NCERT Solutions Class 8 Maths Chapter 11

NCERT Solutions for Class 8 Mathematics Chapter 11 – Mensuration 

Classes 8, 9, and 10 Mathematics is significant for students as it forms the base for higher mathematical concepts, which will be introduced in Classes 11 and 12. Mathematics also plays an imperative role in most competitive examinations, especially for JEE aspirants.

Chapter 11 of Class 8 Mathematics is about Mensuration. In this chapter, students will learn about 2D shapes and 3D shapes, for example, triangle, rectangular, circle, quadrilateral, cupid, cube, cylinder, polygon, etc. They will do some measurements based on area, surface area, volume, and perimeter. The chapter will help them to learn about things that are directly connected to our real world. 

Extramarks is the go-to online learning platform for Mathematics for lakhs of students. Our experienced Mathematics teachers have prepared NCERT Solutions and chapter notes after a lot of research and past question paper analysis. Students can refer to our NCERT Solutions for Class 8 Mathematics Chapter 11 to learn and practise all Mensuration concepts.

Along with the NCERT Solutions, students will find  other study resources, including the CBSE syllabus, important questions with answer solutions, CBSE revision notes, etc. Extramarks provides comprehensive study materials for students to learn and prepare for their exams. 

Key Topics Covered In NCERT Solutions for Class 8 Mathematics Chapter 11

Mensuration is a branch of geometry that deals with the measurement of length, area, and volume.

Our NCERT Solutions for Class 8 Mathematics Chapter 11 will cover topics related to Mensuration. 


In this chapter, the students will learn about closed 2-D, 3-D-shaped figures like triangles, rectangular, circles, quadrilaterals, cubes, cylinders, and polygons. This chapter will also teach how to calculate the perimeter, area, volume, and surface area.

Let us Recall

In the previous chapter, we studied the shapes and their measurements, for example, 

Shape Area
Rectangular L X B 
Square L X L where L is the side of the square
Triangle ½ x B x H
Parallelogram B x H
Circle π R^2

Area of Trapezium 

To find out the area of the trapezium, we will divide the shape into two parts. First is the triangle part. The second is its rectangular part, and the third is its other triangle part. The sum of the area of both shapes is shown below. 

Area of Trapezium = 

Students will get more examples of calculating questions related to this topic  in our NCERT Solutions for Class 8 Mathematics Chapter 11.

Area of a General Quadrilateral

Drawing one of a generic quadrilateral’s diagonals will allow us to divide it into two triangles. By “triangulating,” we can find a formula for any common quadrilateral.

Area of quadrilateral ABCD

= (area of ∆ ABC) + (area of ∆ ADC)

= ( 1/ 2 AC × h1 ) + ( 1/ 2 AC × h2 )

= 1 / 2 AC × ( h1 + h2 )

= 1 / 2 d ( h1 + h2 ) where d denotes the length of diagonal AC

Area of Polygon

Area of polygon can be calculated with the same method we use to calculate the area of the quadrilateral. Split the shapes into triangles and calculate the area quickly.

By constructing two diagonals AC and AD the pentagon ABCDE is divided into three parts. So, area ABCDE = area of ∆ ABC + area of ∆ ACD + area of ∆ AED.

Many students find it difficult to remember the formulas for calculating the area of different shapes. In our NCERT Solutions for Class 8 Mathematics Chapter 11 we have given multiple examples which has helped many students remember the formulas better. Students can register on Extramarks website and get full access to our study materials.

Solid Shapes

In earlier classes, the students had learnt about the 2D and 3D shapes . In this section, we will see some solid shapes such as cuboids, cubes, cylinders, etc.

Our Mathematics faculty members have provided excellent visual demonstrations to show different formats of solid shapes in our NCERT Solutions for Class 8 Mathematics Chapter 11. This helps students to better visualise each shape and remember it properly. 

Surface area of Cube, Cuboid  and Cylinder

  • Surface area of cuboid

It has six rectangular surfaces. To find out the surface area of a cuboid, do a sum of all the surfaces.

Write the dimension of each side. We know that a cuboid has three pairs of identical faces. To the total area of all the faces of the box we calculate the total surface area of a cuboid which is area I + area II + area III + area IV +area V + area VI 

Surface area of a cuboid = h × l + b × l + b × h + l × h + b × h + l × b

So the total surface area = 2 (h × l + b × h + b × l) = 2(lb + bh + hl) where h, l and b are the height, length and width of the cuboid respectively.

