Question

The curved surface area of a right circular cylinder is 44000m2 if the circumference of its base is 110 cm then it's height is

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Answer 1

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

Let assume that

• Radius of cylinder be r m

• Height of cylinder be h m

Now, given that

• Circumference of base =110 cm = 1.1 m

$\rm\implies \:2\pi \: r \: = \: 1.1 - - - (1)$

Further given that, Curved Surface Area of cylinder is 44000 m^2.

$\rm \: CSA_{(cylinder)} = 44000 \: {m}^{2}$

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

$\color{green}\boxed{ \rm{ \:CSA_{(cylinder)} \: = \: 2 \: \pi \: r \: h \: }}$

So, using this result, we get

$\rm \: 2 \: \pi \: r \: h \: = \: 44000$

$\rm \: 1.1 \times h \: = \: 44000$

$\rm\implies \: \: h \: = \: 40000 \: m$

So, height of cylinder, h = 40000 m

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

Additional Information :-

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

Answer 2

Answer:

• Refer the attachment:)

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