A hemispherical bowl is made of steel, $0.25$ cm thick. The inner radius of the bowl is $5$ cm. Find the outer curved surface area of the bowl.

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Solution

Given, Inner radius of the bowl $= 5$ cm
Thickness of steel $= 0.25$ cm
Outer radius of the bowl $= 5 + 0.25 = 5.25$ cm
Outer curved surface area of the bowl = $2 π R^{2}$
$= 2 \times \frac{22}{7} \times (5.25)^{2} = 173.25 c m^{2}$

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A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Given, Inner radius of the bowl =5 cmThickness of steel =0.25 cmOuter radius of the bowl=5+0.25=5.25 cmOuter curved surface area of the bowl = 2πR2=2×227×(5.25)2=173.25 cm2

A hemispherical bowl is made of steel, $0.25$ cm thick. The inner radius of the bowl is $5$ cm. Find the outer curved surface area of the bowl.

Given, Inner radius of the bowl $= 5$ cm
Thickness of steel $= 0.25$ cm
Outer radius of the bowl $= 5 + 0.25 = 5.25$ cm
Outer curved surface area of the bowl = $2 π R^{2}$
$= 2 \times \frac{22}{7} \times (5.25)^{2} = 173.25 c m^{2}$