NCERT Solutions For Class 8 Maths Chapter 11 Mensuration

 

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration- It is a branch of mathematics in which we deal with different geometric shapes both 2D and 3D shapes. It is a process of measurement. In this chapter, you will study different geometrical shapes like circle, quadrilateral, triangle, cube, cuboid, sphere, cylinder, cone and their area and volume. Also, you will learn to find area of polygon, area of trapezium, area of general quadrilaterals and special quadrilaterals. In NCERT solutions for class 8 maths chapter 11 mensuration, you will find questions related to finding area and volume 3D shapes. A 3D shape can be measured by three dimensions-length, width, and depth/height. As the 3D shape is bounded by a number of planes or surfaces, you can measure the total surface area (TSA), lateral surface area (LSA), curved surface area (CSA) and volume (V). For 2-Dimensional shapes, you can measure the area (A) and perimeter(P). 

There are 4 exercises with 34 questions in this chapter. All these questions are explained in solutions of NCERT for class 8 maths chapter 11 mensuration in a detailed manner using diagrams. It is easy for you to visualize and understand the problem. Only using the formulas and finding the answer is not enough, you should know how these formulas are derived so you can find the area or volume of the new shape you may come across. In CBSE NCERT solutions for class 8 maths chapter 11 mensuration, you will find some new ways to solve the problem. It will give you more concept clarity and help you in solving a new problem. Check NCERT solutions from class 6 to 12 to learn science and maths.

Before going in detail of this chapter we need to discuss some important terms like perimeter and area. Let's understand these terms with some scenarios that we come across in our daily life. 

Example:- If you have a field and you want to put the boundary on that field. So, how much raw material is needed for the boundary of the field? Solution: In order to know the required material you need to find out the entire length of the boundary.

  

Let’s take another example-

If you want to laminate the door. So you should know what is the total area of the door so you can arrange the materials to get it laminated.

The concept behind in these two examples is when you are talking about the length of the boundary it means you are talking about perimeter and when you are talking about the entire space that time you are talking about the area. 

The topics that we are covered in this chapter are-

  • 11.1 Introduction
  • 11.2 Let us Recall
  • 11.3 Area of Trapezium
  • 11.4 Area of a General Quadrilateral
  • 11.5 Area of a Polygon
  • 11.6 Solid Shapes
  • 11.7 Surface Area of Cube, Cuboid, and Cylinder
  • 11.8 Volume of Cube, Cuboid, and Cylinder
  • 11.9 Volume and Capacity

NCERT Solutions to the following exercises of Class 8 Maths Mensuration

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.2 let us recall

Question:(a) Match the following figures with their respective areas in the box.

 

Answer:

1. 

Area of above shape = (\pi \times 7^{2})\div 2 =  77 cm^{2}

Area of above shape = 14\times 7 = 98cm^{2}

Area of above shape = 7\times 7 = 49cm^{2}

Area of above triangle = \frac{1}{2}\times 14\times 7=49cm^{2}

Area of above shape = b*h = 14\times 7=98cm^{2}

 

Question:(b) Write the perimeter of each shape.

 

 

Answer:

  

Perimeter of shape = \pi \times r + 2\times r=\pi \times 7+2\times 7=36cm

 

perimeter of shape = 2\left ( l+ b \right )2\left ( 14+ 7 \right ) = 42 cm

perimeter of shape = 4\times side= 4\times 7=28cm

perimeter of shape = 14+11+9=34cm

perimeter of this shape cannot be calculated only with height and breadth,we need slant height or angle.

Solutions of NCERT for class 8 maths chapter 11 mensuration-Exercise: 11.1

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

 

Answer:

Given: 

    Perimeter of square = perimeter of rectangle 

\Rightarrow    4\times side = 2(length + breadth)

\Rightarrow    4\times 60 = 2(80 + breadth)

\Rightarrow      240\div 2=(80+breadth)

\Rightarrow         120=(80+breadth)

 \Rightarrow      (breadth)=120-80

 \Rightarrow         (breadth)=40

 

Area of square = side^{2}= 60^{2}=3600 m^{2}

Area of rectangle = = (length\times breadth)=80\times 40=3200m^{2}

Hence, area of square is greater than area of rectangle.

