Question

A solid cylinder has total surface area of 462 square cm its curved surface area is one third of its total surface area find the volume of the cylinder.

please Solve​

Answer 1

ANsWeR :

$\sf\: 539{cm}^{3}$

EXpLaNaTiOn :

Let r be the radius of the base and h be the height of the cylinder. Then,

$\large \boxed{ \purple{ \sf Total \: surface \: area = 2\pi \: r(r + h)}}$

$\large \boxed{ \purple{ \sf Curved \: surface \: area =2\pi rh}}$

We Have,

$\star \sf \tt \: C.S.A = \dfrac{1}{3} (T.S.A)$

$\rightarrow \sf \: 2\pi \: rh = \dfrac{1}{3}{2\pi \: r (r + h)}$

$\rightarrow \sf 6\pi \: r \: h = 2\pi \: rh + 2\pi {r}^{2}$

$\rightarrow \sf 4\pi \: r \: h = 2\pi {r}^{2}$

$\rightarrow \sf 2h = r$

$\because \tt\: Total \: Surface \: Area = 462 {cm}^{3}$

$\rightarrow \sf \: 2\pi \: r(r + h) = 462$

$\rightarrow \sf \: 2\pi \: r( \dfrac{r}{2} + r) = 462$

$\rightarrow \sf \: 2\pi \: r \times \dfrac{3r}{2} = 462$

$\rightarrow \sf \: 2 \times \dfrac{22}{7} \times \dfrac{3}{2} \times {r}^{2} = 462$

$\rightarrow \sf \: {r}^{2} = 49$

$\rightarrow \sf \: r = \sqrt{49}$

$\rightarrow \sf \: r = \large \boxed{ \red{ \sf \: 7 \: cm}}$

Now,

$\sf \: 2h = r$

$\sf\: h = \dfrac{r}{2}$

$\sf \: h = \large \boxed{ \red{ \sf \dfrac{7}{2} \: cm}}$

Volume of the Cylinder :

$\large \boxed{ \purple{ \sf \: Volume \: = \pi {r}^{2}h }}$

$\rightarrow \sf \: \dfrac{22}{7} \times 7 \times 7 \times \dfrac{7}{2}$

$\rightarrow \sf \: Vol= \large \boxed{ \red{ \sf \: 539 {cm}^{3} }}$

Answer 2

Answer:

${\underline{\boxed{\sf\green{Volume \: of \: cylinder = 539 \: {cm}^{3} }}}}$

Step-by-step explanation:

Given:-

• Total surface area = 462 sq.cm
• Curved surface area= 1/3 of TSA

To find:-

• Volume of cylinder = ?

We know that,

$\star{\boxed{\sf\green{CSA = 2 \pi r h }}}$

A/Q,

$\Rightarrow\sf 2\pi r h = \dfrac{1}{3} [2 \pi r(r + h)] \\\\\Rightarrow\sf 6 \pi r h = 2 \pi r (r + h) \\\\\Rightarrow\sf 6 \pi r h = 2 \pi {r}^{2} + 2 \pi r h \\\\\Rightarrow\sf 6\pi r h - 2 \pi r h = 2 \pi {r}^{2} \\\\\Rightarrow\sf 4\pi r h = 2 \pi {r}^{2} \\\\\Rightarrow\sf 2h = r \\\\\star\sf h = \dfrac{r}{2} ............(i)$

A/Q,

$\star{\boxed{\sf\pink{TSA = 2 \pi r (r + h) }}}$

$\Rightarrow\sf 462 = 2 \pi r (r + h) \\\\\Rightarrow\sf 462 = 2 \times \dfrac{22}{7} \times r (r + \dfrac{r}{2}) \\\\\Rightarrow\sf 462 =\dfrac{44r}{7} (\dfrac{3r}{2}) \\\\\Rightarrow\sf 462 =\dfrac{132{r}^{2}}{14} \\\\\Rightarrow\sf 132{r}^{2} = 6468 \\\\\Rightarrow\sf {r}^{2} = \dfrac{6468}{132} \\\\\Rightarrow\sf {r}^{2} = 49 \\\\\Rightarrow\sf r =\sqrt{49} \\\\\star{\boxed{\bold\green{r = 7 \: cm}}}$

$\rule{100}{2}$

$\sf Now, h = \dfrac{r}{2} \\\\\Rightarrow\sf h =\dfrac{7}{2} \\\\\star{\boxed{\sf\green{h = 3.5 \: cm}}}$

Now,we know that,

$\star{\boxed{\bold\green{Volume \: of \: cylinder = \pi {r}^{2} h }}}$

$\Rightarrow\sf Volume = \dfrac{22}{7} \times {(7)}^{2} \times 3.5 \\\\\Rightarrow{\boxed{\sf\pink{Volume = 539 \: {cm}^{3} }}}$

$\therefore\sf\green{Volume \: of \: cylinder}$ = ${\boxed{\sf\pink{539 \: {cm}^{3} }}}$