• School
  • Engineering and Architecture
  • Management and Business Administration
  • Medicine and Allied Sciences
  • Law
  • Animation and Design
  • Media, Mass Communication and Journalism
  • Finance & Accounts
  • Computer Application and IT
  • Pharmacy
  • Hospitality and Tourism
  • Competition
  • Study Abroad
  • Arts, Commerce & Sciences
  • Learn
  • Online Courses and Certifications
NCERT Solutions For Class 8 Maths Chapter 11 Mensuration

NCERT Solutions For Class 8 Maths Chapter 11 Mensuration

Edited By Ramraj Saini | Updated on Mar 01, 2024 06:36 PM IST

Mensuration Class 8 Questions And Answers provided here. These NCERT Solutions are prepared by expert team considering the latest syllabus and pattern of CBSE 2023-24. 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.

This Story also Contains
  1. Mensuration Class 8 Questions And Answers PDF Free Download
  2. Mensuration Class 8 Solutions - Important Formulae
  3. Mensuration Class 8 NCERT Solutions (Intext Questions and Exercise)
  4. NCERT Solutions For Class 8 Maths Chapter 11 Mensuration - Topics
  5. NCERT Solutions for Class 8 Maths - Chapter Wise
  6. NCERT Solutions for Class 8 - Subject Wise
  7. NCERT Solutions For Class 8 Maths Chapter 11 Mensuration To Remember - Formulae
  8. NCERT Books and NCERT Syllabus
NCERT Solutions For Class 8 Maths Chapter 11 Mensuration
NCERT Solutions For Class 8 Maths Chapter 11 Mensuration

There are 4 exercises with 34 questions in this chapter. All these questions are explained in NCERT solutions 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. Here you will get the detailed NCERT Solutions for Class 8 Maths by clicking on the link.

Mensuration Class 8 Questions And Answers PDF Free Download

Download PDF

Mensuration Class 8 Solutions - Important Formulae

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

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

  • Perimeter of a Square: 4 × Side

  • Perimeter of a Circle (Circumference): 2πr (where r is the radius of the circle)

  • Perimeter of a Parallelogram: 2(Base + Height)

  • Perimeter of a Triangle: a + b + c (where a, b, and c are the side lengths)

  • Perimeter of a Trapezium: a + b + c + d (where a, b, c, and d are the sides of a trapezoid)

  • Perimeter of a Kite: 2a + 2b (where a is the length of the first pair of sides and b is the length of the second pair)

  • Perimeter of a Rhombus: 4 × Side

  • Perimeter of a Hexagon: 6 × Side

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

Volume of a Cuboid: Base Area × Height = Length × Breadth × Height

Volume of a Cone: (1/3)πr2h

Volume of a Sphere: (4/3)πr3

Volume of a Hemisphere: (2/3)πr3

Free download NCERT Solutions for Class 8 Maths Chapter 11 Mensuration for CBSE Exam.

Mensuration Class 8 NCERT Solutions (Intext Questions and Exercise)

Class 8 mensuration ncert solutions - Topic 11.2 Let Us Recall

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

45454578 1643870847806

Answer:

1.

1643870897163

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

1643870896823

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

1643870896187

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

1643870894567

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

1643870911717

Area of above shape = b*h =

14\times 7=98cm^{2}

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

45454578


1643871037982

Answer:

2.

1643871007420

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

1643871007699

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

1643871008242

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

1643871008617

perimeter of shape = 14+11+9=34cm

1643871074104

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

Class 8 maths chapter 11 question answer - 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?

Screenshot%20(75) 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, the area of the square is greater than the area of the rectangle.

Question: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 Rs. 55 per m^{2} .

212

Answer:

212

Area of plot = side^{2}=25^{2}=625m^{2}

Area of house = 20\times 15=300m^{2}

Area of garden around house = 625-300=325m^{2}

the rate = 55 per m^{2} .

the total cost of developing a garden around the house= 55\times 325= Rs 17875

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]. 1643871256807 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 )

=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

Question: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 m^{2}? (If required you can split the tiles in whatever way you want to fill up the corners).