Question: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].

Answer:

Length of rectangle is 20 – (3.5 + 3.5) metres=13 metres

Breadth of rectangle = 7 metres

diameter of circular side = 7metres

Area of garden = area of rectangle + 2 times area of semi circular part

Area of garden

 =\left ( 13\times 7 \right ) + \left ( \pi \times 7^{2}\div 4 \right )

=\left ( 91 ) + \left ( 38.5 )

=129.5 metres^{2}

Perimetre of garden

 =\left ( 2\times \pi \times r \right )+\left ( 2\times 13 \right )

=\left ( 22 \right )+\left ( 26 \right )

=48 metres

CBSE NCERT solutions for class 8 maths chapter 11 mensuration topic 11.3 area of trapezium

Question:1  Nazma’s sister also has a trapezium-shaped plot. Divide it into three parts as shown (Fig 11.4). Show that the area of a trapezium 

        WXYZ=h\frac{(a+b)}{2}.

Answer:

Area of trapezium WXYZ = Area of triangle with base 'c' + area of rectangle + area of triangle with base'd'

                                         =(\frac{1}{2}\times c\times h)+(b\times h)+\left ( \frac{1}{2}\times d\times h \right )

Taking 'h' common, we get

                                          =(\frac{c}{2}+b+\frac{d}{2})h

                                         =h(\frac{c+d}{2}+b)

  Replacing   c+d =a-b          

                                        =h(\frac{a-b}{2}+b)

                                        =h(\frac{a+b}{2})

Hence proved that the area of a trapezium 

WXYZ=h\frac{(a+b)}{2}

Question:2 If h = 10 cm, c = 6 cm, b = 12 cm, d = 4 cm, find the values of each of its parts separetely and add to find the area WXYZ.            Verify it by putting the values of h, a and b in the expression \frac{h(a+b)}{2}.

Answer:

Area of trapezium WXYZ = Area of traingle with base'c'+area of rectangle +area of triangle with base 'd'

                                          =\left ( \frac{1}{2}\times c\times h \right )+\left ( b\times h \right )+\left ( \frac{1}{2}\times d\times h \right )

                                         =\left ( \frac{1}{2}\times 6\times 10 \right )+\left ( 12\times 10 \right )+\left ( \frac{1}{2}\times 4\times 10 \right )

                                        =\left ( 30 \right )+\left ( 120 \right )+\left ( 20 \right )

                                       =170 cm^{2}

the area WXYZ by  the expression 

.\frac{h(a+b)}{2}

=\frac{10 \left ( 22+12 \right )}{2}

=\frac{10 \left ( 34 \right )}{2}

=10\times 17

=170 cm^{2}

Hence,we can conclude area from given expression and calculated area is equal.

Solutions of NCERT for class 8 maths chapter 11 mensuration topic 11.3 area of trapezium

Question:1 Find the area of the following trapeziums (Fig 11.8)

      

 

Answer:

(i) Area of trapezium 

 =\frac{1}{2}\ (Sum\: of\: Parallel\: sides)\times (Distance\: between \:parallel \:sides)

=\frac{1}{2}\times 16\times 3

=27cm^{2}

(ii)

 

Area of trapezium 

 =\frac{1}{2}\ (Sum\: of\: Parallel\: sides)\times (Distance\: between \:parallel \:sides)

=\frac{1}{2}\times (10+5)\times 6

=45 cm^{2}

Solutions of NCERT for class 8 maths chapter 11 mensuration topic 11.4 area of a general quadrilateral

Question:1 We know that parallelogram is also a quadrilateral. Let us also split such a quadrilateral into two triangles, find their areas  and hence that of the parallelogram. Does this agree with the formula that you know already? (Fig 11.12)

Answer:

Area \:of \:quadrilateral \:ABCD = Area \:of( \bigtriangleup ABD \:+\:\bigtriangleup BCD)

                                                           =\left ( \frac{1}{2}\times b\times h \right )+\left ( \frac{1}{2}\times b\times h \right )                                     

                                                            =bh

This agree with the formula that we know already.

