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A building is in the form of a cylinder surmounted by a hemisphere valted dome and contains 411921m341\dfrac{19}{21} m^3 of air. If the internal diameter of dome is equal to the total height of the building, find the height of the building.

Mensuration

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Answer

The below figure shows the building in the form of a cylinder surmounted by a hemisphere valted dome:

A building is in the form of a cylinder surmounted by a hemisphere valted dome and contains 4119/21 m3 of air. If the internal diameter of dome is equal to the total height of the building, find the height of the building. Mensuration, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Let the radius of the dome be r.

∴ Internal diameter = 2r.

Given, internal diameter is equal to height.

∴ Height of building (h) = 2r.

Height of hemispherical area = r.

So, height of cylindrical area, h1 = 2r - r = r.

Volume of building (V) = Volume of cylindrical area + Volume of hemispherical area.

V=πr2h1+23πr3V=πr2.r+23πr3V=πr3+23πr3V=53πr3.\therefore V = πr^2h_1 + \dfrac{2}{3}πr^3 \\[1em] \Rightarrow V = πr^2.r + \dfrac{2}{3}πr^3 \\[1em] \Rightarrow V = πr^3 + \dfrac{2}{3}πr^3 \\[1em] \Rightarrow V = \dfrac{5}{3}πr^3.

Given, V = 411921=88021 m341\dfrac{19}{21} = \dfrac{880}{21} \text{ m}^3

53×227×r3=88021r3=880×3×75×22×21r3=184802310r3=8r=2 m.\therefore \dfrac{5}{3} \times \dfrac{22}{7} \times r^3 = \dfrac{880}{21} \\[1em] \Rightarrow r^3 = \dfrac{880 \times 3 \times 7}{5 \times 22 \times 21} \\[1em] \Rightarrow r^3 = \dfrac{18480}{2310} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r = 2 \text{ m}.

h = 2r = 2(2) = 4 m.

Hence, the height of the building is 4 m.

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