question_answer

If PQR is an equilateral triangle and \[PX\bot QR,\] then \[P{{X}^{2}}\] is equal to:                      (CBSE 2013)

A) \[{{(RX)}^{2}}\]

B) \[2{{(RX)}^{2}}\]

C) \[3{{(RX)}^{2}}\]

D) \[2RX\]

Correct Answer: C

Solution :

[c] Altitude of an equilateral triangle PQR,

i.e.,  \[PX=(Side)\frac{\sqrt{3}}{2}=QR\frac{\sqrt{3}}{2}=2(RX)\frac{\sqrt{3}}{2}\,\,\,\,\,\,\,(QX=RX)\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,PX=\sqrt{3}\,\,(RX)\]

Hence, \[P{{X}^{2}}=3{{(RX)}^{2}}\]