The diameters of a frustum of a cone are 32 , | vedclass

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(A) Given: Diameter $D_1 = 32 \,cm$,so radius $r_1 = 16 \,cm$. Diameter $D_2 = 20 \,cm$,so radius $r_2 = 10 \,cm$. Height $h = 8 \,cm$.$1$. Slant heig

Vedclass · Accepted Answer

Curved Surface Area: $816.4 \,cm^2$,Volume: $4320.64 \,cm^3$

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(A) Given: Diameter $D_1 = 32 \,cm$,so radius $r_1 = 16 \,cm$. Diameter $D_2 = 20 \,cm$,so radius $r_2 = 10 \,cm$. Height $h = 8 \,cm$.$1$. Slant height $l = \sqrt{h^2 + (r_1 - r_2)^2} = \sqrt{8^2 + (16 - 10)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \,cm$.$2$. Curved Surface Area $(CSA)$ = $\pi(r_1 + r_2)l = 3.14 	imes (16 + 10) 	imes 10 = 3.14 	imes 26 	imes 10 = 816.4 \,cm^2$.$3$. Volume $V = \frac{1}{3}\pi h(r_1^2 + r_2^2 + r_1r_2) = \frac{1}{3} 	imes 3.14 	imes 8 	imes (16^2 + 10^2 + 16 	imes 10) = \frac{1}{3} 	imes 3.14 	imes 8 	imes (256 + 100 + 160) = \frac{1}{3} 	imes 3.14 	imes 8 	imes 516 = 3.14 	imes 8 	imes 172 = 4320.64 \,cm^3$.

The diameters of a frustum of a cone are $32 \,cm$ and $20 \,cm$. If its height is $8 \,cm$,find its curved surface area and volume. $(\pi = 3.14)$

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