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.4.1 area of a special quadrilaterals

Question:1 Find the area of these quadrilaterals (Fig 11.14).

Answer:

(i) Area=

=\frac{1}{2}\times d\left ( h1+h2 \right )

=\frac{1}{2}\times 6\times \left ( 3+5 \right )

=3\times \left ( 8 \right )

=24 cm^{2}

(ii) Area=

=\frac{1}{2}\times d1\times \left ( d2 \right )

=\frac{1}{2}\times 7\times \left ( 6 \right )

= 7\times \left (3 \right )

= 21cm^{2}

(iii)Area= 

=\frac{1}{2}\times d\left ( h1+h2 \right )

=\frac{1}{2}\times 8\times \left ( 2+2 \right )

=4\times 4

=16 cm^{2}

Solutions of NCERT for class 8 maths chapter 11 mensuration topic 11.5 area of a polygon

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

            

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

Answer:

(i)

Area of  polygon EFGHI = area of \triangleEFI + area of quadrilateral FGHI

Draw a diagonal FH 

Area of  polygon EFGHI = area of \triangleEFI + area of \triangle FGH + area of \triangleFHI

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

 Draw diagonal NP and NR .

Area of polygon MNOPQR = area of \triangleNOP+area  of \triangleNPQ +area of \triangleNMR +area of \triangle NRQ

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

Answer:

the area of polygon MNOPQR 

= area of \triangleMAN + area of trapezium ACON+area of  \triangleCPO + area of  \triangleMBR+area of trapezium BDQR+area of  \triangleDPQ

=(\frac{1}{2}\times 2\times 2.5)+(4(2.5+3)\div 2)+(\frac{1}{2}\times 3\times 3)+(\frac{1}{2}\times 4\times 2.5)+(3(2.5+2)\div 2)+(\frac{1}{2}\times 2\times 2)

=2.5+11+4.5+5+6.75+2

=31.75 cm^{2}

NCERT solutions for class 8 maths chapter 11 mensuration-Exercise: 11.2

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 perpendicular distance between them is  0.8 m.

 

Answer:

Area of table = 

=\frac{1}{2}\times sum \, of \, parallel\, sides\times distance\, between\, parallel\, sides

=\frac{1}{2}\times\left ( 1+1.2 \right )\times 0.8

= 0.4\times 2.2

=0.88 m^{2}

Question:2 The area of a trapezium is 34 cm^{2} and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.

Answer:

Let the length of the other parallel side be x

area of a trapezium = 

=\frac{1}{2}\times \left ( 10+x \right )\times 4=34

= \left ( 10+x \right )=17

x=17-10

x=7 cm

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

Question:3 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.

 

Answer:

 BC = 48 m, CD = 17 m and AD = 40 m,

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

                                                                                                120=AB+48+17+40

                                                                                                AB=15m

Area of trapezium =

\frac{1}{2}\times sum \, of\, parallel\, sides\times height

=\frac{1}{2}\times (40+48)\times 15

=\frac{1}{2}\times (88)\times 15

= (44)\times 15

=660m^{2}

Question:4 The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

    

Answer:

 The diagonal of a quadrilateral shaped field is 24 m 

the perpendiculars are 8 m and 13 m. 

the area of the field =

\frac{1}{2}\times d\times (h1+h2)

=\frac{1}{2}\times 24\times (13+8)

=12\times 21

=252 m^{2}

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:

Rhombus is a type of parallelogram and area of parallelogram is product of base and height.

So,Area of rhombus = base \times height

                           = 5\times 4.8

                           = 24

Let the other diagonal be x

Area of rhombus =

=\frac{1}{2}\times product\, of\, diagonals

24=\frac{1}{2}\times8\times x

x=6cm

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 m^{2} and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

        

Answer:

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

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

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

So equation becomes :    

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

or                              3x = 210

or                               x = 70

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

Question:9 Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.