Answer:

Area of parallelogram

=base \times height

=24\times10

=240 cm^{2}

a floor of area 1080m^{2} = 1080m^{2}\times 10000 = 10800000cm^{2}

Number of tiles = 10800000\div 240 = 45000 tiles

Question: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 = 2\pi r , where r is the radius of the circle.

222 1115


Answer:

(a) circumference

= 2.8+\left ( \pi \times 1.4 \right )

=7.2 cm

(b)circumference

=1.5+2.8+1.5+\left ( \pi \times 1.4 \right )

=1.5+2.8+1.5+4.4

=10.2cm

(c) circumference

=2 +\left ( \pi \times 1.4 \right )+2

=4+4.4

=8.4

Hence, the perimeter of (b) is highest so ant have to take a longer round in (b).

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} .

2323

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} .

2323

Answer:

2323

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.

NCERT Solutions 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)

1643871315672 252533

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}

NCERT Solutions 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)

1643871374343Answer:

115484

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).

15115165165

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 )

= 21 cm^{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}

NCERT Solutions 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.

16438714089291643871436209

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

Answer:

(i)

Area of polygon EFGHI = area of \triangle EFI + area of quadrilateral FGHI

Draw a diagonal FH

Area of polygon EFGHI = area of \triangle EFI + area of \triangle FGH + area of \triangle FHI

ii)

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

Draw diagonal NP and NR .

Area of polygon MNOPQR = area of \triangle NOP+area of \triangle NPQ +area of \triangle NMR +area of \triangle NRQ

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

Area of \Delta AFB=\frac{1}{2}\times AF\times BF=\frac{1}{2}\times 3\times 2=....

Area of trapezium FBCH=FH\times \frac{(BF+CH)}{2}

=3\times \frac{(2+3)}{2}[FH=AH-AF]

Area of \Delta CHD=\frac{1}{2}\times HD\times CH=....., Area of \Delta ADE=\frac{1}{2}\times AD\times GE=...

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

2525233

Answer:

2525233

Area of Polygon ABCDE = area of \traingle\bigtriangleup AFB + area of trapezium BCHF + area of \traingle\bigtriangleup CDH+area of \traingle\bigtriangleup AED =(\frac{1}{2}\times AF\times BF)+\left ( FH\times \left ( BF+CH \right )\div 2 \right )+(\frac{1}{2}\times HD\times CH)+(\frac{1}{2}\times AD\times EG)

=(\frac{1}{2}\times 3\times2)+\left ( 3\times \left ( 2+3 \right )\div 2 \right )+(\frac{1}{2}\times 2\times 3)+(\frac{1}{2}\times 8\times2.5)

= 3+7.5+3+10

=23.5 cm^{2}

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

1643871560644

Answer:

the area of polygon MNOPQR

= area of \triangle MAN + area of trapezium ACON+area of \triangle CPO + area of \triangle MBR+area of trapezium BDQR+area of \triangle DPQ

=(\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}

Class 8 maths chapter 11 ncert solutions - 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. 1643871584769

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 .

1643871610276

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. 1643871642365

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:5 The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

Answer:

Area of rhombus =

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

=\frac{1}{2}\times 7.5\times 12

= 7.5\times 6

= 45 cm^{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:7 The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m^{2} is 4.

Answer:

Area of 1 tile =

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

=\frac{1}{2}\times45\times 30

=675 cm^{2}

Area of 3000 tiles =

=675 \times 3000

= 202.5 m^{2}

Total cost of polishing =

= 202.5\times 4

= Rs 810

Question:8 Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m^{2} and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

2626

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 . 1643871675480 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.

3636333

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.

356356

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

NCERT Solutions 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):

3563

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

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).

1643871700324

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

365

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

NCERT Solutions 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?

1659338580356

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.

456

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?

1659338695497

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.

1643871784104

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

Question:10 A company packages its milk powder in a cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the area of the label? 1643871806694Answer:

Area of label = 2πrh

= 2\Pi \times(7)\times(16) = 703.71\ cm^2

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

NCERT class 8 maths ch 11 question answer - Topic 11.8.1 Cuboid

Question:1 Find the volume of the following cuboids 1643871835132

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

NCERT Solutions 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 .

