#### 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. ( \[\pi\] = 3.14, \[\sqrt{3}\]= 1.73 )

#### Solution

Draw PQ ⊥ AB.

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

In right ∆APQ,

\[AQ = \sqrt{{AP}^2 - {PQ}^2}\]

\[ \Rightarrow AQ = \sqrt{8^2 - 4^2}\]

\[ \Rightarrow AQ = \sqrt{64 - 16}\]

\[ \Rightarrow AQ = \sqrt{48}\]

\[ \Rightarrow AQ = 4\sqrt{3}\text{ cm} \]

∴ AB = 2AQ = \[2 \times 4\sqrt{3} = 8\sqrt{3} \text{ cm }\]

Also,

\[\sin\angle APQ = \frac{AQ}{AP}\]

\[ \Rightarrow \sin\angle APQ = \frac{4\sqrt{3}}{8}\]

\[ \Rightarrow \sin\angle APQ = \frac{\sqrt{3}}{2} = \sin60°\]

\[ \Rightarrow \angle APQ = 60°\]

\[ \Rightarrow \sin\angle APQ = \frac{4\sqrt{3}}{8}\]

\[ \Rightarrow \sin\angle APQ = \frac{\sqrt{3}}{2} = \sin60°\]

\[ \Rightarrow \angle APQ = 60°\]

Similarly,

\[\angle BPQ = 60°\]

∴ ∠APB = ∠APQ + ∠BPQ = 60º + 60º = 120º

Radius of the circle,

Measure of arc AB,

∴ Area of the shaded portion = Area of the sector ABP − Area of ∆APB

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

\[= \frac{\theta}{360°} \times \pi r^2 - \frac{1}{2} \times AB \times PQ\]

\[ = \frac{120° }{360° } \times 3 . 14 \times 8^2 - \frac{1}{2} \times 8\sqrt{3} \times 4\]

\[ = 66 . 97 - 27 . 68\]

\[ = 39 . 29 {cm}^2\]

\[ = \frac{120° }{360° } \times 3 . 14 \times 8^2 - \frac{1}{2} \times 8\sqrt{3} \times 4\]

\[ = 66 . 97 - 27 . 68\]

\[ = 39 . 29 {cm}^2\]

Thus, the area of the shaded portion is 39.29 cm^{2}.

Is there an error in this question or solution?

Solution 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. ( π = 3.14, √ 3 = 1.73 ) Concept: Surface Area of a Combination of Solids.