Question

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

$PT=QT$

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

Answer 1

Here, $PQRS$ is a square

$\Rightarrow$ $PQ=QR=RS=SP$ [ Sides of square are equal ] ------ ( 1 )

$\Rightarrow$ $\angle SPQ=\angle PQR=\angle QRS=\angle RSP={90}^o$ [ Angles of square ]

$△SRT$ is an equilateral triangle.

$\Rightarrow$ $SR=RT=TS$ [ Sides of equilateral triangle are equal ] ----- ( 2 )

$\Rightarrow$ $\angle TSR=\angle SRT=\angle RTS={60}^o$

From ( 1 ) and ( 2 ),

$\Rightarrow$ $PQ=QR=SP=SR=RT=TS$ ----- ( 3 )

$\Rightarrow$ $\angle TSP=\angle TSR+\angle RSP={60}^o+{90}^o={150}^o$

$\Rightarrow$ $\angle TRQ=\angle TRS+\angle SRQ={60}^o+{90}^o={150}^o$

$\Rightarrow$ $\angle TSP=\angle TRQ={150}^o$ ----- ( 4 )

Now, in $△TSP$ and $△TRQ$

$\Rightarrow$ $TS=TR$ [ From ( 3 ) ]

$\Rightarrow$ $\angle TSP=\angle TRQ$ [ From ( 4 ) ]

$\Rightarrow$ $SP=RQ$ [ From ( 3 ) ]

$\therefore$ $△TSP\cong △TRQ$ [ By SAS congruence rule ]

$\Rightarrow$ $PT=QT$ [ CPCT ] ---- Hence proved

Consider $△TQR,$

$\Rightarrow$ $QR=TR$ [ From ( 3 ) ]

$\therefore$ $△TQR$ is an isosceles triangle.

$\Rightarrow$ $\angle QTR=\angle TQR$ [ Angles opposite to equal sides ]

Now, a sum of angles in a triangle is equal to ${180}^o.$

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

$\Rightarrow$ $2\angle TQR+{150}^o={180}^o$ [ From ( 4 ) ]

$\Rightarrow$ $2\angle TQR={180}^o-{150}^o$

$\Rightarrow$ $2\angle TQR={30}^o$

$\Rightarrow$ $\angle TQR={15}^o$ ---- Hence proved