  • Surface area of cube

Like a cuboid It has six rectangular surfaces. To find out the surface area of the cube, do a sum of all the surfaces.

The surface area of cube = Total Surface Area = 6(side)^2

  • Surface area of Cylinder

It covers two parts of the cylinder. The first one is its top/bottom face, and the second one is its curved surface. To find out the cylinder’s surface area, we sum up both of the parts areas.

Surface Area of Cylinder = Curved Surface + Area of Circular top/base


Surface Area of Cylinder = 2πr (r + h)

Solved examples are given in our NCERT Solutions for Class 8 Mathematics Chapter 11. By practising these examples and related exercises, students will be able to infer different steps for calculating the surface areas of various shapes.

Volume of cube, cuboid and cylinder

  • Volume of cuboid

It is a 3D figure. It has length, breadth, and height. To find out the volume of the cuboid, we will multiply all three dimensions. 

The volume of Cuboid = L x B x H

  • Volume of cube

The cube is a special type of a cuboid, where length, breadth and height are all equal.

Hence, volume of cube = L * L * L

  • Volume of cylinder

It is also a 3d figure. We can determine the volume of any cylinder by multiplying its base area by its height.

The volume of the Cylinder =  πr2 * h

Similar to the surface area examples, our NCERT Solutions for Class 8 Mathematics Chapter 11 also contains a lot of solved examples for calculating volume. These solutions also have a lot of topic-specific exercises which students can practise to become confident with the concepts.

Volume and capacity

  • Volume is the amount of a solid that covers the space
  • Capacity is the quantity of solid that a container can hold.

NCERT Solutions for Class 8 Mathematics Chapter 11: Exercise & Answer Solutions

Extramarks study materials contain a lot of self-practice exercises. These exercises cover different formats of questions from the NCERT Textbook and NCERT Exemplars.

All the questions will make concepts clearer for the students. The questions also cover the past years’ questions from the CBSE board exams. The step-by-step detailed workings given for each answer help students to revise the concepts and also get familiar with the various concepts covered in this chapter, Mensuration.

Click on the below links to view exercise-specific questions and solutions covered in our NCERT Class 8 Mathematics Chapter 11 Solutions:

Along with Class 8 Mathematics solutions, students can explore NCERT Solutions on our Extramarks website for all other primary and secondary classes.

NCERT Exemplar Class 8 Mathematics Chapter 11

NCERT Exemplars are a golden resource that contains many exercises with different question formats. There are many questions to practice related to the chapter, and it covers all the essential topics and concepts of NCERT Class 8 Mathematics Chapter 11. The questions are NCERT-based advanced level questions. Once students become confident of solving questions from these books, they can quickly solve their NCERT textbook questions and hence will start scoring good marks in their examinations.

Students can perform well in CBSE Board and other competitive exams by referring to NCERT Exemplar and NCERT Solution.

Key Features of NCERT Solutions for Class 8 Mathematics Chapter 11

Extramarks study materials are widely trusted by all the stakeholders – students, parents, and teachers. The main reason for this trust is our high-quality research-oriented study materials that are prepared by experienced subject matter experts.

Some key features of our NCERT Solutions for Class 8 Mathematics Chapter 11. 

  • It covers all the concepts and topics of Mensuration.. So, students can confidently rely on our NCERT solutions for complete chapter revision.
  • It also provides many questions to practice on all the topics and concepts.
  • The solutions are built by taking the CBSE syllabus and strictly following the NCERT guidelines.
  • We have given a lot of visual demonstrations about the shapes and their parameters for students to understand these concepts quickly and make learning fun for them.

Q.1 A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?


Perimeter of square=4×side = 4×60 m = 240 mPerimeter of rectangle=2(Length+Breadth) = 2(80 m+Breadth) = 160 m+2×Breadth

Perimeter of the square=Perimeter of the rectangle 160 m+2×Breadth =240 mBreadth of the rectangle=(802)m=40 mArea of square=(Side)2=(60 m)2=3600 m2Area of rectangle=Length×Breadth =(80×40)m2 =3200 m2Therefore, the area of the square field is larger thanthe area of the rectangular field.

Q.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.


