# NCERT Exemplar Solutions Class 10 Mathematics Solutions for Surface Areas and Volumes - Exercise 12.4 in Chapter 12 - Surface Areas and Volumes

Question 4 Surface Areas and Volumes - Exercise 12.4

A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contain 41\ \frac{19}{21}m^3 of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building.

Answer:

Dome (hemisphere) Cylinder
Radius = r Radius = r

H = 2r [[given]]

⇒ h + r = 2r

⇒ h = 2r – r = r

Volume of building =41\ \frac{19}{21}m^3 = \frac{880}{21}m^3

⇒ Vol. of cylinder + Vol. of hemisphere = \frac{880}{21}m^3

\Rightarrow \pi r^{2} h+\frac{2}{3} \pi r^{3}=\frac{880}{21} \\ \Rightarrow \pi r^{2} r+\frac{2}{3} \pi r^{3}=\frac{880}{21}[\because h=r] \\ \Rightarrow \frac{5}{3} \pi r^{3}=\frac{880}{21} \\ \Rightarrow \frac{5}{3} \times \frac{22}{7} \times r^{3}=\frac{880}{21} \\ \Rightarrow r^{3}=\frac{880 \times 3 \times 7}{21 \times 5 \times 22} \\ \Rightarrow r^{3}=8 \\ \Rightarrow r=2 m

Hence, height of the building is 2 × 2 = 4 m. [[∵ H = 2r (Given)]]

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