Question

The curved surface area of a right circular cylinder is 264 cm2 and its volume is 924 cm3 , find the height of the cylinder.

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Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

• Radius of cylinder be r cm.

• Height of cylinder be h cm.

Given that,

• Curved Surface Area of cylinder = 264 $cm^2$

We know,

Curved Surface Area of cylinder of radius r and height h is given by

$\boxed{ \rm{ \:CSA_{(Cylinder)} = 2 \: \pi \: r \: h \: }}$

So,

$\rm \: 2 \: \pi \: r \: h \: = \: 264 \: - - - (1)$

Further, given that

• Volume of cylinder = 924 $cm^3$

We know,

Volume of cylinder of radius r and height h is given by

$\boxed{ \rm{ \:Volume_{(Cylinder)} = \pi \: {r}^{2} \: h \: }}$

So,

$\rm \: \pi \: {r}^{2} \: h \: = \: 924 - - - (2)$

On dividing equation (2) by (1), we get

$\rm \: \dfrac{ \pi \: {r}^{2} \: h \: }{2 \: \pi \: r \: h \:} = \dfrac{924}{264}$

$\rm \: \dfrac{ r }{2} = \dfrac{7}{2}$

$\rm\implies \:r \: = \: 7 \: cm$

On substituting the value of r in equation (1), we get

$\rm \: 2 \times \dfrac{22}{7} \times 7 \times h \: = \: 264$

$\rm \: 44h \: = \: 264$

$\rm\implies \:h \: = \: 6 \: cm$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$