Question

How much ice-cream can be put into a cone with base radius 3.5 cm and slant height 12.5cm?​

Answer 1

$\sf Given \begin{cases} & \sf{Radius\;of\;cone,\;r = \bf{3.5\;cm}} \\ & \sf{Slant\;height\;of\;cone,\;l = \bf{12.5\;cm}} \end{cases}$

To find: How much ice-cream can be put into a cone?

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$\setlength{\unitlength}{1.8mm}\begin{picture}(5,5)\thicklines\put(0,0){\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\put(-0.5,-1){\line(1,2){13}}\put(25.5,-1){\line(-1,2){13}}\multiput(12.5,-1)(2,0){7}{\line(1,0){1}}\multiput(12.5,-1)(0,4){7}{\line(0,1){2}}\put(15,1.6){\sf{3.5 cm}}\put(14.5,10){\sf{h}}\put(21,10){\sf{12.5 cm}}\end{picture}$

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$\dag\;{\underline{\frak{As\;we\;know\;that,}}}$

$\star\;{\boxed{\sf{\pink{(slant\;height)^2 = (Base)^2 + (Height)^2}}}}$

$:\implies\sf l^2 = b^2 + h^2$

$:\implies\sf (12.5)^2 = (3.5)^2 + h^2$

$:\implies\sf h^2 = (12.5)^2 - (3.5)^2$

$:\implies\sf h^2 = 156.25 - 12.25$

$:\implies\sf h^2 = 144$

$:\implies\sf \sqrt{h^2} = \sqrt{144}$

$:\implies{\underline{\boxed{\frak{\purple{h = 12\;cm}}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Height\; of\;cone\;is\; \bf{12\;cm}.}}}$

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$\underline{\sf{\bigstar\:{According\:to\:the\:question\::}}}$

Amount of ice - cream that can be put into a cone = Volume of cone

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Therefore,

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

$:\implies\sf Volume_{\;(cone)} = \dfrac{1}{3} \times \dfrac{22}{7} \times (3.5)^2 \times 12$

$:\implies\sf Volume_{\;(cone)} = \dfrac{1}{ \cancel{3}} \times \dfrac{22}{7} \times 12.5 \times \cancel{12}$

$:\implies\sf Volume_{\;(cone)} = \dfrac{22}{7} \times 12.5 \times 4$

$:\implies{\underline{\boxed{\frak{\purple{Volume_{\;(cone)} = 157.1428\;cm^3}}}}}\;\bigstar$

${\sf{\therefore{Amount\; of\;ice-cream\;that\;can\;be\;put\;into\;a\;cone\;is\; \bf{157.1428\;cm^3}.}}}$

Answer 2

$\Huge \bf {Given:-}$

base radius =3.5cm

Slant height=12.5cm

$\Huge \bf {To \:Find:-}$

How much ice-cream can be put into a cone?

$\Huge \bf {Diagram:-}$

$\setlength{\unitlength}{1.8mm}\begin{picture}(5,5)\thicklines\put(0,0){\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\put(-0.5,-1){\line(1,2){13}}\put(25.5,-1){\line(-1,2){13}}\multiput(12.5,-1)(2,0){7}{\line(1,0){1}}\multiput(12.5,-1)(0,4){7}{\line(0,1){2}}\put(15,1.6){\bf{3.5 cm}}\put(14.5,10){\bf{h}}\put(21,10){\bf{12.5 cm}}\end{picture}$

$\Huge \bf {Solution:-}$

$\sf {\boxed{\purple{(slant \:height)^2=(base)^2+(height)^2}}}$

$\sf (12.5)^2=(3.5)^2+h^2$

$\sf 156.25=12.25+h^2$

$\sf h^2=156.25-12.25$

$\sf h=\sqrt{144}cm$

$\sf h=12cm$

${\text{Quantity\:of\:ice-cream\:=\:volume \:of\:cone}}$

${\bf{\boxed{\pink{Volume\:of\:cone=\dfrac{1}{3}×\pi ×(r)^2×h}}}}$

$\bf \implies{\dfrac{1}{\cancel 3}×\dfrac{22}{7}×(3.5)^2×{\cancel 12}^4}$

$\bf \implies \dfrac{22}{7}×12.25×4$

$\bf \implies \dfrac{22}{7}×49$

$\bf \implies 22×7$

$\bf \implies 154cm^3$

$\;{\underline{\sf{Amount\; of\;ice-cream\;that\;can\;be\;put\;into\;a\;cone\;is\; \bf{154\;cm^3}.}}}$