A joker’s cap is in the form of a right- circular cone whose base radius is 7 cm and height is 24 cm. The area of the sheet required to make 10 such caps is

A) 550 cm2

B) 5500 cm2

C) 55000 cm2

D) None

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2 Answers

Correct option is: B) 5500 cm2

We have r = 7 cm and h = 24 cm.

Area of sheet required to make 1 joker's cap

\(\pi rl = \pi r\sqrt {r^2+h^2}\) (\(\because\) l = \(\sqrt {r^2+h^2}\)))

\(\frac {22}7 \times 7\sqrt {7^2+24^2}\)

\(22\sqrt{49+576} = 22\sqrt{625} = 22 \times 25 = 550 cm^2\)

\(\therefore\) Area of sheet required to make 10 such caps = 550 \(\times\) 10 \(cm^2\)= 5500 \(cm^2\).

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Correct option is: B) 5500 cm2

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