Area of the square plot=(25 )2m2=625 m2Area of the house=(15 m)×(20 m)=300 m2Area of the remaining portion=(Area of square plot)(Area of the house) =625 m2300 m2 = 325 m2Total cost of developing a garden around the house at the rate of Rs 55 per m2=(55×325) = 17,875

Q.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].


Length of the rectangle=[20(3.5 + 3.5)]metres=13 mCircumference of 1 semicircular part=πr=227×3.5 =11 m.

Perimeter of the garden=AB+Length of both semicircular regions+CD = 13 m+22 m+13 m = 48 mArea of the garden=Area of rectangle+2×Area of two semicircular regions =[(13×7)+2×12×227×(3.5)2]m2 =(91+38.5)m2 =129.5m2Area and perimeter of the garden are 129.5m2 and 48 m.

Q.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? (If required you can split the tiles in whatever way you want to fill up the corners).


Area of parallelogram=Base×HeightHence, area of one tile=24 cm×10 cm=240 cm2Required number of tiles=Areaofthe floorArea of eachtile=1080 m2240cm2 Since1m=100cm 1m2=10000cm21080 m2=10000×1080 cm2 =10,800,000cm2Hence,the required number of tiles =10,800,000  cm2240cm2 = 45000

Q.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.


(a) Radius of semi-circular part={2.82}cm =1.4 cm Perimeter of the given figure=2.8 cm + πr =2.8 cm+(227×1.4)cm =2.8 cm+4.4cm =7.2cm

(b) Radius of semi-circular part={2.82}cm=1.4 cm Perimeter of the given figure=1.5 cm+2.8 cm+1.5 cm+π(1.4 cm) =5.8 cm+227×1.4 cm =5.8 cm+4.4cm =10.2cm

(c)Radius of semi-circular part={2.82}cm=1.4 cm Perimeter of the given figure=2 cm+πr+2 cm =4 cm+227×1.4 cm =4 cm+4.4cm =8.4cm

Thus, the ant will have to take a longer round for the food piece ( b ), because the perimeter of the figure given in option ( b ) is the greatest among all. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@BD3F@

Q.6 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.


Area of trapezium=12(Sum of parallel sides)×(Distances between parallel sides) =[12(1+1.2)(0.8)]m2 =0.88m2

Q.7 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.


It is given that,area of trapezium=34 cm2and height=4 cmLet the length of one parallel side be a. We know that,Area of trapezium=12(Sum of parallel sides)×(Distances between parallel sides)

34 cm2=12(10cm+a)×4 cm34 cm2=2(10cm+a)17 cm=10cm+aa=17 cm10cm=7 cmThus, the length of the other parallel side is 7 cm.

Q.8 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.


Length of the fence of a trapezium shaped field ABCD=AB+BC+CD+DA120 m=AB+48 m+17 m+40 mAB=120 m105 m=15 m

Area of the field ABCD=12(AD+BC)×AB                            =12(40+48)×15m2                            =12×88×15m2                            =660m2Area of the field ABCD=660m2

Q.9 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.


It is given that,Length of the diagonal,d= 24 mLength of the perpendiculars,h1and h2, from the opposite verticesto the diagonal are 8 m and 13 mArea of the quadrilateral =12dh1+h2 =12×24×13 m+8 m =12×24×21 =252 m2Thus, the area of the field is 252 m2.

Q.10 The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.


Area of rhombus = 1 2 ( Product of its diagonals ) Therefore, area of the given rhombus= 1 2 ( 7.5×12 ) = 45 cm 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@9114@

Q.11 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.


Let the other diagonal of the rhombus be x.Since rhombus is also a parallelogram.Therefore, area of the given rhombus=Base×Height =5 cm×4.8 cm =24 cm2Also, area of rhombus =12(Product of its diagonals)24 cm2=12(8 cm×x)x=24×28cm=6 cm The length of the other diagonal of the rhombus is 6 cm andarea of the given rhombus is 24 cm2.

Q.12 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.


Area of rhombus = 12Product of its diagonalsArea of each tile = 12×45×30cm2 = 675 cm2Area of 3000 tiles = 675×3000cm2                        = 2025000 cm2                       = 202.5 m2the total cost of polishing the floor, if the cost per m2 is Rs 4= 4×202.5= 810Thus, the cost of polishing the floor is Rs 810.

Q.13 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.


