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NCERT Solutions for Class 8 Maths Chapter 9 Mensuration

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NCERT Solutions for Class 8 Maths Chapter 9 Mensuration - Free PDF Download

Class 8 students have a lot on their plate in terms of academics. A subject like Maths requires them to spend enough time in solving assignments and preparing for exams. Vedantu’s team of scholars understand the needs and understanding level of Class 8 students and have designed NCERT Solution for Class 8 Maths Chapter 9 keeping all of this in mind. NCERT Solution Class 8 Maths Chapter 9 Mensuration Solution is entirely based on the latest CBSE curriculum. It will be able to clarify all the doubts that you might have in the Mensuration chapter. The problems in the Class 8 Maths Chapter 9 are well explained with diagrams and step-by-step explanations to make it simple for you to follow the solution.

Class:

NCERT Solutions For Class 8

Subject:

Class 8 Maths

Chapter Name:

Chapter 9 - Mensuration

Content Type:

Text, Videos, Images and PDF Format

Academic Year:

2024-25

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes


You can also download Class 8 Science to help you to revise complete syllabus ans score more marks in your examinations.


Access Exercise wise NCERT Solutions for Chapter 9 Maths Class 8

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Access NCERT Solutions for Class 8 Maths Chapter 9 – Mensuration

Exercise 9.1

Q1. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?


A square and a Rectangular field


Ans: Let Side of a Square be ‘a’.

         Length and Breadth of Rectangle are ‘l’ and ‘b’ respectively.

         Perimeter of Square =4(side of square) = 4(a)

                                                                             = 4(60m) = 240m         

         Perimeter of Rectangle = 2(Length+Breadth) = 2(l + b)

                                                                                  = 2(80m+ b)

         Now, It is given that the perimeter of the square and the perimeter of the rectangle are equal.

         So, 240m2(80m+ b)

                240m= 160m+2b

                80m= 2 b

                40m= b

        Area of square = (side)2= (60m)2=3600m2

        Area of Rectangle = Length x Breadth = (80x40)m2 = 3200m2

        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.55perm2?


Square shape plot


Ans: Area of the Square Plot = (side)2= (25m)2= 625m2

          Area of the House = Length x Breadth = 15m x 20m= 300m2

          Now, Area of the remaining portion = Area of the square plot – Area of the house

                                                                   =  625m2- 300m2

                                                                   =   325m2

          The cost of developing the garden around the house is Rs.55perm2.

          Therefore, total cost of developing the area325m2 is Rs.(55 x 325) = Rs.17875.


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)m)?


Combination of a rectangle and two semicircles


Ans: As we have given that length of rectangle = [20 - (3.5 + 3.5)]m

                                                                                  =  [20 - 7]m= 13m

         Breadth = 7m.

         Now, we have to find the circumference of both semi circles.

         As, Diameter = 7m, so, Radius(r) = 72m= 3.5m

         Circumference of one semicircle = π r= 227(3.5)m= 11m

         Circumference of both circles = 2 x 11m= 22m


Combination of a rectangle ABCD and two semicircles


         Now, Perimeter of the garden = AB + CD + Length of both semi-circular regions AD & BC

                                                        = (13 + 13 + 22)m

                                                        =  48m

         Area of the garden = Area of the rectangle + 2xArea of two semi-circular regions

                                       = (Length x Breadth) + 2 x 12π r2

                                       = [(13 x 7) + 2 x (12 x 227 x 3.5 x 3.5)]m2

                                       = (91 + 38.5)m2

                                       = 129.5m2


Q4. A flooring tile has the shape of a parallelogram whose base is 24cmand the corresponding height is 10cm.How many such tiles are required to cover a floor of area 1080m2? (If required you can split the tiles in whatever way you want to fill up the corners).  

Ans: Given that the Base of a parallelogram is 24cm and Height of a parallelogram is 10cm.

          Therefore, Area of Parallelogram = Base x Height

                                                               = (24 x 10)cm2

                                                               = 240cm2.

          Therefore, Area of parallelogram = Area of one tile.

          Now, we have to find the number of tiles.

