Question

If the volume of a right circular cone of height 9 cm is 48 π cm³, find the diameter of its base.​

Answer 1

$\red{\boxed{\boxed{ diameter = 12 cm }}}$

Step-by-step explanation:

If the volume of a right circular cone of height 9 cm is 48 π cm³, find the diameter of its base.

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$v = 48\pi \\ \frac{1}{3} \: \pi \: {r}^{2} = 48\pi \: \: \: \: \: \: \: \: \: \: \\ \frac{1}{3} {r}^{2} =48 \: \: \: \: \: \: \: \: \\ {r}^{2} = 144 \: \: \: \\ \: \: \: r = \sqrt{144} \\ \: \: \: r = 12 \: cm$

then :

diameter

= 2 x r

= 2 x 12

= 24 cm

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${\purple{\boxed{ \:  \:  {\mathfrak{\underline{\purple{ @11177420 }}}} }}}$

Answer 2

Given - Volume and height of a right circular cone are 48πcm³ and 9cm respectively.

To Find - Diameter of it's base

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As we know that :

$\: \: \:$

$\underline{\boxed{\tt\purple{\bigstar \: volume \: of \: a \: right \: circular \: cone = \frac{1}{3}\pi \:r {}^{2}h }}}$

$\: \:$

Put the values :

$\: \:$

$\tt:\implies \: \dfrac{1}{3} \pi \: r {}^{2} h = 48\pi \\ \\ \\ \tt:\implies \: \frac{1}{3} \pi \: r {}^{2}9 = 48\pi \\ \\ \\ \tt:\implies \: \frac{1}{\cancel{3}}\cancel{\pi} \: r {}^{2}\cancel{9} = 48\cancel{\pi} \\ \\ \\ \tt:\implies \: 3r {}^{2} = 48 \\ \\ \\ \tt:\implies \: {r}^{2} = \cancel{\frac{48}{3} } \\ \\ \\ \tt:\implies \: {r}^{2} = 16 \\ \\ \\ \tt:\implies \: r = \sqrt{16 } \\ \\ \\ \tt:\implies \: r = \underline{\boxed{\sf\purple{4cm}}}\bigstar$

$\: \:$

Now,

We know that

$\tt:\implies \: D = 2 × R$

$\tt:\implies D = 2 × 4$

$\tt:\implies D = \underline{\boxed{\sf\purple{8cm}}} \bigstar$

$\: \:$

$\tt\therefore{\underline{Hence, \: the \: diameter \: of \: the \: base \: of \: the \: right \: circular \: cylinder \: is \: \bold{8cm}.}}$