picture frame has outer dimensions 24 \mathrm{~cm} \times 28 \mathrm{~cm} and inner dimensions 16 \mathrm{~cm} \times 20 \mathrm{~cm}. Find the area of each section of the frame, if the width of each section is same.

  • Sol. Width of each section =4 cm. Area of trapezium ABQP = Area of trapezium RCDS =21​×4×(16+24)cm=80 cm2 Area of trapezium BQRC= Area of trapezium APSD =21​×4×(28+20)cm2 =96 cm2

Exercise: 9.2

  • There are two cuboidal boxes as shown in the figure below. Which box requires the lesser amount of material to make?

cuboidal box

  • 60 cm

cuboidal box 50cm X 50cm

  • 50 cm (b) Sol. Total surface area of first box =2(ℓ b+bh+ℓh) =2(60×40+40×50+60×50)cm2 =200(24+20+30)cm2 =200×74 cm2=14800 cm2 Total surface area of second box =6( Edge )2=6×50×50 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.
  • A suitcase of 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? Sol. 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 =(961382400​)cm=14400 cm =144 m Thus, 144 m of tarpaulin is required to cover 100 suitcases.
  • Find the side of a cube whose surface area is 600 cm2. Sol. 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.
  • 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.

Rukhsar painted the outside of the cabinet of measure 1 \mathrm{~m} \times 2 \mathrm{~m} \times 1.5 \mathrm{~m}. How much surface area did she cover if she painted all except the bottom of the cabinet.

  • Sol. Here ℓ=2 m, b=1 m and h=1.5 m Area to be painted =2bh+2ℓ h+ℓb =(2×1×1.5+2×2×1.5+2×1)m2 =(3+6+2)m2=11 m2
  • 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? Sol. Here ℓ=15 m, b=10 m and h=7 m Area to be painted =2bh+2ℓ h+ℓb =(2×10×7+2×15×7+15×10)m2 =(140+210+150)m2=500 m2 Since each can of paint covers 100 m2, therefore number of cans required =100500​=5.
  • Describe how the two figures given below are alike and how they are different. Which box has larger lateral surface area?

Describe how the two figures given below are alike and how they are different. Which box has larger lateral surface area?

  • Sol. Similarity between 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 cube =4(7)2 =196 cm2 Lateral surface area of cylinder =2×722​×27​×7 =154 cm2 Hence, the cube has larger lateral surface area.
  • 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 ? Sol. Here, r=7 m and h=3 m. Sheet of metal required to make a closed cylinder = Total surface area of the cylinder =(2πrrh+2πr2) sq. units. =(2×722​×7×3+2×722​×7×7)m2=(132+308)m2=440 m2
  • 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. Sol. 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 Length =33 cm4224 cm2​=128 cm 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
  • 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 .

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 .

  • Sol. In one revolution, the roller will cover an area equal to its lateral surface area. Thus, in 1 revolution, area of the road covered =2π rh =2×722​×42 cm×1 m =2×722​×10042​ m×1 m=100264​ m2 In 750 revolutions, area of the road covered =(750×100264​)m2=1980 m2
  • 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.

A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm .

  • Sol. 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 area of a cylinder of radius 7 cm and height (20-4) cm i.e., 16 cm . This curved surface area =(2×722​×7×16)cm2 =704 cm2

Exercise : 9.3

  • 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. Sol. (a) Volume (b) Surface area (c) Volume
  • 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 by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?

