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Area of Triangle

In the figure...

Question

In the figure, $P Q S O$ is a trapezium in which $P Q$ is parallel to $O S, ∠ P O S = 120^{\circ}$ and $∠ P Q S = 90^{\circ} .$ Points $P, Q$ and $R$ lie on a circle with center $O$ and radius $10.$ Then the area of the shaded region, is

\frac{100 π}{3} - \frac{25 \sqrt{3}}{2}

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\frac{100 π}{3} - \frac{50 \sqrt{3}}{2}

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\frac{100 π}{3} - \frac{75 \sqrt{3}}{2}

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\frac{100 π}{3} - \frac{100 \sqrt{3}}{2}

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Solution

The correct option is C $\frac{100 π}{3} - \frac{75 \sqrt{3}}{2}$

$Area (□ P Q S O) = Area (Δ P Q O) + Area (Δ Q O S) = \frac{\sqrt{3}}{4} \times r^{2} + \frac{1}{2} \times \frac{\sqrt{3} r}{2} \times \frac{r}{2} = \frac{\sqrt{3}}{4} \times (10)^{2} + \frac{1}{2} \times \frac{\sqrt{3} \times 10}{2} \times \frac{10}{2} = \frac{75 \sqrt{3}}{2}$
$Required Area = \frac{1}{2} \times r^{2} \times \frac{2 π}{3} - Area (□ P Q S O) = \frac{100 π}{3} - \frac{75 \sqrt{3}}{2}$

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In the figure, PQSO is a trapezium in which PQ is parallel to OS, ∠POS=120∘ and ∠PQS=90∘. Points P,Q and R lie on a circle with center O and radius 10. Then the area of the shaded region, is

The correct option is C 100π3−75√32 Area (□PQSO)=Area (ΔPQO)+Area (ΔQOS)=√34×r2+12×√3r2×r2=√34×(10)2+12×√3×102×102=75√32 Required Area=12×r2×2π3−Area (□PQSO)=100π3−75√32

In the figure, $P Q S O$ is a trapezium in which $P Q$ is parallel to $O S, ∠ P O S = 120^{\circ}$ and $∠ P Q S = 90^{\circ} .$ Points $P, Q$ and $R$ lie on a circle with center $O$ and radius $10.$ Then the area of the shaded region, is