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Question

A cubical block of side $$7cm$$ is surmounted by a hemisphere of the largest size. Find the surface area of the resultant solid.


Solution

Consider the problem
The hemisphere can occupy whole of the side of cube.
Therefore, Greatest diameter of hemisphere $$=side\;of\;cube$$ $$=7cm$$ 
Here, base of hemisphere falls on cube, so that area should not from part of solid.
$$Surface\,area\,of\,solid = Area\,of\,cube + curved\,surface\,area\,of\,hemisphere - base\,area\,of\,hemisphere$$
Area of cube $$ = 6{a^2}$$
Here, $$a=side=7cm$$
So,
$$\begin{array}{l} =6{ \left( 7 \right) ^{ 2 } } \\ =6\times \left( { 7\times 7 } \right)  \\ =6\times 49 \\ =294c{ m^{ 2 } } \end{array}$$
And,
Curved surface area of hemisphere $$ = 2\pi {r^2}$$
Diameter of hemisphere$$=7cm$$
Therefore, radius is 
$$\begin{array}{l} r=\dfrac { { diameter } }{ 2 }  \\ =\dfrac { 7 }{ 2 } cm \end{array}$$
So,
$$\begin{array}{l} =2\times \dfrac { { 22 } }{ 7 } \times { \left( { \dfrac { 7 }{ 2 }  } \right) ^{ 2 } } \\ =2\times \dfrac { { 22 } }{ 7 } \times \dfrac { 7 }{ 2 } \times \dfrac { 7 }{ 2 }  \\ =77c{ m^{ 2 } } \end{array}$$
And,
Base area of hemisphere $$ = \pi {r^2}$$
$$\begin{array}{l} =\dfrac { { 22 } }{ 7 } \times { \left( { \dfrac { 7 }{ 2 }  } \right) ^{ 2 } } \\ =\dfrac { { 22 } }{ 7 } \times \dfrac { 7 }{ 2 } \times \dfrac { 7 }{ 2 }  \\ =\dfrac { { 77 } }{ 2 } c{ m^{ 2 } } \end{array}$$
Now,
$$Surface\,area\,of\,solid = Area\,of\,cube + curved\,surface\,area\,of\,hemisphere - base\,area\,of\,hemisphere$$
$$\begin{array}{l} =294+77-\dfrac { { 77 } }{ 2 }  \\ =294+77-38.5 \\ =371-38.5 \\ =332.5c{ m^{ 2 } } \end{array}$$
Hence, surface area of solid $$ = 332.5c{m^2}$$



Mathematics

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