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration - Topic 11.8.3 Cylinder

Question:1 Find the volume of the following cylinders. 1643871864048

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.

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

To find the number of smaller tanks that can be filled with water from it.

Answer:

(c) We have to find out the volume.

Question:2 Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and 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?

1655617952866

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 11 Mensuration - Topics

  • Let us Recall

  • Area of Trapezium

  • Area of a General Quadrilateral

  • Area of a Polygon

  • Solid Shapes

  • Surface Area of Cube, Cuboid, and Cylinder

  • The volume of Cube, Cuboid, and Cylinder

  • Volume and Capacity

NCERT Solutions for Class 8 Maths - Chapter Wise

Chapter -1

Rational Numbers

Chapter -2

Linear Equations in One Variable

Chapter-3

Understanding Quadrilaterals

Chapter-4

Practical Geometry

Chapter-5

Data Handling

Chapter-6

Squares and Square Roots

Chapter-7

Cubes and Cube Roots

Chapter-8

Comparing Quantities

Chapter-9

Algebraic Expressions and Identities

Chapter-10

Visualizing Solid Shapes

Chapter-11

Mensuration

Chapter-12

Exponents and Powers

Chapter-13

Direct and Inverse Proportions

Chapter-14

Factorization

Chapter-15

Introduction to Graphs

Chapter-16

Playing with Numbers

NCERT Solutions for Class 8 - Subject Wise

NCERT Solutions For Class 8 Maths Chapter 11 Mensuration To Remember - Formulae

  • 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(lb + bh + lh)

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 = πr 2 h

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.

NCERT Books and NCERT Syllabus

Frequently Asked Questions (FAQs)

1. What are the important topics of chapter Mensuration ?

Area of trapezium, area of quadrilateral, area of the polygon,  area and volume of the solid shapes like cube, cuboid and cylinder are covered in this chapter.

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

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

3. Does CBSE class maths is tough ?

No, CBSE class 8 maths is a basic and damn simple where most of the topics related to the previous classes.

4. Which is the official website of NCERT ?

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

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

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

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

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

Also Read

Articles

Upcoming School Exams

National Institute of Open Schooling 12th Examination

Admit Card Date:04 October,2024 - 29 November,2024

National Institute of Open Schooling 10th examination

Admit Card Date:04 October,2024 - 29 November,2024

Mizoram Board of School Education 12th Examination

Application Date:07 October,2024 - 22 November,2024

Mizoram Board 10th Examination

Application Date:07 October,2024 - 22 November,2024

Punjab Board of Secondary Education 10th Examination

Application Correction Date:08 October,2024 - 27 November,2024

View All School Exams
Get answers from students and experts

Popular Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
Read More

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
Read More

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
Read More

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
Read More

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
Read More

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
Read More

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
Read More

With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
Read More

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
Read More

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
Read More

Colleges After 12th

Popular Course After 12th

Applications for Admissions are open.

VMC VIQ Scholarship Test
Apply

Register for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.

JEE Main Important Physics formulas
Apply

As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

JEE Main Important Chemistry formulas
Apply

As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters

TOEFL ® Registrations 2024
Apply

Accepted by more than 11,000 universities in over 150 countries worldwide

Pearson | PTE
Apply

Register now for PTE & Unlock 20% OFF : Use promo code: 'C360SPL20'. Valid till 30th NOV'24! Trusted by 3,500+ universities globally

JEE Main high scoring chapters and topics
Apply

As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE

View All Application Forms
News and Notifications
Tamil Nadu teacher fatally stabbed in school, violence erupts; assailant detained
Bihar govt puts hold on new teachers transfer policy
Youth stabs teacher on school premises; now detained, says police
Air Pollution: Schools in Rajasthan's Khairthal-Tijara go online for classes 1 to 5
Govt informs SC about NCERT guidelines for development of e-content for children with disabilities
Government school principal in Andhra chops off girl students' hair for coming late
Delhi Air Pollution: DU, JNU, Jamia shift to online classes; Delhi, Gurugram schools suspend physical sessions
Back to top