Question

The volume and curved surface area of a cylinder are 1650 cm and 660 em
respectively. Find the radius and height of cylinder.
CS
Find the
tio of
--​

Answer 1

QUESTION :-

The volume and curved surdace area of cylinder are 1650 cm³ and 660 cm² respectively.Find the radius and height of the cylinder.

GIVEN :-

• Volume of cylinder = 1650 cm³
• CSA of cylinder = 660 cm

TO FIND :-

• Radius and height of the cylinder

SOLUTION :-

$\large {\underline {\boxed {\bigstar {\red {\sf{CSA\: of \: cylinder = 2\pi rh}}}}}}$

We have ,

• CSA = 660 cm²

$\implies \sf \: 2\pi rh = 660 \\ \\ \implies \sf \: 2 \times \frac{22}{7} \times rh = 660 \\ \\ \implies \sf \: rh = \frac{660 \times 7}{2 \times 22} \\ \\ \implies \sf \: rh = 105 \\ \\ \implies {\underline {\boxed {\pink{\sf {\: h = \frac{105}{r} }}}}} \sf \: \longrightarrow \: eq(1)$

$\large {\underline {\boxed {\bigstar {\red {\sf{ \: volume \: of \: cylinder = \pi {r}^{2} h}}}}}}$

We are given ,

• Volume of cylinder = 1650 cm³

$\implies \sf \: \pi {r}^{2} h = 1650 \\ \\ \implies \sf \: \frac{22}{7} \times {r}^{2} \times h = 1650 \\ \\ \implies \sf \: {r}^{2} h = 1650 \times \frac{7}{22} \\ \\ \implies \sf \: {r}^{2} h = 525$

$\implies \sf \: {r}^{2} \bigg( \frac{105}{r} \bigg) = 525 \\ \\ \implies \sf \cancel{ {r}^{2}} \bigg( \frac{105}{ \cancel{r}} \bigg) = 525 \\ \\ \implies \sf \:105r = 525 \\ \\ \implies \sf \: r = \frac{525}{105} \\ \\ \implies {\underline {\boxed {\blue {\sf \: r = 5 \: cm}}}}$

From eq(1) ,

$\implies \sf \: h = \frac{105}{r} \\ \\ \implies \sf \: h = \frac{105}{5} \\ \\ \implies {\underline {\boxed {\blue{\sf {h = 21 \: cm}}}}}$

∴ The radius and height of the cylinder are 5 cm and 21 cm

Answer 2

QUESTION :-

The volume and curved surdace area of cylinder are 1650 cm³ and 660 cm² respectively.Find the radius and height of the cylinder.

GIVEN :-

Volume of cylinder = 1650 cm³

CSA of cylinder = 660 cm

TO FIND :-

Radius and height of the cylinder

SOLUTION :-

$\large {\underline {\boxed {\bigstar {\red {\sf{CSA\: of \: cylinder = 2\pi rh}}}}}}$

We have ,

CSA = 660 cm²

$\implies \sf \: 2\pi rh = 660 \\ \\ \implies \sf \: 2 \times \frac{22}{7} \times rh = 660 \\ \\ \implies \sf \: rh = \frac{660 \times 7}{2 \times 22} \\ \\ \implies \sf \: rh = 105 \\ \\ \implies {\underline {\boxed {\pink{\sf {\: h = \frac{105}{r} }}}}} \sf \: \longrightarrow \: eq(1)$

$\large {\underline {\boxed {\bigstar {\red {\sf{ \: volume \: of \: cylinder = \pi {r}^{2} h}}}}}}$

We are given ,

Volume of cylinder = 1650 cm³

$\implies \sf \: \pi {r}^{2} h = 1650 \\ \\ \implies \sf \: \frac{22}{7} \times {r}^{2} \times h = 1650 \\ \\ \implies \sf \: {r}^{2} h = 1650 \times \frac{7}{22} \\ \\ \implies \sf \: {r}^{2} h = 525$

$\implies \sf \: {r}^{2} \bigg( \frac{105}{r} \bigg) = 525 \\ \\ \implies \sf \cancel{ {r}^{2}} \bigg( \frac{105}{ \cancel{r}} \bigg) = 525 \\ \\ \implies \sf \:105r = 525 \\ \\ \implies \sf \: r = \frac{525}{105} \\ \\ \implies {\underline {\boxed {\blue {\sf \: r = 5 \: cm}}}}$

From eq(1) ,

$\implies \sf \: h = \frac{105}{r} \\ \\ \implies \sf \: h = \frac{105}{5} \\ \\ \implies {\underline {\boxed {\blue{\sf {h = 21 \: cm}}}}}$

∴ The radius and height of the cylinder are 5 cm and 21 cm