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(C) The volume of a cone is given by $V = \frac{1}{3} \pi r^2 h$.Given $r = 50 \ cm$ and $h = 25 \ cm$,the volume is $V = \frac{1}{3} \pi (50)^2 (25)

Vedclass · Accepted Answer

$7857.14$

Vedclass · Answer

(C) The volume of a cone is given by $V = \frac{1}{3} \pi r^2 h$.Given $r = 50 \ cm$ and $h = 25 \ cm$,the volume is $V = \frac{1}{3} \pi (50)^2 (25) = \frac{62500 \pi}{3} \ cm^3$.When the cone is recast into a sphere,the volume remains constant. Let $R$ be the radius of the sphere.Volume of the sphere $= \frac{4}{3} \pi R^3$.Equating the volumes: $\frac{4}{3} \pi R^3 = \frac{62500 \pi}{3}$.$4 R^3 = 62500 \Rightarrow R^3 = 15625$.Taking the cube root,$R = \sqrt[3]{15625} = 25 \ cm$.The surface area of the sphere is $A = 4 \pi R^2$.$A = 4 	imes \frac{22}{7} 	imes (25)^2 = 4 	imes \frac{22}{7} 	imes 625 = \frac{55000}{7} \approx 7857.14 \ cm^2$.

$A$ solid cone is $25 \ cm$ high and the radius of its base is $50 \ cm$. It is melted and recast into a solid sphere. Determine the surface area of the sphere (in $cm^2$).

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