Question

A solid sphere of radius $r$ cm is bisected along two perpendicular planes. The total surface area (in$c{m}^2$) of the $4$ pieces formed, is?

• A. $4\pi r^2$
• B. $6\pi r^2$
• C. $8\pi r^2$
• D. $12\pi r^2$

Answer 1

The correct option is C $8\pi r^2$

A solid sphere when bisected along two perpendicular planes, each piece formed has three surfaces.

Total Surface Area of each piece =Curved Surface area of the piece +2 Semi circles as a plane surface

=$\frac{1}{4}th$ of $4\pi r^2+2\times \frac{\pi r^2}{2}$

$=2\pi r^2$

$\therefore$ Total Surface Area of 4 pieces $=4\times 2\pi r^2$

$=8\pi r^2$