Question

In the given figure, ∆PQR is an equilateral triangle. Point S is on seg QR such that
QS = $\frac{1}{3}$ 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 = $\frac{1}{2}\mathrm{QR}=\frac{x}{2}$

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

$\therefore \mathrm{ST}=\mathrm{QT}-\mathrm{QS}=\frac{x}{2}-\frac{x}{3}=\frac{x}{6}$

According to Pythagoras theorem,
In ∆PQT

$$\begin{gathered} {\mathrm{PQ}}^2={\mathrm{QT}}^2+{\mathrm{PT}}^2 \\ \Rightarrow {\left(x\right)}^2={\left(\frac{x}{2}\right)}^2+{\mathrm{PT}}^2 \\ \Rightarrow x^2=\frac{x^2}{4}+{\mathrm{PT}}^2 \\ \Rightarrow {\mathrm{PT}}^2=x^2-\frac{x^2}{4} \\ \Rightarrow {\mathrm{PT}}^2=\frac{3x^2}{4} \\ \Rightarrow \mathrm{PT}=\frac{\sqrt{3}x}{2} \end{gathered}$$

In ∆PST

$$\begin{gathered} {\mathrm{PS}}^2={\mathrm{ST}}^2+{\mathrm{PT}}^2 \\ \Rightarrow {\mathrm{PS}}^2={\left(\frac{x}{6}\right)}^2+{\left(\frac{\sqrt{3}x}{2}\right)}^2 \\ \Rightarrow {\mathrm{PS}}^2=\frac{x^2}{36}+\frac{3x^2}{4} \\ \Rightarrow {\mathrm{PS}}^2=\frac{x^2+27x^2}{36} \\ \Rightarrow {\mathrm{PS}}^2=\frac{28x^2}{36} \\ \Rightarrow {\mathrm{PS}}^2=\frac{7x^2}{9} \\ \Rightarrow 9{\mathrm{PS}}^2=7{\mathrm{PQ}}^2 \end{gathered}$$

Hence, 9 PS² = 7 PQ².