Question : In an equilateral $\triangle$ PQR, S is the point on the side QR such that QR = 3QS. If PQ = 9 cm, then what will be the length (in cm) of PS?
Option 1: $\sqrt{60} $
Option 2: $\sqrt{63}$
Option 3: $\sqrt{62}$
Option 4: $\sqrt{61}$
Correct Answer: $\sqrt{63}$
Solution : Draw PT perpendicular to QR. PT = $\frac{\sqrt3}{2}\times 9=\frac{9\sqrt3}{2}$ cm QT = $\frac{9}{2}$ cm Given, QR = 3QS ⇒ QS = $\frac{9}{3}$ = 3 cm ST = QT – QS =$\frac{9}{2} -3$ = $\frac{3}{2}$ Using Pythagoras' theorem: Hypotenuse2 = Base2 + Perpendicular2 PS2 = PT2 + ST2 (PS)2 = $(\frac{9\sqrt3}{2})^2 + (\frac{3}{2})^2$ ⇒ $(PS)^2=\frac{243}{4} + \frac{9}{4}$ ⇒ $(PS)^2=\frac{252}{4}$ ⇒ $(PS)^2=63$ ⇒ $PS=\sqrt{63}$ Hence, the correct answer is $\sqrt{63}$ cm.
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Question : In $\triangle$PQR, $\angle$ PQR = $90^{\circ}$, PQ = 5 cm and QR = 12 cm. What is the radius (in cm) of the circumcircle of $\triangle$PQR?
Option 1: 6.5
Option 2: 7.5
Option 3: 13
Option 4: 15
Question : The area of an equilateral triangle is $4 \sqrt{3} \mathrm{~cm}^2$. Find the side (in cm) of the triangle.
Option 1: $2$
Option 2: $4$
Option 3: $\sqrt{3}$
Option 4: $2 \sqrt{3}$
Question : If each side of an equilateral triangle is 12 cm, then its altitude is equal to:
Option 1: $6 \sqrt{3}\ \text{cm}$
Option 2: $3 \sqrt{6}\ \text{cm}$
Option 3: $6 \sqrt{2}\ \text{cm}$
Option 4: $3 \sqrt{2}\ \text{cm}$
Question : $\triangle$ PQR circumscribes a circle with centre O and radius r cm such that $\angle$ PQR = $90^{\circ}$. If PQ = 3 cm, QR = 4 cm, then the value of r is:
Option 1: 2
Option 2: 1.5
Option 3: 2.5
Option 4: 1
Question : The length of each side of an equilateral triangle is $14 \sqrt{3}$ cm. The area of the incircle (in cm2), is:
Option 1: 450
Option 2: 308
Option 3: 154
Option 4: 77
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