Chapter 9 Mensuration Ex 9.3 NCERT Solutions for Class 8 Maths

If you are in search of NCERT solutions for Class 8 Maths Chapter 9 Mensuration, then you are at the right place. When it comes to Mathematics, it is a subject that holds value for students across various academic streams, be it science, biology, or commerce and having a very good understanding of basic math concepts will help you in achieving your goal and make it much more easier.

Here we provide detailed NCERT solutions for Class 8 Maths of all the chapters, exercise wise with updated syllabus by NCERT. These NCERT solutions for Class 8 Maths have been carefully crafted to provide detailed explanations, ensuring that students can grasp the concepts effectively. We also provide extra practice questions for the same, so feel free to bookmark to boost your preparation.

NCERT Solutions for Class 8 Maths Chapter 9 Mensuration Ex 9.3

NCERT Solutions for Class 8 Maths Chapter 9 Mensuration Ex 9.3

Class 8 Maths Chapter 9 Mensuration Exercise 9.3


We find area when a region covered by a boundary, such as outer and inner surface area of a cylinder, a cone, a sphere and surface of wall or floor.
When the amount of space occupied by an object such as water, milk, coffee, tea, etc., then we have to find out volume of the object.
(a) Volume
(b) Surface area
(c) Volume


Yes, we can say that volume of cylinder B is greater as the radius of cylinder B is greater than that of cylinder A. Therefore greater radius will give us a greater volume.

Volume of A-
Diameter of cylinder A = 7 cm
Radius of cylinder A = 7/2 cm
Height of cylinder A = 14 cm

\[Volume\ of\ cylinder\ A=\ \pi r^2h\]

\[=\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\times14=539\ cm^3\]

\[Volume\ of\ cylinder\ A\ is\ 539\ cm^3\]

Volume of B-
Diameter of cylinder B = 14 cm
Radius of cylinder B = 14/2 = 7 cm
Height of cylinder B = 7 cm

\[Volume\ of\ cylinder\ B=\ \pi r^2h\]

\[=\frac{22}{7}\times7\times7\times7=1078\ cm^3\]

\[Volume\ of\ cylinder\ B\ is\ 1078\ cm^3\]

Surface area of cylinder A-

\[Surface\ area\ of\ cylinder\ A=2\pi r(r+h)\]

\[=2\times\frac{22}{7}\ \times\frac{7}{2}\times\left(\frac{7}{2}+14\right)=385\ cm^2\]

Surface area of cylinder B-

\[Surface\ area\ of\ cylinder\ B=2\pi r(r+h)\]

\[=2\times\frac{22}{7}\times7(\ 7+7)=616\ cm^2\]

Yes, the cylinder with greater volume also has a greater surface area.


Base area of cuboid = 180 cm2 [Given]
Volume of cuboid = 900 cm3 [Given]

We know that, Volume of cuboid = lbh

900 = 180×h

\[h\ =\ \frac{900}{180}\ =\ 5\ cm\]

Hence the height of the cuboid is 5 cm.


Length of cuboid (l) = 60 cm [Given]
Breadth of cuboid (b) = 54 cm [Given]
Height of cuboid (h) = 30 cm [Given]

\[We\ know\ that,\ Volume\ of\ cuboid=lbh\]

\[=60\times54\times30=97200\ cm^3\]

\[Volume\ of\ cube=(Side)^3\]

\[=6\times6\times6=216\ cm^3\]

\[Also\ ,the\ Number\ of\ small\ cubes=\frac{volume\ of\ cuboid}{volume\ of\ cube}\]

\[=\frac{97200}{216}=\ 450\ cubes\]

Hence , the required number of cubes is 450.


\[Volume\ of\ cylinder=1.54\ m^3\ \ \left[Given\right]\]

Diameter (d) of cylinder = 140 cm [Given]
Radius ( r )= d/2 = 140/2 = 70 cm [Given]

\[Volume\ of\ cylinder=\pi r^2h\]

\[1.54=\frac{22}{7}\times0.7\times0.7\times h\]

\[h=\frac{\left(1.54\times7\right)}{(22\times0.7\times0.7)}\ =\ 1\ m\]

Hence, the height of the cylinder is 1 m.


Radius of cylindrical tank (r) = 1.5 m
Height of cylindrical tank (h) = 7 m

\[Volume\ of\ cylindrical\ \tan k\ \ \ (V)\ \ =\pi r^2h\]

\[=\frac{22}{7}\times1.5\times1.5\times7\ =49.5\ cm^3\]

\[=49.5\times1000\ liters=49500\ \ liters\ \ \ \ [∵\ 1\ m^3=1000\ litres]\]

Hence, the required quantity of milk is 49500 litres.


(i) Let the edge of the cube be “ l” .

\[Surface\ area\ of\ the\ cube\ \left(A\right)\ =6l^2\]

\[When\ the\ edge\ of\ the\ cube\ 'l'\ is\ doubled,\ then\]

\[Surface\ area\ of\ the\ cube,\ say\ A’=6(2l)^2=6\times4l^2=4\times(6l^2)\]

\[A’=4\ \times A\]

\[Hence,\ the\ surface\ area\ will\ increase\ by\ four\ times.\]

\[(ii)\ Volume\ of\ a\ cube\ \left(V\right)=l^3\]

\[When\ the\ edge\ of\ the\ cube\ 'l'\ is\ doubled,\ then\]

\[Volume\ of\ the\ cube\ ,\ say\ V\ ’=(2l)^3=8(l^3)\]

\[V\ ’=8\times V\]

\[Hence,\ the\ volume\ will\ increase\ 8\ times.\]


\[The\ \ volume\ of\ the\ reservoir=108\ m^3\]

\[Rate\ of\ pouring\ water\ into\ cuboidal\ reservoir=60\ litres/\min\]

\[=\frac{60}{1000}m^3\ per\min\ \ \ \ \ \left[∵\ 1\ liter=\frac{1}{1000}\ m^3\right]\]

\[=\frac{\left(60\times60\right)}{1000}\ m^3\ per\ hour\]

\[Therefore,\frac{\left(60\times60\right)}{1000}\ m^3\ water\ filled\ in\ reservoir\ will\ take=1\ hour\]

\[Therefore\ 1\ m^3\ water\ filled\ in\ reservoir\ will\ take=\frac{1000}{60\ \times60}\ hours\]

\[Therefore,\ 108\ m^3\ water\ filled\ in\ reservoir\ will\ take=\frac{\left(108\times1000\right)}{(60\times60)}hours=30\ hours\]

Hence, It will take 30 hours to fill the reservoir.

  1. Accurate Answers: NCERT solutions for Class 8 Maths are prepared by expert educators and thoroughly reviewed to ensure quality and accurate answers.
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