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 )

Answer 1

Draw PQ ⊥ AB.



∴ AQ = QB (Perpendicular from the centre of the circle to the chord bisects the chord)

In right ∆APQ,

$$\begin{gathered} \mathrm{AQ}=\sqrt{{\mathrm{AP}}^2-{\mathrm{PQ}}^2} \\ \Rightarrow \mathrm{AQ}=\sqrt{8^2-4^2} \\ \Rightarrow \mathrm{AQ}=\sqrt{64-16} \\ \Rightarrow \mathrm{AQ}=\sqrt{48} \\ \Rightarrow \mathrm{AQ}=4\sqrt{3}\mathrm{cm} \end{gathered}$$
∴ AB = 2AQ = $2\times 4\sqrt{3}=8\sqrt{3}\mathrm{cm}$

Also,
$$\begin{gathered} \sin \angle \mathrm{APQ}=\frac{\mathrm{AQ}}{\mathrm{AP}} \\ \Rightarrow \sin \angle \mathrm{APQ}=\frac{4\sqrt{3}}{8} \\ \Rightarrow \sin \angle \mathrm{APQ}=\frac{\sqrt{3}}{2}=\sin 60° \\ \Rightarrow \angle \mathrm{APQ}=60° \end{gathered}$$
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
$$\begin{gathered} =\frac{\theta }{360°}\times \mathrm{\pi }{r}^2-\frac{1}{2}\times \mathrm{AB}\times \mathrm{PQ} \\ =\frac{120°}{360°}\times 3.14\times 8^2-\frac{1}{2}\times 8\sqrt{3}\times 4 \\ =66.97-27.68 \\ =39.29{\mathrm{cm}}^2 \end{gathered}$$
Thus, the area of the shaded portion is 39.29 cm².