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: Hypotenuse 2 = Base 2 + Perpendicular 2 PS 2 = PT 2 + ST 2 (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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