Question

the height of cone is15cm . its volume is 1570 find the radious​

Answer 1

Question -

Find the radius of the cone if the height of the cone is 15 cm and its volume is 1570cm²

Solution -

Given,

height of the cone (h) = 15cm

We know that,

$volume \: of \: \: the \: cone = \frac{1}{3} \pi {r}^{2} h$

Putting the values we will get -

$1570 = \frac{1}{3} \times \frac{22}{7} \times {r}^{2} \times 15$

$1570 = \frac{110}{7} \times {r}^{2}$

${r}^{2} = 1570 \times \frac{7}{110}$

${r}^{2} = 99.909090......$

${r}^{2} = 100(approx)$

$r = \sqrt{100}$

$r = 10$

Therefore the radius of the cone is 10 cm

Related Formulae -

$1)volume \: of \: cone = \frac{1}{3} \pi {r}^{2} h$

$2)curved \: surface \: area \: of \: \: cone = \pi \: rl$

$3)total \: \: surface \: \: area \: \: of \: \: cone = \pi \: r(l + r)$

Answer 2

$\setlength{\unitlength}{30} \begin{picture}(10,6) \linethickness{1.2} \qbezier(1,1)(3., 0)(5,1)\qbezier(1,1)(3.,2)(5,1)\put(3,1){\circle*{0.15}}\put(3,1){\line(0,1){3}}\qbezier(1,1)(1,1)(3,4)\qbezier(5,1)(3,4)(3,4)\put(3,1){\line(1,0){2}}\put(3.2,1.1){ \bf r \: cm }\put(1.9,1.9){ \bf 15 \: cm }\put(4,3.5){\boxed{ \bf @ValiantPrinceiss</p><p> }}\end{picture}$

$\bf GivEn\begin{cases} & \sf{Height\;of\;cone = \bf{15\;cm}} \\ & \sf{Volume\;of\;cone = \bf{1570\;cm}} \end{cases}$

We have to find, radius of the cone.

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$\star\;{\underline{\frak{We\;know\;that,}}}$

$\star\;{\boxed{\sf{\purple{Volume_{\;(cone)} = \dfrac{1}{3} \pi r^2 h}}}}$

$\star\;{\underline{\frak{Putting\;values,}}}$

$:\implies\sf \frac{1}{3 } \times \frac{22}{7} \times {x}^{2} \times 15 \: = 1570$

$:\implies\sf {x}^{2} = 100$

$:\implies\sf x = \sqrt{100}$

$:\implies\sf x=10$

$:\implies{\boxed{\frak{\pink{10\;cm}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;Radius\;of\;cone\;is\; \bf{10\;cm}.}}}$

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$\qquad\qquad\qquad\boxed{\underline{\underline{\purple{\bigstar \: \bf\:More\:to\:know\:\bigstar}}}}$

$\boxed{\begin{minipage}{6 cm}\bigstar \:\underline{\textbf{Formulae Related to Cone :}}\\\\\sf {\textcircled{\footnotesize\textsf{1}}} \:Area\:of\:Base =\pi r^2 \\\\ \sf {\textcircled{\footnotesize\textsf{2}}} \:\:CSA = \pi rl\\\\\sf{\textcircled{\footnotesize\textsf{3}}} \:\:TSA = Area\:of\:Base + CSA\\{\quad\:\:\:\qquad=\pi r^2+\pi rl}\\ \\{\textcircled{\footnotesize\textsf{4}}} \: \:Volume=\dfrac{1}{3}\pi r^2h\end{minipage}}$