Question

In the given figure, ∆PQR is an equilateral triangle. Point S is on seg QR such that QS = n\[\frac{1}{3}\] QR.

Prove that: 9 PS² = 7 PQ²

Answer 1

Let the side of equilateral triangle ∆PQR be x.

PT be the altitude of the ∆PQR.

We know that, in equilateral triangle, altitude divides the base in two equal parts.

∴ QT = TR = \[\frac{1}{2}QR = \frac{x}{2}\]

Given: QS = \[\frac{1}{3}\] QR = \[\frac{x}{3}\]

\[\therefore ST = QT - QS = \frac{x}{2} - \frac{x}{3} = \frac{x}{6}\]

According to Pythagoras theorem,

In ∆PQT

\[{PQ}^2 = {QT}^2 + {PT}^2 \]

\[ \Rightarrow \left( x \right)^2 = \left( \frac{x}{2} \right)^2 + {PT}^2 \]

\[ \Rightarrow x^2 = \frac{x^2}{4} + {PT}^2 \]

\[ \Rightarrow {PT}^2 = x^2 - \frac{x^2}{4}\]

\[ \Rightarrow {PT}^2 = \frac{3 x^2}{4}\]

\[ \Rightarrow PT = \frac{\sqrt{3}x}{2}\]

In ∆PST \[{PS}^2 = {ST}^2 + {PT}^2 \]

\[ \Rightarrow {PS}^2 = \left( \frac{x}{6} \right)^2 + \left( \frac{\sqrt{3}x}{2} \right)^2 \]

\[ \Rightarrow {PS}^2 = \frac{x^2}{36} + \frac{3 x^2}{4}\]

\[ \Rightarrow {PS}^2 = \frac{x^2 + 27 x^2}{36}\]

\[ \Rightarrow {PS}^2 = \frac{28 x^2}{36}\]

\[ \Rightarrow {PS}^2 = \frac{7 x^2}{9}\]

\[ \Rightarrow 9 {PS}^2 = 7 {PQ}^2\]

Hence, 9 PS² = 7 PQ².