Question

If both the radius and the height of a right circular cone are increased by 20%, find the percentage of increase in the volume.

Answer 1

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Answer 2

$Let \:r \:and \:h \:are \: radius \:and \: height \\of \: a \:right \:circular \:cone$

$Volume \: of \: the \: cone (V_{1}) = \frac{1}{3}\pi r^{2} h \: --(1)$

/* If radius and height increased by 20% then new measures are */

$New \: radius = r\Big( \frac{100+20}{100}\Big) \\= r\Big(\frac{120}{100}\Big) \\= 1.2r$

$New \: height = h\Big( \frac{100+20}{100}\Big) \\= h\Big(\frac{120}{100}\Big) \\= 1.2h$

$Now, Volume \: of \: new \: cone (V_{2}) \\= \frac{1}{3} \pi \Big(1.2r\Big)^{2} \times (1.2h) \\= 1.44 \times 0.4 \pi r^{2} h \\= 0.576 \pi r^{2}h\:--(2)$

$Increase \: in \:the \: volume \: of \:cone \\= V_{2} - V_{1} \\= 0.576 \pi r^{2} h - \frac{1}{3}\pi r^{2} h \\= \pi r^{2} h \Big( 0.576 - \frac{1}{3}\Big) \\= \frac{1.728-1}{3}\pi r^{2}h\\= \frac{0.728}{3} \pi r^{2}h \: --(3)$

$Percentage \: of \: increase \:in \: the \: volume \\= \frac{Increased\: volume }{V_{1}} \times 100 \\= \frac{\frac{0.728}{3}\pi r^{2}h }{\frac{1}{3} \pi r^{2} h }\times 100\\= 0.728 \times 100 \\= 72.8\%$

Therefore.,

$\red {Percentage \: of \: increase \:in \: the \: volume}\\\green {= 72.8\%}$

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