Chapter 9 Mensuration Ex 9.2 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 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

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.

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.

Solution:

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

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

Substituting the values, we get

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

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

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

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

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

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

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

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

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.

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 = 2πrh

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

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

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

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

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

Solution:

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

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

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

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

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

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.

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 = 2πrh

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

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

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

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

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 = 2πrh

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

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

  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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