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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