          Given that Area of floor = 1080m2

          So, Number of tiles = AreaoffloorAreaofonetile

                                         = 1080m2240cm21080 x 10000240cm2           (1m=100cm)

                                         = 45000tiles.

          Hence, 45000tiles are required to cover a floor of area 1080m2


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 =2π r, where r is

the radius of the circle.

Ant


Semicircle

Combination of two shapes


Combination of a triangle and semicircle


Ans: (a) Diameter(d) = 2.8cm, so radius(r) = 2.82cm= 1.4cm

                Therefore, Perimeter = d + π r

                                                  = 2.8cm + 227(1.4cm)

                                                  = [2.8 + (0.2 x 22)]cm

                                                  = (2.8 + 4.4)cm

                                                  = 7.2cm

          (b) Diameter( d) = 2.8cm, so radius(r) = 2.82cm= 1.4cm

                Therefore, Perimeter = 1.5cm + 2.8cm + 1.5cm + π (1.4cm)

                                                     = 5.8cm + 227(1.4cm)

                                                     = 5.8cm + 4.4cm

                                                     = 10.2cm.

           (c) Radius(r) = 2.82cm= 1.4cm

                Perimeter = 2cm + π r + 2cm

                                =  4cm + 227(1.4cm)

                                = 4cm + 4.4cm

                                = 8.4cm.

Thus, the ant will have to take a longer round for the food piece (b) because its perimeter is  10.2cm which is the greatest among all.


Exercise 9.2

Q1. The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1mand 1.2mand perpendicular distance between them is0.8m.


Trapezium shaped Table


Ans: Area of Trapezium = 12 ( Sum of parallel sides)  x  (Distances between parallel sides) 

                                           = 12[(1m + 1.2m) x 0.8m]

                                           =  12(1.76m2)= 0.88m2


Q2. The area of a trapezium is 34cm2and the length of one of the parallel sides is 10cmand its height is 4cm. Find the length of the other parallel side.

Ans: It is given that Area of Trapezium =  34cm2.

          Length of one parallel side =  10cm

          Height =  4cm

          Now, let length of other parallel side = ‘a’ cm

Therefore, Area of Trapezium =  12 ( Sum of parallel sides)  x  (Distances between parallel sides) 

                       34cm2              = 12[(10 + a) x 4]cm

                       34cm2              = (20 + 2a)cm

                        142cm              = a cm

  Thus, Length of the other parallel side (a) = 7cm.


Q3. Length of the fence of a trapezium shaped field ABCD is 120m. If BC = 48m, CD =17m and AD =40m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.


Trapezium


Ans: Length of the fence of Trapezium ABCD = AB+BC+CD+DA

                                                                    120m =(AB + 48 + 17 + 40)m

                                                                    120m  =AB + 105m

          Therefore, AB = 15m

          Area of the field ABCD = 12(AD + BC) x AB

                      =12(40 + 48) x 15

                      =12(88) x 15

                          = 660m2


Q4. The diagonal of a quadrilateral shaped field is 24m and the perpendiculars dropped on it from the remaining opposite vertices are 8m and 13m. Find the area of the field. 


Quadrilateral


Ans: It is given that Length of diagonal =  24m

         Length of the perpendiculars h1and h2 from the opposite vertices to the diagonal are 

          h1= 8m and  h2= 13m

         Area of Quadrilateral = 12d(h1 + h2)

                                               = 12(24m) x (13m + 8cm)

                                               = 12(24m)(21m)

                                               = 252m2

          Thus, Area of field =  252m2


Q5. The diagonals of a rhombus are 7.5cmand 12cm. Find its area.

Ans: Area of Rhombus = 12 (Product of its diagonals) 

                                          = 12 x 7.5cm x 12cm

                                        = 45cm2


Q6. Find the area of a rhombus whose side is 6cmand whose altitude is 4cm. If one of its diagonals is 8cmlong, find the length of the other diagonal.

