A cylinder with radius 8 , cm and height 9 , | vedclass

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Question

(A) Given: Radius of cylinder $(r)$ = $8 \, cm$,Height of cylinder $(h_1)$ = $9 \, cm$,Height of cone $(h_2)$ = $15 \, cm$.Slant height of cone $(l)$

Vedclass · Accepted Answer

$880 \, cm^2, 2816 \, cm^3$

Vedclass · Answer

(A) Given: Radius of cylinder $(r)$ = $8 \, cm$,Height of cylinder $(h_1)$ = $9 \, cm$,Height of cone $(h_2)$ = $15 \, cm$.Slant height of cone $(l)$ = $\sqrt{r^2 + h_2^2} = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \, cm$.Total Surface Area = Curved surface area of cylinder + Curved surface area of cone + Base area of cylinder.$= 2\pi rh_1 + \pi rl + \pi r^2 = \pi r(2h_1 + l + r) = \frac{22}{7} 	imes 8 	imes (2 	imes 9 + 17 + 8) = \frac{176}{7} 	imes (18 + 17 + 8) = \frac{176}{7} 	imes 43 \approx 1081.14 \, cm^2$.(Note: Using $\pi = 3.14$,Area $\approx 3.14 	imes 8 	imes 43 = 1079.68 \, cm^2$).Volume = Volume of cylinder + Volume of cone = $\pi r^2 h_1 + \frac{1}{3} \pi r^2 h_2 = \pi r^2 (h_1 + \frac{h_2}{3}) = \frac{22}{7} 	imes 8^2 	imes (9 + \frac{15}{3}) = \frac{22}{7} 	imes 64 	imes 14 = 22 	imes 64 	imes 2 = 2816 \, cm^3$.

Part $I$	Part $II$
$1.$ Volume of a cuboid	$a.$ $\frac{1}{3} \pi r^{2} h$
$2.$ Volume of a cone	$b.$ $\pi r^{2} h$
$3.$ Volume of a cylinder	$c.$ $\frac{4}{3} \pi r^{3}$
$4.$ Volume of a sphere	$d.$ $lbh$

$A$ cylinder with radius $8 \, cm$ and height $9 \, cm$ is surmounted by a cone of height $15 \, cm$. Find the total surface area and the volume of the combined solid.

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Which of the following correctly matches the information given in Part $I$ and Part $II$?

Part $I$ Part $II$

$1.$ Volume of a cuboid $a.$ $\frac{1}{3} \pi r^{2} h$

$2.$ Volume of a cone $b.$ $\pi r^{2} h$

$3.$ Volume of a cylinder $c.$ $\frac{4}{3} \pi r^{3}$

$4.$ Volume of a sphere $d.$ $lbh$

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