Math, asked by panav5660, 1 year ago

Curved surface area of a cylinder is 528 sq cm. If circumference of its base is 44 cm, find the height of the cylinder?

Answers

Answered by sumit9695
0
here is your answers my dear friend
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Answered by Anonymous
59

Step-by-step explanation:

Answer:

\setlength{\unitlength}{1.5 cm} \thicklines \begin{picture}(2,0)\qbezier(0,0)(0,0)(0,2.5)\qbezier(2,0)(2,0)(2,2.5)\qbezier(0,0)(1,0.6)(2,0)\qbezier(0,0)( 1, - 0.6)(2,0) \put(2.3,1){\vector(0,1){1.5}}\put(2.3,1){\vector(0, - 1){1.2}}\put(2.5,1){ $\bf 8  \: cm$}\put(0.4,0.1){ $\bf r \: cm$}\put(0,0){\vector(1,0){1}}\qbezier(0,2.5)(1,1.9)(2,2.5)\qbezier(0,2.5)(1,3.1)(2,2.5)\end{picture}

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Total surface area of cylinder = 528 cm²

Height of cylinder = 8 cm

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☯ Let Radius of the base of cylinder be r cm.

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

:\implies\sf 2 \pi r(h + r) = 528\\ \\

:\implies\sf 2 \times \dfrac{22}{7} \times r(8 + r) = 528\\ \\

:\implies\sf \dfrac{44}{7} \times r(8 + r) = 528\\ \\

:\implies\sf r(8 + r) = \cancel{528} \times \dfrac{7}{ \cancel{44}}\\ \\

:\implies\sf 8r + r^2 = 12 \times 7\\ \\

:\implies\sf 8r + r^2 = 84\\ \\

:\implies\sf r^2 + 8r - 84 = 0\\ \\

:\implies\sf r^2 - 6r + 14r - 84 = 0\\ \\

:\implies\sf r(r - 6) + 14(r - 6) = 0\\ \\

:\implies\sf (r - 6)(r + 14) = 0\\ \\

:\implies\sf (r - 6) = 0\quad or \quad(r + 14) = 0\\ \\

:\implies\sf r = 6 \quad or \quad r = - 14\\ \\

We know that, radius of a cylinder can't be negative.

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\therefore\;{\underline{\textsf{Hence,\;Radius\;of\;cylinder\;is\; \textbf{6 cm}.}}}

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\boxed{\begin{minipage}{6 cm}\bigstar$\:\underline{\textbf{Formulae Related to Cylinder :}}\\\\\sf {\textcircled{\footnotesize\textsf{1}}} \:Area\:of\:Base\:and\:top =\pi r^2 \\\\ \sf {\textcircled{\footnotesize\textsf{2}}} \:\:CSA =2 \pi rh\\\\\sf{\textcircled{\footnotesize\textsf{3}}} \:\:TSA = Area \: of \: Base + CSA  + area \: of \: top \\{\quad\:\:\quad=\pi r^2  +  2\pi rh+\pi r^2 = 2 \pi r(h + r)}\\ \\{\textcircled{\footnotesize\textsf{4}}} \: \:Volume=\pi r^2h\end{minipage}}

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