Ans: Let the length of other diagonal of Rhombus be ‘X

          A Rhombus is a special case of Parallelogram

          The area of Parallelogram is given by its base and height

          Thus, Area of Rhombus = Base x Height

                                                     = 6cm x 4cm = 24cm2

          So, 

          Area of Rhombus = 12 (Product of its diagonals) 

                           24cm2  = 12(8cm x X)

                            X = 6cm

          Therefore, Length of other diagonal of Rhombus = 6cm


Q7. The floor of a building consists of 3000tiles which are rhombus shaped and each of its diagonals are 45cmand 30cmin length. Find the total cost of polishing the floor, if the cost per m2is Rs4.

Ans: Given that each diagonals of Rhombus are 45cmand 30cm

          Area of Rhombus = 12 (Product of its diagonals) 

                                          =  (12 x 45 x 30)cm2

                                        =  675cm2

          So, Area of 3000 tiles = (675x3000)cm2=2025000cm2=202.5m2

          Now, it is given that cost of polishing is Rs.4perm2

          So, Cost of Polishing 202.5m2 area = Rs(4 x 202.5) =  Rs 810

          Hence, 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 10500m2and the perpendicular distance between the two parallel sides is 100m, find the length of the side along the river.


Trapezium shaped field


Ans: Let the length of the side along the road = ‘l

          And Let the length of the side along the river = ‘2l

         It is given that distance between two parallel sides = 100m

         and Area of Trapezium =   10500m2

         Area of Trapezium = 12 (Sum of parallel sides) (Distance between the parallel sides) 

                     10500m2    = 12(l + 2l) x (100m)

                     3l = (2 x 10500100)m = 210m

                       l = 2103m = 70m

         Therefore, Length of the side along the river ‘2l’ =140m


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.


Octagonal surface


Ans: 


Octagonal shape


It is given in the figure that side of octagon = 5cm

Area of Trapezium ABCH = Area of Trapezium DEFG

Area of Trapezium =  12 (Sum of parallel sides) (Distance between the parallel sides) 

                               = [12(4)(11 + 5)]m2

                               = (12 x 4 x 16)m2 = 32m2

In rectangle HCDG, Length(l) = 11mand Breadth(b) = 5m

So, Area of rectangle = (11 x 5)m2= 55m2

Therefore, Area of octagon = Area of Trapezium ABCH + Area of Trapezium DEFG +

                                               Area of Rectangle

                                            = (32 + 32 + 55)m2

                                            = 119m2


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?


Pentagonal shape


Ans: From Jyoti’s Way of finding area , 


Jyoti’s Diagram of Pentagonal shape


       Area of Pentagon = 2 (Area of Trapezium ABCF)

                                      = 2[ 12 (Sum of parallel sides) (Distance between the parallel sides) ]

                                      =[2 x 12(15 + 30)(152)]m2

                                      = 337.5m2

      From Kavita’s Way of finding area , 


Kavita’s Diagram of Pentagonal shape


Area of Pentagon = Area of Triangle ABE + Area of Square BCDE

                             = [ 12(basexheight)] + (sidexside)

                             = [12 x 15 x (30 - 15) + (15)2]m2

                             = (12 x 15 x 15 + 225)m2

                             = (112.5 + 225)m2

                             =  337.5m2


Q11. Diagram of the adjacent picture frame has outer dimensions = 24cm× 28cmand inner dimensions 16cm× 20cm Find the area of each section of the frame, if the width of each section is same.


Adjacent Picture frames


Ans: 


Adjacent Picture frames with dimensions


Given that, the width of each section is the same. 

Therefore,

IB = BJ = CK = CL = DM = DN = AO = AP

IL = IB + BC + CL

28 = IB + 20 + CL

2820 = IB+ CL8 = IB+ CL

IB = CL = 4cm

Hence, IB = BJ = CK = CL = DM = DN = AO = AP = 4cm

Area of section BEFC = Area of section DGHA = Area of Trapezium

  [12(base1 + base2)(h)]=[12(20 + 28)(4)]cm2=96 cm2

Therefore, Area of section ABEH = Area of section CDGF. 

Hence area of each section of frame is 96 cm2


Exercise 9.3

Q1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?


Cuboidal Boxes


Ans: From the given figure; Length(l), Breadth(b) and Height(h) of the Cuboid is

                                                   60cm,40cm, 50cm respectively.

          And, Side of Cube is  50cm.

