# 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.

Solution:

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

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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