Question

In the figure, PQRS is a square and SRT is an equilateral triangle. Prove that

(i) PT = QT (ii) $\angle TQR={15}^{\circ }$

Answer 1

Given: PQRS is a square and SRT is an equilateral triangle. PT and QT are joined.

To prove : (i) PT = QT; (ii) $\angle TQR={15}^{\circ }$

Proof : In $\Delta TSPand\Delta TQR$

ST = RT (Sides of equilateral triangle)

SP = RQ (Sides of square)

and $\angle TSP=\angle TRQ$ $(Each={60}^{\circ }+{90}^{\circ })$

$\therefore \Delta TSP\cong \Delta TQR$ (SAS axiom)

$\therefore PT=QT$ (c.p.c.t.)

In $\Delta TQR,$

$\therefore RT=RQ$ (Square sides)

$\therefore \angle RTQ=\angle RQT$

But $\angle TRQ={60}^{\circ }+{90}^{\circ }={150}^{\circ }$

$\therefore \angle RTQ+\angle RQT={180}^{\circ }-{150}^{\circ }={30}^{\circ }$

$\therefore \angle PTQ=\angle RQT$ (Proved)

$\therefore \angle RQT=\frac{{30}^{\circ }}{2}={15}^{\circ }$

$\Rightarrow \angle TQR={15}^{\circ }$