Question

In fig., $PQRS$ is square and $SRT$ is an equilateral triangle. Prove that
(i) $PT=QT$
(ii) $\angle TQR={15}^{\circ }$

Answer 1

(i)

As PQRS is a square, so,

$\angle PSR=\angle QRS$ (each ${90}^{\circ }$)

Also, $\Delta SRT$ is an equilateral triangle, then,

$\angle TSR=\angle TRS$ (each ${60}^{\circ }$)

Now,

$\angle PSR+\angle TSR=\angle QRS+\angle TRS$

$\angle TSP=\angle TRQ$

In $\Delta TSP$ and $\Delta TRQ$,

$TS=TR$ (sides of equilateral triangle)

$\angle TSP=\angle TRQ$

$PS=QR$ (sides of square)

So, by SAS congruence rule,

$\Delta TSP\equiv \Delta TRQ$

By CPCT,

$PT=QT$

Hence proved.

(ii)

In $\Delta TQR$,

$\angle TQR=\angle QTR$

It is known that the sum of the angles of a triangle is ${180}^{\circ }$, so,

$\angle TQR+\angle QTR+\angle TRQ={180}^{\circ }$

$\angle TQR+\angle TQR+\angle TRS+\angle SRQ={180}^{\circ }$

$2\angle TQR+{60}^{\circ }+{90}^{\circ }={180}^{\circ }$

$2\angle TQR={30}^{\circ }$

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

Hence proved.