          Now, Total Surface area of Cuboid(a) = 2(l h + b h +l b)

                                                                       = [2{ (60)(40) + (40)(50) + (50)(60)} ]cm2

                                                                       = [2(2400 + 2000 + 3000)]cm2

                                                                       = (2 x 7400)cm2

                                                                       = 14800cm2

                       Total Surface area of Cube(b) = 6l2

                                                                        = 6(50cm)2=15000cm2

          Therefore, Cuboidal box(a) requires a lesser amount of material for making.


Q2. A suitcase with measure 80cm x 48cm x 24cmis to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96cmis required to cover 100 such suitcases? 

Ans: Given that length(l),breadth(b),height(h) of suitcase is (80,48,24)cmrespectively.

         Total surface area of suitcase =  2(l h + b h + l b)

                                                         = 2[(80)(48) + (48)(24) + (24)(80)]

                                                         = 2[3840 + 1152 + 1920]

                                                         = 13824m2

         Total surface area of  100 suitcases = 100 x 13824cm2

                                                                    = 1382400cm2

                                                                    = required Tarpaulin

         We have given that breadth of Tarpaulin is  96cm and we have to find Length 

         of tarpaulin.

         Required Tarpaulin = (Length x Breadth) of Tarpaulin

                         1382400cm2 = Length x   96cm

                      Length = (138240096)cm = 14400cm

       Therefore, Length = 144m(1m = 100cm)

       Thus, 144m of tarpaulin is required to cover 100 suitcases.


Q3. Find the side of a cube whose surface area is 600cm2 .

Ans: It is given that the surface area of the cube is  600cm2 .

          We have to find the side of the cube (a). 

          Surface area of cube = 6a2

                                600cm26a2

                                      a2 = 100cm2

                                          a = 10cm

          Thus, the side of the cube is  10cm .


Q4. Rukhsar painted the outside of the cabinet of measure1m x 2m x 1.5m. How much surface area did she cover if she painted all except the bottom of the cabinet?


Cuboidal Shape


Ans: It is given that length (l), breadth(b), height(h) of the cabinet is 

         2m,1m,1.5mrespectively. 

         Area of the surface = 2h(1 + b) + lb

                                        = [2 x 1.5 x (2 + 1) + (2)(1)]m2

                                        = [3(3) + 2]m2

                                        = 11m2


Q5. Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15m,10m and 7mrespectively. From each can of paint  100m2 of area is painted. How many cans of paint will she need to paint the room? 

Ans: Given that length (l), breadth(b), height(h) of Cuboid is

          15m,10m and 7m respectively.

          Area of the hall to be painted = Area of the wall + Area of ceiling

                                                               = 2 h (l + b) + lb

                                                            = [2 x 7(15 + 10) + (15 x 10)]m2

                                                            = [(2 x 175) + 150]m2

                                                            = 500m2

          It is given that  100m2 area is to be painted from each can.

          Therefore, Number of cans required to paint an area of  500m2 

                                      = 500100=5

         Hence, 5 cans of paint are required to paint the room.


Q6. Describe how the two figures at the right are alike and how they are different. Which box has a larger lateral surface area?


Solid shapes (Cylinder and cube)


Ans: The above given two figures alike for same height(h)= 7cm

          And the difference between in these two figures is that one is cylinder and the other 

          One is a cube.

          Now, we have to find the lateral surface area for both of the given figures.

          Given that Side of cube(l)= 7cm 

                           Height and Diameter of cylinder =  7cmeach

                           Radius of cylinder = Diameter2 = 72cm = 3.5cm

          First, Lateral surface area of Cube = 4l2=4(72)cm2 = 4 x 49cm2 = 196cm2

          Second, Lateral surface area of cylinder = 2π rh

                                                                           = 2 x 227 x 3.5cm x 7cm = 154cm2

           Therefore, the Cube has a larger lateral surface area.


Q7. A closed cylindrical tank of radius 7m and height 3m is made from a sheet of metal. How much sheet of metal is required?

Ans: Given that the radius and height of Cylinder is  7m and 3m respectively.

         Therefore, Total surface area of Cylinder = 2π r (r + h)

                                                                                 = 2 x 227 x 7m(7m + 3m)

                                                                                 = 440m2

          Thus, 440m2 of metal sheet is required.