Let the length of the field along the road belm. Hence, the length of the field along the river will be 2lm. Area of trapezium = 1 2 ( Sum of parallel sides ) ( Distance between the parallel sides ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@E24E@

10500 m2=12(l+2l) (100 m)3l=(2×10500100)m3l=210 ml =70 mThus, length of the field along the river=(2×70) m=140 m

Q.14 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.


Side of regular octagon = 5 cm
Area of trapezium ABCH = Area of trapezium DEFG

Area of trapezium ABCH = 12×4×11+5m2                               = 12×4×16m2                               =  32 m2

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
Therefore, the area of octagonal surface is 119 m2

Q.15 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?


Jyoti way of calculating area:

Area of pentagon=2( Area of trapezium ABCF ) =[ 2× 1 2 ×(15+30)( 15 2 ) ] m 2 =337.5 m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@78C6@

Kavita’s way of calculating area:

Area of pentagon ABCDE=Area of ΔABE+Area of square BCDE =[ 1 2 ×15×( 3015 )+ ( 15 ) 2 ] m 2 =[ 1 2 ×15×15+225 ] m 2 =337.5 m 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@9817@

No, there is no other way to find its area.

Q.16 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.


It is given that the width of each section is same.


LM = BM = CN = NE = OF = OH = PK = PI

CH = CN + NO+ OH

28 = 2CN+20

2CN = 28 – 20

2CN = 8

CN = 4 cm

Hence, LM = BM = NE = OF = OH = PK = PI = 4 cm

Area of section NDGO = Area of section AMPJ =[12(20+28)(4)]cm2 =96 cm2Area of section NDGO = Area of section AMPJ=96 cm2Also,Area of section AMND=Area of section PJGO =[12(16+24)(4)] =80 cm2Area of section AMND=Area of section PJGO=80 cm2

Q.17 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?


Jyoti’s way of calculating area:

Area of pentagon=2(Area of trapezium ABCF) =[2×12×(15+30)(152)]m2 =337.5 m2

Kavita’s way of calculating area:

Area of pentagon ABCDE=Area of ΔABE+Area of square BCDE =[12×15×(3015)+(15)2]m2 =[12×15×15+225]m2 =337.5 m2

No, there is no other way to find its area.

Q.18 There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to


We know that,Total surface area of the cuboid=2 (lb+bh+hl)Total surface area of cuboid=[2{(60)(40)+(40)(50)+(50)(60)}]cm2 =[2(2400+2000+3000)] cm2 =[2(2400+2000+3000)](2×7400) cm2 =[2(2400+2000+3000)]14800 cm2Total surface area of the cube=6 (l)2 =6 (50 cm)2 =15000 cm2Since the total surface area of cuboid is less thanthe total surface area of cube,therefore, the cuboidal box willrequire lesser amount of material.

Q.19 A suitcase with 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?


A suitcase is in the shape of a cuboid.Total surface area of a cuboid= 2(lb+bh+hl)Hence,Total surface area of suitcase=2[(80)(48)+(48)(24)+(24)(80)] cm2 =2[3840 + 1152 + 1920] cm2 =13824 cm2Total surface area of 100 suitcases=(13824 × 100)cm2 =1382400 cm2Required tarpaulin=Length×Breadth 1382400 cm2 =Length×96 cmLength=(138240096)cm=14400 cm=144 mThus, 144 m of tarpaulin is required to cover 100 suitcases.

Q.20 Find the side of a cube whose surface area is 600 cm2.


Given that, surface area of cube=600 cm2Let the length of each side of cube bex.Surface area of cube=6 (Side)2

600 cm2=6x2 x2=100 cm2 x=10 cm The side of the cube is 10 cm.

Q.21 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?


Length of the cabinet = 2 mBreadth of the cabinet = 1 mHeight of the cabinet = 1.5 m

Area of the cabinet that was painted=2h(l+b)+lb = [2×1.5×(2+1)+(2)(1)] m2 = [3(3)+2] m2 = (9+2) m2 = 11 m2 Area of the cabinet that was painted is 11 m2.

Q.22 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?


Given that,Length=15 m breadth=10 mheight=7 mArea of the hall to be painted=(Area of the walls)+(Area of the ceiling)

= 2h(l+b)+lb=2(7)(15+10)+15×10 m2= 14(25)+150 m2= 500 m2It is given that 100 m2area can be painted from each can.Number of cans required to paint an area of 500 m2=500100=5Hence, 5 cans are required to paint the walls and the ceiling of the cuboidal hall.

Q.23 Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area?