           

Answer:

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

Area of recatangular surface = 11\times5 = 55 m^2

Area of trapezium surface = 

                                                  \frac{1}{2}\times4(11+5) = 2\times16 = 32 m^2

So total area of octagonal surface = 55 + 2(32) = 55 + 64 = 119 m^2

Question:10 There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways.

            

Find the area of this park using both ways. Can you suggest some other way of finding its area?

Answer:

Area of pentagonal park according to Jyoti's diagram :-

                       = 2(Area of trapezium)   =  2(\frac{1}{2}\times\frac{15}{2}\times45) = \frac{15\times45}{2} = 337.5 m^2

Area of pentagonal park according to Kavita's diagram :- 

                      = Area of triangle + Area of square.

                     = \frac{1}{2}\times15\times15 + 15\times15 = 112.5 + 225 = 337.5 m^2

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

        

Answer:

Area of opposite sections will be same.

So area of horizontal sections,

                                                     =\frac{1}{2}\times4(16+24) = 2(40) = 80\ cm^2

And area of vertical sections,

                                                     = \frac{1}{2}\times8(20+28) = 4(48) = 96\ cm^2

Solutions of NCERT for class 8 maths chapter 11 mensuration topic 11.7.1 cuboid

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

Answer:

(i)  Total surface area = 2 (h × l + b × h + b × l) = 2(lb + bh + hl)

                                   =2(6\times4 + 4\times2+6\times2) = 2(24+8+12)

                                    = 2(24+8+12) = 88\ cm^2

(ii) Total surface area = 2 (h × l + b × h + b × l) = 2(lb + bh + hl)

                                  = 2(4\times4+4\times10+10\times4) = 2(16 + 40+40)

                                   = 2(96) = 192\ cm^2

CBSE NCERT solutions for class 8 maths chapter 11 mensuration topic 11.7.2 cube

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

Answer:

Surface area of cube A = 6l^2

                                      =  6(10)^2 = 600\ cm^2

Lateral surface area of cube B = 4l^2

                                                 =4(8)^2 = 4\times64 = 256\ cm^2

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.7.3 cylinders

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

Answer:

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

(i)  Area = 

                      2\Pi \times14(14+8) = 2\Pi\times 14(22) = 1935.22\ cm^2

(ii) Area = 

                     2\Pi \times1(1+2) = 2\Pi\times 1(3) = 18.84\ cm^2

Solutions of NCERT for class 8 maths chapter 11 mensuration-Exercise: 11.3

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

 

Answer:

Surface area of cuboid (a)   = 2(60\times40 + 40\times50 + 50\times60) =14800\ cm^2

Surface area of cube (b)   = 6(50)^2 = 15000\ cm^2

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

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

Answer:

Surface area of suitcase    = 2(80\times48+48\times24+24\times80) = 13824\ cm^2.

Area of such 100 suitcase will be 1382400\ cm^2

So lenght of tarpaulin cloth   = \frac{1382400\ cm^2 }{96\ cm} = 14400\ cm\ or\ 144m

Question:3 Find the side of a cube whose surface area is 600 cm^{2}.

Answer:

Surface area of cube   =\ 6l^2

So,                       6l^2 = 600

or                           l^2 = 100

or                           l = 10\ cm.

Thus side of cube is 10 cm.  

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

        

Answer:

Required area = Total area - Area of bottom surface

Total area    = 2(1\times2+ 2\times1.5 + 1.5\times1 ) = 13\ m^2

Area of bottom surface  = 1\times2 = 2\ m^2

So required area = 13-2\ m^2 = 11\ m^2

Question:5  Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint  100 m^{2} of area is painted. How many cans of paint will she need to paint the room?

Answer:

Total area painted by Daniel : 

                                                     = 2(15\times10+10\times7+7\times15) - 15\times10                                   (\because  Bottom surface is excluded.)

So,  Area                     

                           = 650 - 150\ m^2 = 500\ m^2

No. of cans of paint required   

                                                  =\frac{ 500\ m^2}{100\ m^2} = 5

Thus 5 cans of paint are required.

