Question : In the given figure, $PQRS$ is a quadrilateral. If $QR = 18$ cm and $PS = 9$ cm, then, what is the area (in cm2) of quadrilateral $PQRS$?
Option 1: $\frac{(64\sqrt{3})}{3}$
Option 2: $\frac{(177\sqrt{3})}{2}$
Option 3: $\frac{(135\sqrt{3})}{2}$
Option 4: $\frac{(98\sqrt{3})}{3}$
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Correct Answer: $\frac{(135\sqrt{3})}{2}$
Solution :
Given: $PQRS$ is a quadrilateral, $QR = 18$ cm and $PS = 9$ cm.
Construction: Extend the lines $QP$ and $RS$ as they meet at $T$ and form $\triangle QRT$ an equilateral triangle.
Take the angles as shown in the figure.
The area of an equilateral $\triangle QRT$ = $\frac{\sqrt{3}}{4}×(18)^2=81\sqrt{3}$ cm
2
And in the $\triangle TSP$ is a right-angled triangle.
$\tan 30^{\circ}=\frac{TS}{PS}=\frac{TS}{9}$
⇒ $\frac{1}{\sqrt3}=\frac{TS}{9}$
⇒ $TS=\frac{9}{\sqrt3}$
The area of right-angled $\triangle TSP$ = $\frac{1}{2}×TS×PS$
The area of right-angled triangle = $\frac{1}{2}×\frac{9}{\sqrt3}×9=\frac{27\sqrt3}{2}$ cm
2
The required area of quadrilateral $PQRS$ = The area of an equilateral $\triangle QRT$ – The area of right-angled $\triangle TSP$
The required area of quadrilateral $PQRS$ = $81\sqrt{3} $ – $\frac{27\sqrt3}{2}$ = $\frac{162\sqrt{3}–27\sqrt3}{2}$ = $\frac{135\sqrt{3}}{2}$ cm
2
Hence, the correct answer is $\frac{135\sqrt{3}}{2}$.
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