Question

If the radius of base of a cone is tripled and height is doubled, then find the volume of a new cone.
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Answer 1

$\underline{\bigstar\:\textsf{According to the Given Question :}}$

Radius of Base of a cone is tripled and Height of a cone is double means r is 3r and h is 2h.

$\begin{gathered}:\implies\sf Volume_{\:(cone)} = \dfrac{1}{3} \pi r^2 h \\\\\\:\implies\sf Volume_{\:(cone)} = \dfrac{1}{3} \pi (3r)^2 \times (2h) \\\\\\:\implies\sf Volume_{\:(cone)} = \dfrac{1}{\cancel{\;3}} \pi \; \cancel{\;9}\;r^2 \times 2h\\\\\\:\implies\sf Volume_{\:(cone)} = 6 \pi r^2 h\\\\\\:\implies{\underline{\boxed{\sf{\mathcal{V}olume_{\:(cone)} = 18\bigg( \dfrac{1}{3}\pi r^2h\bigg)}}}}\end{gathered}$

⠀⠀

$\therefore{\underline{\sf{Hence, \;it\; becomes\;\bf{18\;times}\sf{\;more\;than\;old\;volume.}}}}$

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$\begin{gathered}\qquad\qquad\boxed{\bf{\mid{\overline{\underline{\purple{\bigstar\: Formulae\:related\:to\:cone :}}}}}\mid}\\\\\end{gathered}$

$\sf Area\:of\:base = \bf{\pi r^2}$

$\sf Curved\:surface\:area\:of\:cone = \bf{\pi rl}$

$\sf Total\:surface\:area\:of\:cone = Area\:of\:base + CSA = \pi r^2 + \pi rl = \bf{\pi r(r + l)}$

$\sf Volume\:of\:cone = \bf{\dfrac{1}{3} \pi r^2 h}$