Question

the circumference of a base of cylinder is 44 cm and height is 15 cm what is the volume of cylinder

Answer 1

Given:-

• Circumference of the base of cylinder is 44 cm.
• Height of the cylinder is 15 cm.

To find:-

• Find the Volume of cylinder ?

Solutions:-

• Let the radius of the cylinder be 'r' cm.

We know that,

The circumference of base of cylinder = 2πr

According to question:-

=> 44 = 2πr

=> 44 = 2 × 22/7 × r

=> 44 = 44/7 × r

=> 44 × 7/44 = r

=> 7 = r

=> r = 7cm

• The radius of the cylinder is 7cm.

Now,

Volume of the cylinder = πr²h

According to the questions:-

=> πr²h

=> 22/7 × (7)² × 15

=> 22/7 × 49 × 15

=> 22 × 7 × 15

=> 2310cm³

Hence, the volume of cylinder is 2310 cm³.

Some Important:-

Volume of cylinder ( Area of base × height ).

= (πr²) × h

= πr²h

Curved surface = ( Perimeter of base ) × height.

= (2πr) × h

= 2πrh

Total surface are = Area of circular ends + curved surface area.

= 2πr² + 2πrh

= 2πr(r + h)

Where,

r = radius of the circular base of the cylinder.

h = height of cylinder.

Answer 2

Answer:

$\large \underline{\sf \pmb{Given}}$

• The circumference of a base of cylinder is 44
• The height of cylinder is 15 cm.

  ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\large \underline{\sf \pmb{To \: Find }}$

• what is the volume of cylinder

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

Firstly Finding Radius of Cylinder .

As we know that.

$\underline{\boxed{ \sf{The \: circumference \: of \: base \: of \: cylinder = 2{\pi}r}}}$

• Substituting the values

$: \implies \sf 44 = 2{ \pi}r$

$: \implies\sf {44 = 2 × \dfrac{22}{7} × r}$

$: \implies\sf{ 44 = \dfrac{44}{7} × r}$

$: \implies \sf \cancel{44} × \dfrac{7} {\cancel{44}} = r$

$: \implies\sf{7 = r}$

$\: \: \underline{\boxed {\sf{radius ={\purple{ 7cm}}}}}$

• The radius of cylinder is 7 cm

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Now Finding the volume of Cylinder.

As we know that

$\underline {\boxed{ \sf{{Volume_{(cylinder)}} = {\pi} {r}^{2}h}}}$

• Substituting the values

${ : \implies \sf{{Volume_{(cylinder)}} = {\ \dfrac{22}{7} } \times {7}^{2} \times 15}}$

${ : \implies \sf{{Volume_{(cylinder)}} = {\ \dfrac{22}{ \cancel{7} }} \times \cancel {7} \times {7} \times 15}}$

${ : \implies \sf{{Volume_{(cylinder)}} = {{22}} \times {7} \times 15}}$

${ : \implies \sf{{Volume_{(cylinder)}} = 2310 {cm}^{3} }}$

$\underline{ \boxed{ \sf{Volume \: of \: Cylinder = \purple{2130\: {cm}^{3} }}}}$

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$\large \underline{\sf \pmb{Therefore}}$

• The volume of cylinder is 2130 cm³

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$\large \underline{\sf \pmb{Additional \: Information }}$

$\begin{gathered}\begin{gathered}\bigstar \: \bf\underline{More \: Useful \: Formulae } \: \bigstar  \\ \begin{gathered}{\boxed{\begin{array} {cccc}{\sf{{\leadsto TSA \: of \: cube \: = \: 6(side)^{2}}}} \\  \\{\sf{{\leadsto LSA \: of \: cube \:= \: 4(side)^{2}}}}  \\  \\{\sf{{\leadsto Volume \: of \: cube \: = \: (side)^{3}}}} \\  \\ {\sf{{\leadsto Diagonal \: of \: cube \: = \: \sqrt(l^{2} + b^{2} + h^{2}}}} \\  \\ {\sf{{\leadsto Perimeter \: of \: cube \: = \: 4(l+b+h)}}} \\  \\ {\sf{{\leadsto CSA \: of \: sphere \: = \: 2 \pi r^{2}}}} \\  \\ {\sf{{\leadsto SA \: of \: sphere \: = \: 4 \pi r^{2}}}} \\  \\{\sf{{\leadsto TSA \: of \: sphere \: = \: 3 \pi r^{2}}}} \\  \\ {\sf{{\leadsto Diameter \: of \: circle \: = \: 2r}}} \\  \\ {\sf{{\leadsto Radius \: of \: circle \: = \: \dfrac{d}{2}}}} \\  \\ {\sf{{\leadsto Volume \: of \: sphere \: = \: \dfrac{4}{3} \pi r^{3}}}} \\  \\ {\sf{{\leadsto Area \: of \: rectangle \: = \: Length \times Breadth}}} \\  \\ {\sf{{\leadsto Perimeter \: of \: rectangle \: = \:2(length+breadth)}}} \\  \\{\sf{{\leadsto Perimeter \: of \: square \: = \: 4 \times sides}}}\end{array}}}\end{gathered}\end{gathered}\end{gathered}$