Chapter 9 Mensuration Ex 9.2 NCERT Solutions for Class 8 Maths

Latest Updated : November 2023

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 ncertforclass8.com to boost your preparation.

NCERT Solutions for Class 8 Maths Chapter 9 Mensuration Ex 9.2

NCERT Solutions for Class 8 Maths Chapter 9 Mensuration Ex 9.2

Class 8 Maths Chapter 9 Mensuration Exercise 9.2

Exercise 9.2

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

Solution:

(a) Length of cuboidal box (l) = 60 cm
Breadth of cuboidal box (b) = 40 cm
Height of cuboidal box (h) = 50 cm

Total surface area of cuboidal box = 2×(lb+bh+hl)
= 2×(60×40 + 40×50 + 50×60)
= 2×(2400+2000+3000)
= 14800 cm2

(b) Length of cubical box (l) = 50 cm
Breadth of cubicalbox (b) = 50 cm
Height of cubicalbox (h) = 50 cm

Total surface area of cubical box = 6×side×side
= 6 (50×50)
= 6×2500
= 15000 cm2

Surface area of the cubical box is 15000 cm2

∴ Cuboidal box requires the lesser amount of material to make.

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

Solution:

Length of suitcase box (l) = 80 cm
Breadth of suitcase box (b)= 48 cm
Height of cuboidal box (h) = 24 cm

Total surface area of suitcase box (cuboid)= 2×(lb+bh+hl)
= 2(80×48+48×24+24×80)
= 2 (3840+1152+1920)
= 2×6912
= 13824 cm2

\[Total\ surface\ area\ of\ suit\ case\ box\ is\ 13824\ cm^2\]

Area of Tarpaulin cloth = Surface area of suitcase
l×b = 13824
l ×96 = 13824
l=13824/96
l = 144

Required tarpaulin for 100 suitcases = 144×100 = 14400 cm = 144 m

∴ Tarpaulin cloth required to cover 100 suitcases is 144 m.

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

Solution:

\[Surface\ area\ of\ cube=600\ cm^{2\ \ }(Given)\]

\[Surface\ area\ of\ a\ cube=6\times(side)^2\]

Substituting the values, we get

\[6\times\ (side)^2=600\]

\[(side)^2=\frac{600}{6}\]

\[(side)^2=\ 100\ \]

\[side\ =\sqrt{100}=\ side\ =\ \pm10\ cm\]

\[Since\ side\ cannot\ be\ negative,\ the\ measure\ of\ each\ side\ ofa\ cube\ is\ 10cm.\]

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

Solution:

Length of cabinet, l = 2 m, Breadth of cabinet, b = 1 m and Height of cabinet, h = 1.5 m

Area painted = Total surface area of the cabinet – Area of bottom surface
Total surface area of the cabinet = 2×(lb+bh+hl)
= 2×(2×1 + 1×1.5 + 1.5×2)
= 2×(2+1.5+3.0)
= 13 m2

Area of bottom = Length × Breadth
= 2 × 1
= 2 m2

\[Area\ painted=13-2=11\ m^2\]

\[The\ required\ surface\ area\ of\ the\ cabinet\ is\ 11\ m^2.\]

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 m2 of area is painted. How many cans of paint will she need to paint the room?

Solution:

Length of wall (l) = 15 m
Breadth of wall (b) = 10 m
Height of wall (h) = 7 m

Total Surface area of classroom = lb+2(bh+hl ) [Floor will not be painted ∴ only lb ]

= 15×10 + 2(10×7 + 7×15)
= 150+2(70+105)
= 150+350
= 500

\[Required\ number\ of\ cans=\frac{Area\ of\ hall}{Area\ of\ one\ can}\]

= 500/100 = 5

Therefore, 5 cans are required to paint the room.

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

Solution:

Similarities : Both figures have the same length and the same height

Differences-

  • The first figure has circular bottom and top
  • The second figure has square bottom and top
  • The first figure is a cylinder and the second figure is a cube

Diameter of cylinder = 7 cm (Given)
Radius of cylinder, r = 7/2 cm
Height of cylinder, h = 7 cm

Lateral surface area of cylinder = 2Ï€rh

\[=2\times\frac{22}{7}\times\frac{7}{2}\times7=154\ cm^2\]

\[\ Lateral\ surface\ area\ of\ cylinder\ is\ 154\ cm^2\]

\[Lateral\ surface\ area\ of\ cube=4\times(side)^2=4\times7^2=4\times49=196\ cm^2\]

\[∴\ Lateral\ surface\ area\ of\ cube\ is\ 196\ cm^2\]

\[Hence,\ the\ cube\ has\ a\ l\arg er\ lateral\ surface\ area.\]

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?