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

        

Answer:

The two figures have same height.The diference between them is one is cylinder and another is cube.

lateral surface area of cylinder = 2\times \pi \times r\times h

                                                  =2\times \pi \times 3.5\times 7

                                                  =154 cm^{2}

lateral surface area of cube = 4\times side^{2}

                                           =4\times 7^{2}   

                                           =4\times 49

                                          =196 cm^{2}

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)

                                              = 2\Pi \times7(7+3) = 14\Pi \times10 = 440\ m^2

Question:8 The lateral surface area of a hollow cylinder is 4224 cm^{2}. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?

Answer:

Lenght of rectangular sheet 

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

So perimeter of rectangular sheet = 2(l + b)

                                                      = 2(128 + 33) = 322\ cm.

Thus perimeter of 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 length is 1 m.

        

Answer:

Area in one complete revolution of roller =  2πrh.

                                                               = 2\Pi \times0.42\times1 = 2.638\ m^2

So area of road  = 2.638\ m^2\times 750 = 1979.20\ m^2

CBSE NCERT solutions for class 8 maths chapter 11 mensuration topic 11.8.1 cuboid

Question:1 Find the volume of the following cuboids

Answer:

(i) Volume of cuboid is given as:

Volume\ of\ cuboid = length\times breadth\times height,

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

its volume will be = 8cm\times 3cm\times 2cm = 48cm^3.

Aslo for Given Surface area of cuboid 24m^2 and height = 3 cm we can easily calculate the volume:

Volume = Surface\ area\times height;

So, Volume = Volume = 24m^2\times \frac{3m}{100} = 0.72m^3

Solutions of NCERT for class 8 maths chapter 11 mensuration topic 11.8.2 cube

Question: Find the volume of the following cubes

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

Answer:

(a) Volume of cube having side equal to 4cm will be

Volume = Side\times side\times side    or    Volume = 4cm\times 4cm\times 4cm= 64cm^3.

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

Volume = 1.5m\times 1.5m\times 1.5m= 3.375m^3.

Solutions of NCERT for class 8 maths chapter 11 mensuration topic 11.8.3 cylinder

Question:1 Find the volume of the following cylinders.

Answer:

(i) The volume of a cylinder given as =  \Pi \times r^2\times length.

or given radius of cylinder = 7cm and length of cylinder = 10cm.

So, we can calculate the volume of the cylinder = \Pi \times (7cm)^2\times 10cm          (Take the value of  \Pi = \frac{22}{7})

The volume of cylinder = 1,540 cm^3.

 

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

we have .

Volume\ of\ cylinder = 250m^2\times 2m = 500m^3.

Solutions for NCERT class 8 maths chapter 11 mensuration-Exercise: 11.4

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

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

    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:2 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?

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:

As volume is directly proportional to the square of the radius of 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 :  \pi\times r^2\times height = \pi\times (\frac{7cm}{2})^2\times 14cm =539cm^3

and Volume of B :  \pi\times r^2\times height = \pi\times (\frac{14cm}{2})^2\times 7cm =1,078cm^3.

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\times \pi\times r(r+h) = 2\times \pi\times \frac{14cm}{2}(\frac{14}{2}+7) = 616cm^2.

and the surface area of cylinder A = 2\times \pi\times r(r+h) = 2\times \pi\times \frac{7cm}{2}\times (\frac{7}{2}+14) = 385cm^2.

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 that the height of a cuboid whose base area is 180cm^2 and volume is 900cm^3;

As Volume\ of\ cuboid = Base\ area\times height

So, we have relation: 900cm^3= 180cm^2\times height

or, Height = 5cm.

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

Answer:

So given the dimensions of cuboid 60 cm \times 54 cm \times 30 cm hence it's the volume is equal to = 97,200cm^3

We have to make small cubes with side 6cm which occupies the volume  = 6cm\times 6cm\times 6cm = 216cm^3

Hence we have now one cube having side length = 6cm volume = 216cm^3.

So, total numbers of small cubes that can be placed in the given cuboid = \frac{97,200}{216} =450

Hence 450 small cubes can be placed in that cuboid.

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

Answer:

Given that the volume of the cylinder is 1.54m^3 and having its diameter of base = 140cm.

