Question

A cubical block of side 7 cm is surmounted by a hemisphere. Find the surface area of the solid(in $c{m}^2$ ).

• A. 320 $c{m}^2$
• B. 312.5 $c{m}^2$
• C. 332.5 $c{m}^2$
• D. 346.8 $c{m}^2$

Answer 1

The correct option is C

332.5 $c{m}^2$

The hemisphere surmounts the cube, the maximum diameter the hemisphere can have = side of cube as shown in the figure below. The hemisphere will just touch the sides of top face of the cube.

Therefore, maximum diameter Hemisphere can have = 7 cm

$\Rightarrow \text{Radius of Hemisphere = r =}\frac{7}{2}=3.5cm$

Surface area of solid = Surface area of 5 Faces of cube + Surface area of Hemisphere + Surface area left uncovered on the top face of the cube (shown in blue)

$=5l^2+2.\pi .r^2+\text{Area of Square ABCD - Area of inscribed circle in square ABCD}$

$=(5\times 7\times 7)+(2\times \frac{22}{7}\times 3.5\times 3.5)+(49-(\frac{22}{7}\times 3.5\times 3.5\left)\right)$

$=245+77+(49-38.5)=322+10.5=332.5c{m}^2$

$\therefore$ The surface area of combined solid = $332.5c{m}^2$