Solution:

Radius of cylindrical tank, r = 7 m
Height of cylindrical tank , h = 3 m

Total surface area of cylindrical tank = 2Ï€r(h+r)

\[=2\times\frac{22}{7}\times7\times\ (3+7)\]

\[=\ 2\ \times22\ \times10\ \ =44\times10=440\ m^2\]

\[Therefore,\ 440\ m^2\ metal\ sheet\ is\ required.\]

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

Solution:

Lateral surface area of hollow cylinder = 4224 cm2
Width of rectangular sheet = 33 cm and lets consider (l) be the length of the rectangular sheet
Lateral surface area of cylinder = Area of the rectangular sheet

4224 = l × b

4224 = l ×33

l = 4224/33 = 128 cm

So the length of the rectangular sheet is 128 cm.

Also, Perimeter of rectangular sheet = 2(l+b)
= 2(128+33)
= 322 cm

The perimeter of the rectangular sheet is 322 cm.

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.

Solution:

Given: Diameter(d) of road roller = 84 cm
∴ Radius of road roller (r)= d/2 = 84/2 = 42 cm

Length of road roller (h) = 1 m = 100 cm
Curved surface area of road roller = 2Ï€rh

\[=\ 2\ \times\frac{22}{7}\times42\ \times100=\ 26400\ cm^2\]

∴ Area covered by road roller in 750 revolutions = 26400 x 750

\[=1,98,00,000\ cm^2=1980\ m^{2\ }[1\ m^2=10,000\ cm^2]\]

\[Hence\ ,the\ area\ of\ the\ road\ is\ 1980\ m^2.\]

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

Solution:

Diameter of the cylindrical container (d) = 14 cm
Radius of cylindrical container (r) = d/2 = 14/2 = 7 cm
Height of cylindrical container = 20 cm

Height of the label, let’s say h = 20 – 2 – 2 (from the figure)
= 16 cm

Curved surface area of label = 2Ï€rh

\[=2\times\frac{22}{7}\times7\times16=704\ cm^2\]

\[Hence,\ the\ area\ of\ the\ label\ is\ 704\ cm^2.\]

Using Class 8 NCERT solutions offers several key advantages for students:

  1. Accurate Answers: NCERT solutions for Class 8 Maths are prepared by expert educators and thoroughly reviewed to ensure quality and accurate answers.
  2. Aligned with Curriculum: NCERT Solutions for Class 8 Maths are specifically designed to go along with the CBSE curriculum, making them the ideal resource for students studying in CBSE affiliated schools.
  3. Crisp an Clear Explanation: The NCERT solutions for Class 8 Maths provide step by step explanations, making complex concepts easier to understand. This helps students to grasp the concepts easily and build a better foundation.
  4. Time-Saver: Instead of spending extra time searching for answers, students can quickly access these NCERT solutions for Class 8 Maths , saving time and allowing them to focus on other aspects of their studies.
  5. Extra Questions: NCERT Solutions for Class 8 Maths also has extra questions that offer additional opportunities for students to build better problem solving skills.
  6. Free Forever: We provide answers which will always be free to access and share.
  7. Self-Study Support: These are beneficial for students who prefer self study. They can independently work through the solutions and improve their understanding of the subject at their own pace.

Summary

On our platform ncertforclass8.com, you will get all the NCERT Solutions for Class 8 Maths expertly solved, ensuring utmost care and accuracy. The NCERT book for Class 8 Maths is strictly followed while creating the NCERT solutions for Class 8 Maths, adhering to the latest CBSE guidelines. We make sure to provide you quality content which will help you to understand the concepts in the best way possible. Also our solutions are completely free for you to access. You have the freedom to copy the NCERT Solutions for Class 8 Maths, download them and even create PDF’s, making it convenient for your offline use.

Leave a Comment

Scroll to Top