Question

★ELLOH ㋡



$\huge \mathfrak \orange{❥QUESTION}$



★ Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere?


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Answer 1

Answer:

9 units

Hence, diameter of the hemisphere is equal to 9 units.

Step-by-step explanation:

. here u go.

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Answer 2

${ \underline{ \underline{\bf \pink❊QueStiOn \purple❊} }}$

➡️Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere?

${ \underline{ \underline{\bf \pink❊GiVeN\purple❊} }}$

$\small \sf \: Volume \: of \: a \: hemisphere⇝Surface \: area \: of \: a \: hemisphere$

(⚝ That means the Volume of a hemisphere is equal to the surface area of a hemisphere . ⚝)

${ \underline{ \underline{\bf \pink❊To \: FiNd\purple❊} }}$

• $\small \sf \: Diameter \: of \: the \: hemisphere$

${ \underline{ \underline{\bf \pink❊ForMulA\purple❊} }}$

• $\small \sf \: Volume \: of \: a \: hemisphere⇝ \frac{2}{3} \pi {r}^{3}$
• $\small \sf \: Surface \: area \: of \: a \: hemisphere ⇝3\pi {r}^{2}$

${ \underline{ \underline{\bf \pink❊SoLuTion \purple❊} }}$

$\small \bf According \: to \: the \: Question,$

$\small \sf \implies \frac{2}{3} \pi {r}^{3} = 3\pi {r}^{2}$

$\small \sf \implies \frac{2}{3} = 3{r}^{}$

$\small \sf \implies 3r = \frac{2}{3}$

$\small \sf \implies r= \frac{9}{2}$

$\small \sf \: We \: know,$

$\small \sf \boxed{ \bf Diameter=2 \times radius}$

$\small \sf \implies \: Diameter= \cancel2 \times \frac{9}{ \cancel2}$

$\small \sf \implies \: Diameter=9$

$\underline{ \sf \: More \: Information:-}$

• $\small \sf \: Volume \: of \: a \: Cuboid \large\leadsto \small \boxed{ \bf l \times b \times h}$

• $\small \sf \: Volume \: of \: a \: Cube \large\leadsto \small \boxed{ \bf {a}^{3} }$

• $\small \sf \: Volume \: of \: a \: Cylinder \large\leadsto \small \boxed{ \bf \pi {r}^{2}h }$

• $\small \sf \: Volume \: of \: a \:Cone\large\leadsto \small \boxed{ \bf \frac{1}{3}\pi r {}^{2} h }$

• $\small \sf \: Volume \: of \: a \: Sphere \large\leadsto \small \boxed{ \bf \frac{4}{3} \pi r {}^{3} }$

• $\small \sf \: Volume \: of \: a \: Hemisphere \large\leadsto \small \boxed{ \bf \frac{2}{3}\pi {r}^{3} }$