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Question

# In the given figure, seg AB is a chord of a circle with centre P. If PA = 8 cm and distance of chord AB from the centre P is 4 cm, find the area of the shaded portion. ( $\mathrm{\pi }$ = 3.14, $\sqrt{3}$= 1.73 )

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Solution

## Draw PQ ⊥ AB. ∴ AQ = QB (Perpendicular from the centre of the circle to the chord bisects the chord) In right ∆APQ, $\mathrm{AQ}=\sqrt{{\mathrm{AP}}^{2}-{\mathrm{PQ}}^{2}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{AQ}=\sqrt{{8}^{2}-{4}^{2}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{AQ}=\sqrt{64-16}\phantom{\rule{0ex}{0ex}}⇒\mathrm{AQ}=\sqrt{48}\phantom{\rule{0ex}{0ex}}⇒\mathrm{AQ}=4\sqrt{3}\mathrm{cm}$ ∴ AB = 2AQ = $2×4\sqrt{3}=8\sqrt{3}\mathrm{cm}$ Also, $\mathrm{sin}\angle \mathrm{APQ}=\frac{\mathrm{AQ}}{\mathrm{AP}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\angle \mathrm{APQ}=\frac{4\sqrt{3}}{8}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\angle \mathrm{APQ}=\frac{\sqrt{3}}{2}=\mathrm{sin}60°\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{APQ}=60°$ Similarly, $\angle \mathrm{BPQ}=60°$ ∴ ∠APB = ∠APQ + ∠BPQ = 60º + 60º = 120º Radius of the circle, r = 8 cm Measure of arc AB, θ = 120º ∴ Area of the shaded portion = Area of the sector ABP − Area of ∆APB $=\frac{\theta }{360°}×\mathrm{\pi }{r}^{2}-\frac{1}{2}×\mathrm{AB}×\mathrm{PQ}\phantom{\rule{0ex}{0ex}}=\frac{120°}{360°}×3.14×{8}^{2}-\frac{1}{2}×8\sqrt{3}×4\phantom{\rule{0ex}{0ex}}=66.97-27.68\phantom{\rule{0ex}{0ex}}=39.29{\mathrm{cm}}^{2}$ Thus, the area of the shaded portion is 39.29 cm2.

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