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