Question

If the volume of a hemisphere is four times its total surface area, what is its radius?​

Answer 1

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

$\: \: \: \: \: \: \: \: \: \bull \: \sf \: Volume_{(hemisphere)} = 4 \times TSA_{(hemisphere)}$

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

$\: \: \: \: \: \: \: \: \: \bull \: \sf \:radius_{(hemisphere)}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\underline{ \boxed{ \bf \: Volume_{(hemisphere)} = \dfrac{2}{3} \pi \: {r}^{3} }}$

$\underline{ \boxed{ \bf \: TSA_{(hemisphere)} = 3\pi \: {r}^{2}}}$

where

$\: \: \: \: \: \: \: \: \: \bull \: \sf \:r \: = \: radius_{(hemisphere)}$

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

$\begin{gathered}\begin{gathered}\bf \: Let- \begin{cases} &\sf{radius \: of \: hemisphere \: = \:r \: units }\end{cases}\end{gathered}\end{gathered}$

According to statement,

$\rm :\longmapsto\:Volume_{(hemisphere)} = 4 \times TSA_{(hemisphere)}$

$\rm :\longmapsto\:\dfrac{2}{3} \pi \: {r}^{3} = 4 \times 3\pi \: {r}^{2}$

$\bf\implies \:r \: = \: 18 \: units$

$\overbrace{ \underline { \boxed { \bf \therefore \: The \: radius \: of \: hemisphere \: = \:18 \: units }}}$

Additional Information :-

Perimeter of rectangle = 2(length× breadth)

Diagonal of rectangle = √(length² + breadth²)

Area of square = side²

Perimeter of square = 4× side

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²