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 $\triangle QRT$ = $\frac{\sqrt{3}}{4}×(18)^2=81\sqrt{3}$ cm2 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 = $\frac{1}{2}×\frac{9}{\sqrt3}×9=\frac{27\sqrt3}{2}$ cm2 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} $ – $\frac{27\sqrt3}{2}$ = $\frac{162\sqrt{3}–27\sqrt3}{2}$ = $\frac{135\sqrt{3}}{2}$ cm2 Hence, the correct answer is $\frac{135\sqrt{3}}{2}$.
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Question : In the given figure, PQR is a triangle, and quadrilateral ABCD is inscribed in it. QD = 2 cm, QC = 5 cm, CR = 3 cm, BR = 4 cm, PB = 6 cm, PA = 5 cm and AD = 3 cm. What is the area (in cm2) of the quadrilateral ABCD?
Option 1: $\frac{(23\sqrt{21})}{4}$
Option 2: $\frac{(15\sqrt{21})}{4}$
Option 3: $\frac{(17\sqrt{21})}{5}$
Option 4: $\frac{(23\sqrt{21})}{5}$
Question : In the given figure, PQR is a quadrant whose radius is 7 cm. A circle is inscribed in the quadrant as shown in the figure. What is the area (in cm2) of the circle?
Option 1: $385-221\sqrt{2}$
Option 2: $308-154\sqrt{2}$
Option 3: $154-77\sqrt{2}$
Option 4: $462-308\sqrt{2}$
Question : In the given figure, the radius of a circle is $14\sqrt{2}$ cm. PQRS is a square. EFGH, ABCD, WXYZ, and LMNO are four identical squares. What is the total area (in cm2) of all the small squares?
Option 1: 31.36
Option 2: 125.44
Option 3: 62.72
Option 4: 156.8
Question : The height of an equilateral triangle is $7 \sqrt{3}$ cm. What is the area of this equilateral triangle?
Option 1: $36 \sqrt{3}$ cm2
Option 2: $25 \sqrt{3}$ cm2
Option 3: $49 \sqrt{3}$ cm2
Option 4: $32 \sqrt{3}$ cm2
Question : The perimeter of an equilateral triangle is 48 cm. Find its area (in cm2).
Option 1: $81\sqrt{3}$
Option 2: $8\sqrt{3}$
Option 3: $25\sqrt{3}$
Option 4: $64\sqrt{3}$
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