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Class 8 Maths Mensuration Solutions

This document contains the solutions to 17 maths questions about mensuration from a Class 8 textbook. It includes step-by-step workings and calculations to find volumes, surface areas, dimensions, costs and ratios for various shapes like aquariums, halls, cubes, cylinders and hollow cylinders. The final high-level summary is that this provides fully worked out solutions to 17 different mensuration word problems involving volumes, surface areas and dimensional analysis of basic 3D shapes.

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ANUSHKA SINGH
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116 views5 pages

Class 8 Maths Mensuration Solutions

This document contains the solutions to 17 maths questions about mensuration from a Class 8 textbook. It includes step-by-step workings and calculations to find volumes, surface areas, dimensions, costs and ratios for various shapes like aquariums, halls, cubes, cylinders and hollow cylinders. The final high-level summary is that this provides fully worked out solutions to 17 different mensuration word problems involving volumes, surface areas and dimensional analysis of basic 3D shapes.

Uploaded by

ANUSHKA SINGH
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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SWARAJ INDIA PUBLIC SCHOOL

CLASS: 8 SUBJECT: MATHS


TOPIC: Mensuration EXERCISE: 18.4
SOLUTION OF CLASSWORK SUMS

Q3. External dimensions of the aquarium = 70 cm x 28 cm x 35 cm


Area of the paper needed to cover the base, side faces and back
faces = Total surface area – (Area of top face + Area of front face)
= 2 𝒍b + 2bh + 2h𝒍 – 𝒍b – 𝒍h
= 𝒍b + 2bh + h𝒍
= 70 x 28 + 2 x 28 x 35 + 35 x 70
= 6370 cm2
Ans. The area of the paper needed is 6370 cm2.

Q6. Perimeter of a floor of the rectangular hall = 236 m


Height = 4.5 m
Area of 4 walls = 2 (𝒍 + b) h
= 236 x 4.5
= 1062 m2
The cost of painting 1 m2 area = 8.40
∴ Total cost of painting = 8.40 x 1062
= 8920.80

Q9. Ratio of dimensions of a rectangular solid = 5 : 4 : 2


Let the dimensions be 5𝒙 , 4𝒙 & 2𝒙 respectively
Acc. to the question
2 (𝒍b + bh + h𝒍) = 1216
2 (5𝒙 x 4𝒙 + 4𝒙 x 2𝒙 + 2𝒙 x 5𝒙) = 1216
2 (20𝒙2 + 8𝒙2 + 10𝒙2) = 1216
𝟏𝟐𝟏𝟔
38𝒙2 = 𝟐
38𝒙2 = 608
𝟔𝟎𝟖
𝒙2 = 𝟑𝟖
𝒙2 = 16
𝒙 = √𝟏𝟔
𝒙=4
∴ Length = 5𝒙
= 5 x 4 ⇒ 20 cm
Breadth = 4𝒙
= 4 x 4 ⇒ 16 cm
Height = 2𝒙
=2x4
= 8 cm
Volume = 𝒍 x b x h
= 20 x 16 x 8
= 2560 cm3

Q11. Dimensions of the cuboidal block = 36 cm x 32 cm x 0.25 m


Volume = 𝒍 x b x h
= (36 x 32 x 25) cm3
The cuboidal block is melted & recast into cubes
Edge of a cube = 4 cm
Volume of cube = (side)3
= (4 x 4 x 4) cm3
𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒄𝒖𝒃𝒐𝒊𝒅
No. of cubes formed = 𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝟏 𝒄𝒖𝒃𝒆
𝟑𝟔 𝐱 𝟑𝟐 𝐱 𝟐𝟓
= 𝟒𝐱𝟒𝐱𝟒
= 450 (Ans.)
Total surface area of a cube = 6a2
=6x4x4
= 96 cm2
Total surface area of 450 cubes = 96 x 450
= 43200 cm2
Cost of silver coating the surface of the cube @ 0.75 per cm2
= 0.75 x 43200
= 32400

Q13. The curved surface area of a cylinder = 4375 m2


It is cut along the height & formed a rectangular sheet
Width of sheet = 35 cm = height of the cylinder
Acc. to the question
Curved surface area = 4375 m2
2𝝅rh = 4375
2𝝅r x 35 = 4375
𝟒𝟑𝟕𝟓
2𝝅r = 𝟑𝟓
2𝝅r = 125
Circumference of cylinder = length of the rectangle
= 125 cm
Perimeter = 2 (𝒍 + b)
= 2 (125 + 35)
= 2 x 160
= 320 cm
Q17. Ratio of the curved surface area to total surface area = 1 : 2
𝟐𝝅𝒓𝒉 𝟏
=
𝟐𝝅𝒓 (𝒉 + 𝒓) 𝟐
𝒉 𝟏
=
𝒉+𝒓 𝟐
2h = h + r
2h – h = r
h=r
Total surface area = 616
𝟐𝝅𝒓 (h + r) = 616
𝟐𝝅𝒉 (h + h) = 616
𝟐𝝅𝒉 x 2h = 616
𝟒𝝅h2 = 616
𝟐𝟐
4x x h2 = 616
𝟕
𝟔𝟏𝟔 𝐱 𝟕
h2 = 𝟒 𝐱 𝟐𝟐
h2 = 49
h = √𝟒𝟗
h = 7 cm
∴ r = 7 cm
Volume = 𝝅r2h
𝟐𝟐
= x7x7x7
𝟕
= 1078 cm3 (Ans.)
Q18. Outer diameter = 4.4 cm
𝒅 𝟒.𝟒 r
Outer radius (R) = 𝟐 = ⇒ 2.2 cm 2cm
𝟐
mm
Outer curved surface area = 𝟐𝝅𝑹𝒉 mm
𝟐𝟐 mm
=2x x 2.2 x 77 77 cm
𝟕 mm
h
= 1064.8 cm2 mm
mm
Inner diameter = 4 cm mm
R
𝒅 𝟒
Inner radius (r) = 𝟐 = 𝟐 ⇒ 2 cm 2.2cm

Inner curved surface area = 𝟐𝝅𝒓𝒉


𝟐𝟐
=2x x 2 x 77
𝟕
= 968 cm2
Total surface area
= Outer surface area + Inner Surface area + 2 (Area of ring)
= 1064.8 + 968 + 𝟐𝝅 (R2 – r2)
𝟐𝟐
= 1064.8 + 968 + 2 x (2.22 – 22)
𝟕
𝟐𝟐
= 1064.8 + 968 + 2 x (4.84 – 4)
𝟕
𝟐𝟐
= 1064.8 + 968 + 2 x x 0.84
𝟕
= 1064.8 + 968 + 5.28
= 2038.08 cm2

H.W.- Do Q 2,8,12 and 16 of ex. 18.4

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