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

  • Sol. The heights and diameters of these cylinders A and B are interchanged. We know that, Volume of cylinder =πr2h Radius of cylinder A=27​ cm Radius of cylinder B(214​)cm=7 cm 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=πr2 h =(722​×27​×27​×14)cm3=539 cm3 Volume of cylinder B=πr2h =(722​×7×7×7)cm3=1078 cm3 Volume of cylinder B is greater. = Surface area of cylinder A=2πr(r+h) =[2×722​×27​(27​+14)]cm2 =[22×(27+28​)]cm2=(22×235​)cm2 =385 cm2 Surface area of cylinder B =2πr(r+h) =[2×722​×7×(7+7)]cm2 =(44×14)cm2 =616 cm2 Thus, the surface area of cylinder B is also greater than the surface area of cylinder A.
  • Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3 ? Sol. Volume of the cuboid =900 cm3 ⇒ (Area of the base) × Height =900 cm3 ⇒180× Height =900 ⇒ Height =180900​=5 Hence, the height of the cuboid is 5 cm .
  • A cuboid is of dimensions 60 cm×54 cm× 30 cm . How many small cubes with side 6 cm can be placed in the given cuboid? Sol. Volume of cuboid =(60×54×30)cm3 =97200 cm3 Volume of cube =(6×6×6)cm3 =216 cm3 No. of cubes = Volume of cube  Volume of cuboid ​ =21697200​=450
  • Find the height of the cylinder whose volume is 1.54 m3 and diameter of the base is 140 cm ? Sol. Let h be the height of cylinder whose radius, r=2140​ cm=70 cm=10070​ m=0.7 m and volume =1.54 m3. ∵ volume =πr2h ⇒πr2 h=1.54 ⇒722​×0.7×0.7×h=1.54 ⇒h=22×0.7×0.71.54×7​=1 m Hence, the height of cylinder is 1 metre.
  • 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.

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.

  • Sol. Quantity of milk that can be stored in the tank = Volume of the tank =πr2h, where r=1.5 and h=7 m =(722​×1.5×1.5×7)m3=49.5 m3 =(49.5×1000) litres [∵1 m3=1000ltr.] =49500 litres.
  • 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? Sol. 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 =6×4x2=24x2 ⇒ 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.
  • 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.

water is pouring into a cuboid reservoir

  • Sol. Volume of the reservoir =108 m3 =108×1000 litres =108000 litres Since water is pouring into reservoir at the rate of 60 litres per minute. ∴ Time taken to fill the reservoir =60×60108000​ hours =30 hours

NCERT Solutions for Class 8 Maths Other Chapters:-

Chapter 1: Rational Numbers

Chapter 2: Linear Equations in One Variable

Chapter 3: Understanding Quadrilaterals

Chapter 4: Data Handling

Chapter 5: Squares and Square Roots

Chapter 6: Cubes and Cube Roots

Chapter 7: Comparing Quantities

Chapter 8: Algebraic Expressions and Identities

Chapter 9: Mensuration

Chapter 10: Exponents and Powers

Chapter 11: Direct and Inverse Proportions

Chapter 12: Factorisation

Chapter 13: Introduction to Graphs


CBSE Notes for Class 8 Maths - All Chapters:-

Class 8 Maths Chapter 1 - Rational Numbers Notes

Class 8 Maths Chapter 2 - Linear Equations In One Variable Notes

Class 8 Maths Chapter 3 - Understanding Quadrilaterals Notes

Class 8 Maths Chapter 4 - Data Handling Notes

Class 8 Maths Chapter 5 - Squares and Square Roots Notes

Class 8 Maths Chapter 6 - Cubes and Cube Roots Notes

Class 8 Maths Chapter 7 - Comparing Quantities Notes

Class 8 Maths Chapter 8 - Algebraic Expressions and Identities Notes

Class 8 Maths Chapter 9 - Mensuration Notes

Class 8 Maths Chapter 10 - Exponents and Powers Notes

Class 8 Maths Chapter 11 - Direct and Inverse Proportions Notes

Class 8 Maths Chapter 12 - Factorisation Notes

Class 8 Maths Chapter 13 - Introduction to Graphs Notes

Frequently Asked Questions

ALLEN's experts provide step-by-step answers to NCERT Solutions for Class 8 Maths Chapter 9. This helps the students learn all the concepts in detail and clear their doubts. Regular practice also helps them score high on math exams.

In Chapter 9, Mensuration of Class 8 Maths, students will study: Introduction Area of Polygon Solid Shapes Surface Area of Cube, Cuboid, and Cylinder Volume of Cube, Cuboid, and Cylinder Volume and Capacity

Chapter 9 of class 8th Maths can be challenging for some students as it involves understanding and applying formulas for area, surface area, and volume. However, with consistent practice and a clear grasp of the basic concepts, it becomes manageable and easier to solve.

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