Question

A hemispherical bowl is made of 0.25 cm thick steel. Inner radius of this bowl is 5 cm. Find the external curved surface of the bowl.

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Answer 1

Hi there ^_^

⬇SOLUTION⬇

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Internal radius r = 5 cm

So, external radius R = internal radius + thickness of steel = 5 cm + 0.25 cm = 5.25 cm

So, external curved surface are (CSA) of the bowl = 2 × pi × R^2

= 2 × 22/7 × (5.25)^2 cm^2

= 2 × 22/7 × 525/100 × 525/100 cm^2

= 2 × 22/7 × 21/4 × 21/4 cm^2

= 693/4 cm^2

= 173.25 cm^2
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Hope it helps ^_^