Question

9. The C.S.A of a right circular cylinder of height 14 cm is 88 cm find the diameter of the base of the cylinder​

Answer 1

❍ Let's Consider Base Diameter of Cylinder be x cm .

$\begin{gathered}\frak {\underline {\dag As \:We\:know\:that\::}}\\\end{gathered}$

$\begin{gathered}\star\boxed{\pink{\bf{ Curved \:Surface \:Area\:_{(Cylinder)} = 2 \pi \times \dfrac{d}{2}\times h \:sq.units}}}\\\end{gathered}$

Where,

d is the Diameter of Cylinder in cm and h is the Height of Cylinder in cm and

$\pi = \dfrac{22}{7}$

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$⠀\begin{gathered}\underline {\frak{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}\\\end{gathered}$

$\begin{gathered}:\implies \sf{ 88cm^{2} = 2 \times \pi \times \dfrac{x}{2}\times 14}\\\\\end{gathered}$

$\begin{gathered}:\implies \sf{ 88cm^{2} = \cancel {2} \times \pi \times \dfrac{x}{\cancel {2}}\times 14}\\\\\end{gathered}$

$\begin{gathered}:\implies \sf{ 88cm^{2} = \pi \times x\times 14}\\\\\end{gathered}$

$\begin{gathered}:\implies \sf{ 88cm^{2} = \dfrac{22}{\cancel {7}} \times x\times \cancel {14}}\\\\\end{gathered}$

$\begin{gathered}:\implies \sf{ 88cm^{2} = 22 \times x \times 2 }\\\\\end{gathered}$

$\begin{gathered}:\implies \sf{ 88cm^{2} = 44 \times x }\\\\\end{gathered}$

$\begin{gathered}:\implies \sf{ \dfrac{\cancel {88}}{\cancel {44}} = 44 \times x }\\\\\end{gathered}$

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$\begin{gathered}\underline {\boxed{\pink{ \mathrm { x = 2\: cm}}}}\:\bf{\bigstar}\\\end{gathered}$

Therefore,

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$\begin{gathered}\underline {\therefore\:{ \mathrm { Hence,\:Diameter \:of\:Cylinder \:is\:\bf{2\: cm}}}}\\\end{gathered}$

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$\begin{gathered}\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}\\\\\end{gathered}$

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$

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