Question

In the given figure, PQR is an equilateral triangle and QRST is a square. Prove that (i) PT = PS, (ii) ∠PSR = 15°.

Answer 1

Given:
Triangle PQR is an equilateral triangle, i.e., each angle is $60°$.
QRST is a square, i.e., each angle is $90°$.

$\angle PQT=\angle PQR+\angle TQR=60+90=150°$
Similarly, $\angle PRS=150°$
Consider the triangles PQT and PRS.

$$\begin{gathered} \mathrm{PQ}=\mathrm{PR}\hspace{0.17em}(\mathrm{Sides}\mathrm{of}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}) \\ \mathrm{QT}=\mathrm{RS}(\mathrm{Sides}\mathrm{of}\mathrm{a}\mathrm{square}) \\ \angle \mathrm{PQT}=\angle \mathrm{PRS}=150°\left(\mathrm{Proved}\right) \\ △\mathrm{PQT}\cong △\mathrm{PRS}(\mathrm{SAS}\mathrm{criterion}) \\ \mathrm{PT}=\mathrm{PS}\left(\mathrm{CPCT}\right) \end{gathered}$$

Further, the side of the square coincides with the triangle, PR = RS, which makes PRS isosceles.
Let $\angle PSR=\angle SPR=x$.
Using angle sum property of a triangle, we get:
$$\begin{gathered} 150°+x+x=150°+2x=180° \\ \Rightarrow 2x=30° \\ \Rightarrow x=15° \end{gathered}$$

∴ $\angle PSR=15°$
Hence, proved.