Q8. The lateral surface area of a hollow cylinder is 4224cm2. It is cut along its height and formed a rectangular sheet of width 33cm. Find the perimeter of the rectangular?

Ans: It is given that Hollow cylinder is cut along its height and formed a 

          Rectangular sheet.

          Area of cylinder =  4224cm2

         And, Breadth of rectangular sheet =  33cm 

          So, Area of Cylinder = Area of Rectangular Sheet

                        4224cm2 = Length x Breadth

                        4224cm2 = Length x 33cm

                        Length = 422433cm = 128cm

          Now, Perimeter of Rectangle = 2(length + breadth)

                                                         = 2(128 + 33)cm = 2(161cm) = 322cm


Q9. A road roller takes 750complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84cm and length is 1m.


Road roller


Ans: In one revolution, the roller will cover an area equal to its lateral surface area.

          Here, Radius = diameter2 = 842cm = 42cm = 42100m(1m = 100cm)

                    Height = 1m

          Thus, In One Revolution, 

          Area of the road covered = 2π rh

                                                      = 2 x 227 x 42100 x 1m2

                                                      = 264100m2

           In 750revolutions, area of road covered = 750 x 264100m2

                                                                            = 1980m2


Q10. A company packages its milk powder in a cylindrical container whose base has a diameter of 14cmand height 20cm. Company places a label around the surface of the container (as shown in the figure). If the label is placed 2cmfrom top and bottom, what is the area of the label?


Cylindrical Container


Ans: It is given that Radius = diameter2 = 142cm = 7cm

                                   Height of label = (20 - 2 - 2)cm(2cmdeductedfromtopbottomeach)

                                                           = 16cm

          As shown in the figure, the label is in the shape of a cylinder.

          So, Area of label(cylinder)= 2π rh

                                                   = 2 x 227 x 7cm x 16cm = 44 x 16cm2

                                                   = 704cm2


Exercise 9.4

Q1. Given a cylindrical tank, in which situation will you find surface area and in which situation volume.


Cylindrical Tank


(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.

Ans: (a) In this situation, we will find the volume.

         (b) Number of cement bags required to plaster cylindrical bags so for that 

              situation, we will find the surface area.

         (c) Number of smaller tanks that can be filled with so for that situation, we will

              find the volume.


Q2. Diameter of cylinder A is 7cm, and the height is 14cm. Diameter of cylinder B is  14cm and height is 7cm. 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?


Two cylindrical shapes


 Ans: The heights and diameters of these cylinders A and B are interchanged.

         We know that,

         Volume of cylinder = π r2h

         If measures of radius(r) and height(h) are same, then the cylinder with greater

         radius will have greater area.

        Here, Radius of cylinder A = 72cm

                   Radius of cylinder B = 142cm = 7cm

       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 =  π r2h

                                                = 227 x 72 x 72 x 14cm3 = 11 x 49cm3 = 539cm3

         Volume of Cylinder B = π r2h

                                                =  227 x 7 x 7 x 7cm3 = 22 x 49cm3 = 1078cm3

         Therefore, the volume of cylinder B is greater.

          Now, Surface area of cylinder A = 2π r(r + h)

                                                              = 2 x 227 x 72 x (72 + 14)cm2

                                                              = 22 x 352cm2 = 385cm2

                   Surface area of cylinder B = 2π r(r + h)

                                                              = 2 x 227 x 7 x (7 + 7)cm2

                                                              = 44 x 14cm2 = 616cm2


Q3. Find the height of a cuboid whose base area is 180cm2 and volume is 900cm3?

Ans. Here, we have given that Base Area of Cuboid = Length x Breadth

                                                                                              = 180cm2

          Volume of Cuboid = Length x Breadth x Height

                         900cm3 =             180cm2 x Height

                        Height = 900180cm = 5cm


Q4. A cuboid is of dimensions 60cm x 54cm x 30cm. How many small cubes with side6cmcan be placed in the given cuboid?

