In the diagram above are two concentric circles... - JAMB Mathematics 2002 Question

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