So, as Volume\ of\ cylinder = Base\ area\times height;

hence putting in the relation we get;

1.54m^3= (\pi\times (\frac{1.4m}{2})^2)\times height

The height of the cylinder would be = 1metre.

Question:6  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?

Answer:

Volume of the cylinder = V =   \Pi r^2h

                                            \Pi (1.5)^2\times7 = 49.48\ m^3

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

\left ( \because 1\ m^3 = 1000\ litres \right )

Question:7(i) If each edge of a cube is doubled,

how many times will its surface area increase?

Answer:

The surface area of cube  = 6l^2

 So if we double the edge l becomes 2l.

New surface area = 6(2l)^2 = 24l^2

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 = l^3

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

Hence 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 reservoir is  108\ m^3 , 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.

Volume of the reservoir is 108\ m^3, then

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

As we know 1m^3 = 1000L.

Then 108\ m^3 = 108,000 Litres ;

Time taken to fill the tank will be:

\frac{1,08,000litres}{60litres\ per\ minute} = 1,800 minutes

or   \frac{1,800hours}{60} =30hours.

NCERT solutions for class 8 maths: Chapter-wise

Chapter -1

NCERT solutions for class 8 maths chapter 1 Rational Numbers             

Chapter -2 

Solutions of NCERT for class 8 maths chapter 2 Linear Equations in One Variable

Chapter-3

CBSE NCERT solutions for class 8 maths chapter 3 Understanding Quadrilaterals

Chapter-4

NCERT solutions for class 8 maths chapter 4 Practical Geometry

Chapter-5

Solutions of NCERT for class 8 maths chapter 5 Data Handling

Chapter-6

CBSE NCERT solutions for class 8 maths chapter 6 Squares and Square Roots

Chapter-7

NCERT solutions for class 8 maths chapter 7 Cubes and Cube Roots

Chapter-8

Solutions of NCERT for class 8 maths chapter 8 Comparing Quantities

Chapter-9

NCERT solutions for class 8 maths chapter 9 Algebraic Expressions and Identities

Chapter-10

CBSE NCERT solutions for class 8 maths chapter 10 Visualizing Solid Shapes

Chapter-11

NCERT solutions for class 8 maths chapter 11 Mensuration

Chapter-12

Solutions of NCERT for class 8 maths chapter 12 Exponents and Powers

Chapter-13

CBSE NCERT solutions for class 8 maths chapter 13 Direct and Inverse Proportions

Chapter-14

NCERT solutions for class 8 maths chapter 14 Factorization

Chapter-15

Solutions of NCERT for class 8 maths chapter 15 Introduction to Graphs

Chapter-16

CBSE NCERT solutions for class 8 maths chapter 16 Playing with Numbers

NCERT solutions for class 8: Subject-wise

Some important formulas from NCERT solutions for class 8 maths chapter 11 mensuration to remember-

  • Area of a trapezium 

                        \frac{1}{2}[\text{sum of the lengths of parallel sides} \ \times \ \text{ perpendicular distance between them}]

  • Area of a rhombus 

                     \frac{1}{2}[\text{product of its diagonals} ]

  • The surface area of a cuboid

                        2(l \times b + b \times h + h \times l)​​

                  l- length of the cuboid 

              b- breadth of the cuboid 

              h- height of the cuboid

  • The surface area of a cube

            6l ^2 \\ \text{ l- side of the cube}

  • The Surface area of a cylinder 

                2\pi r(r+h)

           r- radius of the cylinder

           h- height of the cylinder

  • The volume of a cuboid

                l \times b \times h

                  l- length of the cuboid 

              b- breadth of the cuboid 

              h- height of the cuboid

  • The volume of a cube = l3
  • The volume of a cylinder = πr2h

           r- radius of the cylinder

           h- height of the cylinder

Tip- If you have derived the formulas and know how to drive it, then you can solve any problem of this chapter easily. You can take help from NCERT solutions for class 8 maths chapter 11 mensuration if you are not able to solve the problem. It will make your task easy.  

Happy Reading!!!

 

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