Both the figures have the same heights and the difference between the two figures is that one is a cylinderand the other is a cube.

Now,Lateral surface area of the cube=4l2=4(7 cm)2=196 cm2Lateral surface area of the cylinder=2πrh sq. units =(2×227×72×7)cm2 =154 cm2Hence, the cube has larger lateral surface area.

Q.24 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?


Total surface area of cylinder=2πr(r+h) sq. units =[2×227×7(7+3)]m2 =440 m2Thus, 440 m2 sheet of metal is required.

Q.25 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.


Surface area of hollow cylinder = Area of rectangular sheet4224 cm2=Length×33 cmLength =422433=128 cm 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 cmPerimeter of the rectangular sheet=322 cm.

Q.26 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.


Given Diameter= 84 cmr = 42 cm=42100 mLength=1 m

In one revolution, the roller will cover an area equal to itslateral surface area.

Lateral surface area of cylinder=2πrh =2×227×42 cm×1 m =2×227×42 100m×1 m =264100 m2In 750 revolutions, area of the road covered=750×264100 m2 = 1980 m2

Q.27 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?


Height of the label=20 cm2 cm2 cm=16 cmDiameter=14 cmRadius of the label=142 cm= 7 cmLabel is in the form of a cylinder having its radius and height as 7 cm and 16 cm.

Area of the label=2πrh =(2×227×7×16)cm2 =704 cm2Area of the label=704 cm2

Q.28 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.


(a) We will find the volume.

(b) We will find the surface area.

(c) We will find the volume.

Q.29 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?


The heights and diameters of these cylinders A and B areinterchanged.If measures of ‘r’ and ‘h’ are same, then the cylinder withgreater radius will have greater volume.Radius of cylinder A =72cm

Radius of cylinder B =142cm=7 cmAs 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=πr2h=(227×72×72×14)cm3=539 cm3Volume of cylinder B=πr2h=(227×7×7×7)cm3=1078 cm3Hence,volume of cylinder B is greater.

Surface area of cylinder A=2πr(h+r)=[227×72(72+14)]cm2=[22×(7+282)]cm2=385 cm2Surface area of cylinder B=2πr(h+r)=[227×7(7+7)]cm2=[44×14]cm2=616 cm2Thus, the surface area of cylinder B is also greater thanthe surface area of cylinder A.

Q.30 Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3?


Base area of the cuboid=180 cm2 Length×Breadth=180 cm2

Volume of cuboid=Length×Breadth×Height 900 cm3=180 cm2×Height Height=900180=5 cmThus, the height of the cuboid is 5 cm.

Q.31 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?


Volume of cuboid=60 cm×54 cm×30 cm=97200 cm3Side of the cube=6 cmVolume of the cube=(6)3cm3=216 cm3Required number of cubes =Volume of the cuboidVolume of the cube =97200 cm3216 cm3=450450 cubes can be placed in the given cuboid.

Q.32 Find the height of the cylinder whose volume is 1.54 m3 and diameter of the base is 140 cm?


Given,Volume of cylinder=1.54 m3Diameter of the base = 140 cmRadius of the base= (1402) cm=70 cm = 70100mVolume of cylinder=πr2h1.54 m3=227×70100×70100×hh=(1.54×10022×7)m=1 mThus, the height of the cylinder is 1 m.

Q.33 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?


Radius of cylinder = 1.5 mLength of cylinder = 7 mVolume of cylinder=πr2h =227×1.5×1.5×7 =49.5 m3We know that ,1m3= 1000 LThe quantity of milk in liters=(49.5 × 1000)L = 49500 LTherefore, 49500 L of milk can be stored in the tank.

Q.34 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?


Let initially the edge of the cube be ‘x’.(i) Initial surface area=6x2If each edge of the cube is doubled, then it becomes 2x.

New surface area=6(2x)2 =24x2 =4×6x2Hence, the surface area will be increased by 4 times.(ii) Initial volume of the cube =x3When each edge of the cube is doubled, it becomes 2x.New volume=(2x)3=8x3=8×x3 The volume of the cube will be increased by 8 times.

Q.35 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.


Volume of cuboidal reservoir=108 m3 =(108×1000) L =108000 LIt is given that water is being poured at the rate of 60 L per min.i.e.(60×60) L=3600 L per hourRequired number of hours=1080003600= 30 hoursIt will take 30 hours to fill the reservoir.

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