Question on: JAMB Mathematics - 2002
A
21/25πR2
B
9/25πR2
C
21/23πR2
D
5/9πR2
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Correct Option: D
r = \(\frac{2}{3}\)R
∴R = \(\frac{3}{3}\)R
Area of small circle = πr2
= π(\(\frac{2R}{3}\))2
Area of the big circle πr2 = π\(\frac{(3R)^2}{3}\)
Area of shaded portion = π(\(\frac{3R}{3}\))2 - π(\(\frac{2R}{3}\))2
= π[(\(\frac{3R}{3}\))2 - (\(\frac{2R}{3}\))2]
= π[(\(\frac{3R}{3}) + (\frac{2R}{3}) - (\frac{3R}{3}\)) - (\(\frac{2R}{3}\))]
= π[(\(\frac{5R}{3}\)) (\(\frac{R}{3}\))]
= π x \(\frac{5R}{3}\) x \(\frac{R}{3}\)
= 5/9πR2
∴R = \(\frac{3}{3}\)R
Area of small circle = πr2
= π(\(\frac{2R}{3}\))2
Area of the big circle πr2 = π\(\frac{(3R)^2}{3}\)
Area of shaded portion = π(\(\frac{3R}{3}\))2 - π(\(\frac{2R}{3}\))2
= π[(\(\frac{3R}{3}\))2 - (\(\frac{2R}{3}\))2]
= π[(\(\frac{3R}{3}) + (\frac{2R}{3}) - (\frac{3R}{3}\)) - (\(\frac{2R}{3}\))]
= π[(\(\frac{5R}{3}\)) (\(\frac{R}{3}\))]
= π x \(\frac{5R}{3}\) x \(\frac{R}{3}\)
= 5/9πR2
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