Ans. From given condition, 

          Volume of Cuboid =  60cm x 54cm x 30cm

                                     = 97200cm3

         Given that side of cube = 6cm

         So, Volume of cube = (6 x 6 x 6)cm3 = 216cm3

         Required number of cubes = volumeofcuboidvolumeofcube = 97200216 = 450

         Therefore, 450 cubes can be placed in the given Cuboid.


Q5. Find the height of the cylinder whose volume is 1.54m3and diameter of the base is 140cm?

Ans. It is given that Radius of Cylinder = 1402cm = 70cm

                                   Volume of Cylinder = π r2h

                                       1.54m3                =227 x 70100m x 70100m x h

                                                        Height =1.54 x 10022 x 7m = 1m

          Hence, the height of cylinder = 1m


Q6. A milk tank is in the form of cylinder whose radius is 1.5mand length is 7m. Find the quantity of milk in litres that can be stored in the tank?

Ans. It is given that Radius and height of the cylinder is 1.5m and 7m respectively.

         Therefore, Volume of Cylinder =  π r2h

                                                                            = 227 x 1.5m x 1.5m x 7m

                                                                         = 22 x 2.25m3

                                                                         = 49.5m3

           As, 1m3 = 1000litre

           So, Required Quantity = 49.5 x 1000litre = 49500litre

          Therefore, 49500litre can be stored in the tank.


Q7. 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?

Ans. (i) Let the edge of the cube be ‘a’.

             Surface area of cube = 6a2

            If each edge of the cube is doubled, then it becomes 2a

            Therefore, New surface area = 6(2a)2 = 24a2 = 4(6a2)

           Clearly, the surface area will be increased by 4 times.


        (ii) Let Volume of the cube = a3

            When each edge of the cube is doubled, it becomes2a.

           New volume = (2a)3 = 8a3 = 8 x a3

           Clearly, the volume of the cube will be increased by 8 times.


Q8. Water is pouring into a cubiodal reservoir at the rate of 60litres per minute. If the volume of reservoir is 108m3, find the number of hours it will take to fill the reservoir.

Ans. Volume of cuboidal reservoir =  108m3  = (108x1000) L = 108000 L

      It is given that water is being poured at the rate of 60 L per minute.

      That is, (60x60) L = 3600 L per hour

      Required number of hours = 30 hours

      Thus, it will take 30 hours to fill the reservoir.


NCERT Solutions for Class 8 Maths Chapter 9 Mensuration - PDF Download

It is not compulsory to be connected to the internet to access our solutions as Vedantu has made NCERT Solutions for Class 8 Maths Mensuration available in the PDF format on its official website. Now download the NCERT Solutions Class 8 Maths Chapter 9 PDF on your devices or get a printout and access it from anywhere, anytime. This mode of revision is quick and easy and can be done at your pace during a crucial exam period.

All Topics of NCERT Class 8 Maths Chapter 9 - Mensuration

The topics covered under chapter 9 Mensuration are given below.

S.No.

Topic Name

9.1

Introduction

9.2

Let Us Recall

9.3

Area of Trapezium

9.4

Area of a General Quadrilateral

9.5

Area of Polygon

9.6

Solid Shapes

9.7

Surface Area of Cube, Cuboid and Cylinder

9.8

Volume of Cube, Cuboid and Cylinder

9.9

Volume and Capacity


Table of Important Formulas

Below given are the list of important formulas that you must remember to solve the exercise of chapter 9 NCERT Maths Class 8th.

S.No.

Shape of the Object

Formula

1

Triangle

Area of triangle = 12 x Base x Height

Perimeter of Triangle = Sum of all three sides

2

Rectangle

Area of rectangle = Length x Breadth

Perimeter of rectangle = 2( length + breadth)

3

Square

Area of square = side x side

Perimeter of square = 4 x side

4

Circle

Area of circle = π x (radius)2

Perimeter of circle = 2π x radius

5

Parallelogram

Area of parallelogram = Base x Height

Perimeter of parallelogram = 2 (sum of two consecutive sides)

6

Trapezium

Area of trapezium = height (sum of parallel sides2)

7

Rhombus

Area of rhombus = 12 x diagonal1 x diagonal2

8

Cuboid 

Total surface area = 2 (length x breadth + breadth x height + height x length)

Volume of Cuboid = Area of base x height 

= length x breadth x height

9

Cube

Total surface area = 6 x (side)2

Volume of cube = (side)3

10

Cylinder

Curved (lateral) surface area = 2πrh

Total surface area =  2πr (r + h)

Volume of cylinder = area of base x height 

= πr2 x h 

where r is the radius of base and h is the height of cylinder.


