The base of a cone with radius 6 , cm and hei | vedclass

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Question

(A) Given: Radius of cone and hemisphere $r = 6 \, cm$,Height of cone $h = 8 \, cm$. First,find the slant height $l$ of the cone: $l = \sqrt{r^2 + h^2

Vedclass · Accepted Answer

$414.48 \, cm^2, 753.6 \, cm^3$

Vedclass · Answer

(A) Given: Radius of cone and hemisphere $r = 6 \, cm$,Height of cone $h = 8 \, cm$. First,find the slant height $l$ of the cone: $l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, cm$. Curved Surface Area $(CSA)$ of the solid = $CSA$ of cone + $CSA$ of hemisphere. $CSA$ = $\pi rl + 2\pi r^2 = \pi r(l + 2r) = 3.14 	imes 6 	imes (10 + 2 	imes 6) = 18.84 	imes (10 + 12) = 18.84 	imes 22 = 414.48 \, cm^2$. Volume of the solid = Volume of cone + Volume of hemisphere. Volume = $\frac{1}{3}\pi r^2h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi r^2(h + 2r) = \frac{1}{3} 	imes 3.14 	imes 6^2 	imes (8 + 2 	imes 6) = \frac{1}{3} 	imes 3.14 	imes 36 	imes (8 + 12) = 3.14 	imes 12 	imes 20 = 37.68 	imes 20 = 753.6 \, cm^3$.

The base of a cone with radius $6 \, cm$ and height $8 \, cm$ is hemispherical. Find the curved surface area and the volume of the solid. $(\pi = 3.14)$

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