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 problemThe 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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