Question

the area of curved surface of a cone is 60πcm2.if the slant height of the cone is 8cm,find the diameter of the base.​

Answer 1

Given:

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• Curved surface area of cone = 60π cm²
• Slant height of cone = 8 cm

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To find:

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• Diameter of base of cone?

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Solution:

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☯ Let radius of cone be r cm.

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$\setlength{\unitlength}{1.5mm}\begin{picture}(5,5)\thicklines\put(0,0){\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\put(-0.5,-1){\line(1,2){13}}\put(25.5,-1){\line(-1,2){13}}\multiput(12.5,-1)(2,0){7}{\line(1,0){1}}\multiput(12.5,-1)(0,4){7}{\line(0,1){2}}\put(18,1.6){\sf{r}}\put(22,10){\sf{8 cm}}\end{picture}$

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We know that,

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$\star\;{\boxed{\sf{\purple{CSA_{\;(cone)} = \pi rl}}}}$

$:\implies\sf \cancel{\pi} \times r \times 8 = 60 \cancel{\pi}$

$:\implies\sf r \times 8 = 60$

$:\implies\sf r = \cancel{ \dfrac{60}{8}}$

$:\implies{\boxed{\sf{\pink{r = 7.5\;cm}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Thus,\;Radius\;of\;cone\;is\; \bf{7.5\;cm}.}}}$

We know that,

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• Diameter = 2 × Radius

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Therefore,

Diameter of base of cone is 15 cm.

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

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$\sf (i)\;Total\;surface\;area\;of\;cone\; = \; \red{\pi r(r + l)}$

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$\sf (ii)\;Volume\;of\;cone\; = \; \purple{ \dfrac{1}{3} \pi r^2 h}$

Answer 2

$\:\:\: \huge{\underline{\boxed{\sf{\red{Solution}}}}}$

Given :

• Curved surface area of cone = 60π cm²
• Slant height of cone = 8 cm

To Find :

• Diameter of the base of cone

Formula used :-

$\:\:\: \star \: {\boxed{\sf{Curved\:Surface\:Area\:of\:cone\:=\: \pi rl}}}$

• Now , let us substitute the given values in the above formula :

$\:\:\: \implies \: \sf 60 \pi \:=\: \pi (r)8$

$\:\:\: \implies \: \sf 60 \pi \:=\:8 \pi (r)$

$\:\:\: \implies \: \sf r\:=\: \cancel \dfrac{60 \pi }{8 \pi }$

$\:\:\: \implies \: \sf r\:=\:7.5\:cm$

$\star \: {\sf{\underline{Radius\:of\:base\:=\:7.5\:cm}}}$

• Now , let us find diameter

$\:\:\: \star \: {\boxed{\sf{Diameter\:=\:2 \times Radius}}}$

$\:\:\: \implies \: \sf Diameter\:=\: 2 \times 7.5$

$\:\:\: {\red \bigstar\: \large {\sf{\underline{\pink{Diameter\:of\:base\:=\:15\:cm}}}}}$

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