A semicircular sheet of metal of diameter 28 | vedclass

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(B) The diameter of the semicircular sheet is $28 \, cm$,so its radius $R = 14 \, cm$.When this sheet is bent into a cone,the radius of the semicircle

Vedclass · Accepted Answer

$622.36$

Vedclass · Answer

(B) The diameter of the semicircular sheet is $28 \, cm$,so its radius $R = 14 \, cm$.When this sheet is bent into a cone,the radius of the semicircle becomes the slant height $l$ of the cone,so $l = 14 \, cm$.The circumference of the base of the cone is equal to the arc length of the semicircle.Circumference of base $= \pi R = \frac{22}{7} 	imes 14 = 44 \, cm$.Let $r$ be the radius of the base of the cone. Then $2 \pi r = 44$.$2 	imes \frac{22}{7} 	imes r = 44 \implies r = 7 \, cm$.The height $h$ of the cone is given by $h = \sqrt{l^2 - r^2} = \sqrt{14^2 - 7^2} = \sqrt{196 - 49} = \sqrt{147} = 7\sqrt{3} \, cm$.Using $\sqrt{3} \approx 1.732$,$h \approx 7 	imes 1.732 = 12.124 \, cm$.The capacity (volume) of the cup is $V = \frac{1}{3} \pi r^2 h$.$V = \frac{1}{3} 	imes \frac{22}{7} 	imes 7 	imes 7 	imes 7\sqrt{3} = \frac{1078\sqrt{3}}{3} \approx 622.36 \, cm^3$.

$A$ semicircular sheet of metal of diameter $28 \, cm$ is bent into an open conical cup. The capacity of the cup (taking $\pi = \frac{22}{7}$) is (in $cm^3$):

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