You should also remember some of the basic conversion parameters. For your ease, the most important parameters are given below.


1 m3 = 1000000 cm3 = 1000 L


1 cm3 = 1 mL


1 L = 1000 cm3


Chapter 9 – Mensuration

Introduction

Mensuration is the process applied to different 2-D and 3-D solids of various shapes and figures to measure their lengths, volumes, area, heights, perimeters and several other dimensions. In this section of Ch 9 Maths Class 8, you will recall the areas of plane figures like triangles, circles, rectangles, etc., that you learned in the previous chapter. In NCERT Solutions for Class 8 Maths Ch 9, you would learn how to calculate perimeter and areas of other closed figures like Quadrilaterals.


Let us Recall

In this section of Mensuration Class 8 NCERT, students would recall the formula for calculating the perimeter and area of a park, flower bed, and amount of cement required to cover a given area. All the problems are based on areas of the following shape:

  • Rectangle - The area of a rectangle is x*y where x is the length and y is the breadth of the rectangle.

  • Square - The area of a square is x2 where x is the length of one side of a square.

  • Triangle - The area of a triangle is ½ *b*h where b is the length of the base and h is the height of the triable

  • Circle - The area of a circle is πr2 where r is the radius of the circle.


Area of a Trapezium

A trapezium is a quadrilateral where two of the sides are parallel to each other. In this portion of Mensuration Class 8 NCERT Solutions, students will learn how to derive the area of a trapezium which is given by:


Area of a trapezium = h * (x + y)/2, where h is the height of the trapezium, x and y are the lengths of its two sides.


Area of a general Quadrilateral

A simple definition of a general quadrilateral is a closed 2-D shape having 4 straight sides. If we break the word Quadrilateral, Quad means 4, and lateral means sides. The area of a quadrilateral in Class 8th Maths Chapter 9 is calculated by splitting it into two triangles and then calculating and adding the areas of the two triangles. 

(Image to be added soon)

So if PQRS is a quadrilateral, then its area = (area of triangle PQR) + (area of triangle PRS) = ½ * d * (h1 + h2), where d is the length of the diagonal from P to R and h1 and h2 are heights of perpendiculars dropped from Q and Pr and S or PR respectively.


9.8 Volume of Cube, Cuboid, and Cylinder

The amount of space that a 3-D object occupies gives the volume of that object. To take examples from real life, the volume of a cupboard in a room is less than the volume of the room where it is placed. 

  • 9.8.1 - Volume of a Cuboid - A 3-D structure with 6 rectangular faces is a cuboid. Its volume is given by “l * h* b”, here l = length, b = breadth, and h= height.

  • 9.8.2 - Volume of a Cube - A cube is a special type of cuboid where its length, breadth, and height are the same. Hence the volume of a cube = length 3.

  • 9.8.3 - Volume of a Cylinder - A cylinder has two circular bases that are parallel to each other and separated by a distance. To measure the volume of a cylinder we use the formula πr2 * h. Here r is the radius of the circular base and h is the distance between the bases. 


Key Features of NCERT Solutions for Class 8 Maths Chapter 9

You will find the NCERT Solutions of Class 8 Maths Chapter 9 by Vedantu extremely beneficial for your exams. The key features are:


  • Comprehensive explanations for each exercise and questions, promoting a deeper understanding of the subject.

  • Clear and structured presentation for easy comprehension.

  • Accurate answers aligned with the curriculum, boosting students' confidence in their knowledge.

  • Visual aids like diagrams and illustrations to simplify complex concepts.

  • Additional tips and insights to enhance students' performance.

  • Chapter summaries for quick revision.

  • Online accessibility and downloadable resources for flexible study and revision.


