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

Question 18 Mensuration Exercise 17.4

A building is in the form of a cylinder surmounted by a hemisphere valted dome and contains 41\frac{19}{21} m3 of air.

If the internal diameter of dome is equal to its total height above the floor, find the height of the building.



As the building is the combination of cylinder and hemisphere so its volume will be the sum of volume of both the individual solids.

Let the radius of the dome be r.

Internal diameter = 2r

Given internal diameter is equal to total height.

Total height of the building = 2r

Height of the hemispherical area = r

So height of cylindrical area, h = 2r-r = r

Volume of the building = 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 Volume of the building = 41\frac{19}{21} =880/21

5/3)π r3= 880/21

(5/3)×(22/7)×r3= 880/21

r3 = 880×3×7/(5×22×21)

r3 = 880/110

r3 = 8

Taking cube root

r = 2 m

Height of the building = 2r = 2×2 = 4m

Hence the height of the building is 4m. M L Aggarwal - Understanding ICSE Mathematics - Class 10 chapter Mensuration Question 18 Solution image

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