Question

In the given figure shown $PQRS$ is a square and $SRT$ is an equilateral triangle then state whether $\angle QTR={15}^{\circ }$ is true/false.

• A. True
• B. False

Answer 1

The correct option is A True

$PQRS$ is a square.

$\therefore$ $\angle PSR=\angle QRS$ [ Each ${90}^o$ ]

$△SRT$ is an equilateral triangle.

$\Rightarrow$ $\angle TSR=\angle TRS$ [ Each ${60}^o$ ]

Now,

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

$\Rightarrow$ $\angle TSP=\angle TRQ$

Now, In $△TSP$ and $△TRQ,$ we have

$\Rightarrow$ $TS=TR$ [ Sides of equilateral triangle ]

$\Rightarrow$ $\angle TSP=\angle TRQ$ [ Proved above ]

$\Rightarrow$ $PS=QR$ [ Sides of square ]

$\therefore$ $△TSP\cong △TRQ$ [ By $SAS$ Congruence test ]

$\Rightarrow$ $PT=QT$ [ C.P.C.T ]

Now, In $△TQR,$ we have

$\Rightarrow$ $TR=QR$ [ $QR=RS=TR$ ]

$\Rightarrow$ $\angle TQR=\angle QTR$ [ Base angles of equal sides are equal ]

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

$\Rightarrow$ $\angle TQR+\angle QTR+\angle TRS+\angle SRQ={180}^o$

$\Rightarrow$ $2\left(\angle QTR\right)+{60}^o+{90}^o={180}^o$

$\Rightarrow$ $2\left(\angle QTR\right)={180}^o-{150}^o$

$\Rightarrow$ $2\left(\angle QTR\right)={30}^o$

$\therefore$ $\angle QTR={15}^o$ ------ Hence proved