Other Study Material for CBSE Class 8 Maths Chapter 9


Conclusion

NCERT Solutions play a crucial role in Class 8 exam prep. Start by thoroughly reading the textbook chapter. After that, solve the NCERT questions for Class 8 Chapter 9. You can find detailed solutions on Vedantu, aligning with CBSE guidelines. Download the free NCERT Solutions for Class 8 Chapter 9 to guide your exam preparation with expert-reviewed answers.


Chapter-wise NCERT Solutions for Class 8 Maths


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FAQs on NCERT Solutions for Class 8 Maths Chapter 9 Mensuration

1. What are some of the applications of mensuration in our everyday lives?

In real life, mensuration is applied in many fields like:

  • Measuring floor and site areas required for purchasing or selling land.

  • Measuring agricultural fields.

  • Measuring surface areas of a house required for estimating painting cost.

  • To know the volume of the level of water in rivers and tanks.

  • To find out the amount of carpet required for covering a specific room.

  • The volume of soil that is needed to fill a ditch.

2. What are solid shapes and what is the method of measuring the surface area of a solid shape?

Any 3-D shapes occupying some space are called solid shapes like cube, sphere, cylinder, etc. To measure the surface area of a solid shape, we need to draw the net of that solid shape. From the net, we can see all the faces of the solid clearly. Then, we calculate the areas of each of the faces and add them up to get the total surface area of the solid shape. Surface area is measured in a square unit.

3. What are the concepts covered under Chapter 9 of Class 8 Maths?

The ideas or topics that are included in Chapter 9 “Mensuration” of Class 8 Maths are given below:

  • Introduction

  • Let Us Recall

  • Area of Trapezium

  • Area of a General Quadrilateral

Area of Special Quadrilaterals

  • Area of a Polygon

  • Solid Shapes

  • Surface Area of Cylinder, Cube and Cuboid

Cuboid

Cube

Cylinders

  • Volume of Cuboid, Cube and Cylinder

Cuboid

Cube

Cylinder

  • Volume and Capacity

  • What Have We Discussed?

4. The base area of the cuboid is 25cm sq. Its volume is 275 cubic cm. What will be the height of the cuboid?

In the question, we are given the base area of the cuboid which is equal to 25cm sq.


The volume of the cuboid is 275 cubic cm.


We know that according to the formula,


The volume of a cuboid = Height × Base Area


Therefore, the height of the cuboid will be = Volume of cuboid/ Base Area


Height = 275/25 = 11cm


Thus, 11cm is the height of the cuboid.

5. The distance between two parallel sides is 15m and the length of one parallel side is 20m. 480m sq. Is the area of the trapezium-shaped field. What is the length of the other parallel side?

Let one parallel side be a = 20m and the other parallel side is ‘b’.


The height of the field is 15m.


Given, the area of the trapezium is 480m sq.


The formula of trapezium is,


Area of trapezium = ½ (a + b) * h


480 = ½ (20 + b) * 15


20 + b = 480 × 2/ 15


64 = 20 + b


b = 44m


Thus, the length of the other parallel side is 44m.

6. What are the perks of NCERT Solutions of Chapter 9 of Class 8 Maths?

The perks of the NCERT Solutions of Chapter 9 of Class 8 Maths are given below:

  • The NCERT Solutions of Chapter 9 of Class 8 Maths offers comprehensive learning.

  • It enables students to develop their reasoning and logical skills.

  • These solutions assist students in understanding the difficult concepts.

  • By practising these, students will have a strong grip over the chapter.

  • You will get a hint of how to answer the questions in the proper format.

  • It helps in scoring good marks in Chapter 9 of Class 8 Maths.

The NCERT Solutions for  Chapter 9 of Class 8 Maths are available free of cost on the Vedantu website and on the Vedantu app.

7. How can I make the best study plan for Chapter 9 of Class 8 Maths?

Keep the following points in mind while making an effective study plan for Chapter 9 of Class 8 Maths:

  • Have a timetable or schedule to manage your time.

  • Centralize the NCERT book to read Chapter 9 of Class 8 Maths.

  • Practice the NCERT Solutions to comprehend the Chapter. 

  • Give yourself a break.

  • Do meditation and exercise to keep your body and mind fit.

  • Attend all your school lectures.