ML Aggarwal Solutions Class 10 Mathematics Solutions for Mensuration Exercise 17.4 in Chapter 17 - Mensuration

Question 17 Mensuration Exercise 17.4

A circular hall (big room) has a hemispherical roof. The greatest height is equal to the inner diameter, find the area of the floor, given that the capacity of the hall is 48510 m³.



As the hall comprises of cylinder and hemisphere so its capacity or volume will be the sum of volume of cylinder and hemisphere.

M L Aggarwal - Understanding ICSE Mathematics - Class 10 chapter Mensuration Question 17 Solution image

Let the radius of the hemisphere be r.

Inner diameter = 2r

Given greatest height equal to inner diameter.

So total height of the hall = 2r

Height of the hemispherical part = r

Height of cylindrical area, h = 2r-r = r

Capacity of the hall = Volume of cylindrical area + volume of hemispherical area

=π r2h + (2/3)π r3

=π r3+ (2/3)π r3 [∵h = r]

=π r3 (1+2/3)

=π r3 (3+2)/3

= (5/3)π r3

Given capacity of hall = 48510 m3

(5/3)π r3= 48510

(5/3)×(22/7)r3= 48510

r3 = 48510×3×7/(22×5)

r3 = 9261

Taking cube root on both sides,

r = 21

Area of the floor = r2

= (22/7)×212

= 22×21×3

= 1386 m2

Hence the Area of the floor is 1386 m2 .

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