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rishabhmishra
Q. The surface area of sphere is \(4\pi r^2\). If a sphere is halved into two hemispheres, what is the percentage change in the total surface area of the solid(s) before and after?
A. 0%
B. -25%
C. -33.33%
D. 25%
E. 50%
Attachment:
surfarea.png
surfarea.png [ 17.62 KiB | Viewed 4528 times ]
• Surface area of sphere = \(4\pi r^2\)

• Surface area TOTAL of ONE hemisphere =
(sphere S.A.*\(\frac{1}{2}\)) PLUS (S.A. of circular face created by the cut)
(\(4\pi r^2\)*\(\frac{1}{2}\)) = \(2\pi r^2\) PLUS extra circle face area= \(\pi r^2\)
\((2\pi r^2 + \pi r^2)\)\(= 3\pi r^2\)

S.A. of TWO hemispheres: \((2 * 3\pi r^2)\) = \(6\pi r^2\)
S.A. of original sphere: \(4\pi r^2\)

• Percent Change? \(\frac{New-Old}{Old} * 100\)

\(\frac{6\pi r^2 - 4\pi r^2}{4\pi r^2} * 100\)


\((\frac{2\pi r^2}{4\pi r^2}*100)=(\frac{2}{4}*100)=(.5*100)=50\) %

Answer E
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