rishabhmishra wrote:

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 [ 17.62 KiB | Viewed 325 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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