Question

PQRSTU is a regular hexagon. Determine each angle of $\Delta PQT$.

Answer 1

The sum of interior angles of a polygon $=(n-2)\times {180}^o$

$\Rightarrow$ The sum of interior angles of a hexagon $=(6-2)\times {180}^o$

$=4\times {180}^o$

$={720}^o$

$\Rightarrow$ Measure of each angle of hexagon $=\frac{{720}^o}{6}={120}^o$

In $△PUT,$

$\Rightarrow$ $PU=UT$ [ Sides of regular hexagon are equal ]

$\Rightarrow$ $\angle UTP=\angle UPT$ [ Equal sides have equal angles opposite to them ]

Now, $\angle PUT+\angle UTP+\angle UPT={180}^o.$

$\Rightarrow$ ${120}^o+2\angle UPT={180}^o.$

$\Rightarrow$ $2\angle UPT={60}^o$

$\Rightarrow$ $\angle UPT={30}^o$

Now, $\angle QPU={120}^o.$

$\Rightarrow$ $\angle QPT+{30}^o={120}^o.$

$\Rightarrow$ $\angle QPT={90}^o.$

$\Rightarrow$ $△PUT\cong △TSR$ [ By SAS congruence theorem ]

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

In $△PTQ$ and $△RTQ,$

$\Rightarrow$ $PQ=QR$ [ Sides of regular hexagon ]

$\Rightarrow$ $PT=TR$ [ Proved ]

$\Rightarrow$ $TQ=TQ$ [ Common side ]

$\therefore$ $△PTQ\cong △RTQ$ [ SSS Congruence rule ]

$\Rightarrow$ $\angle PQT=\angle RQT$ [ CPCT ]

$\Rightarrow$ $\angle PQR={120}^o$

$\Rightarrow$ $\angle PQT+\angle RQT={120}^o$

$\Rightarrow$ $2\angle PQT={120}^o$

$\therefore$ $PQT={60}^o$

In $△PQT,$

$\Rightarrow$ $\angle PQT+\angle QPT+\angle PTQ={180}^o$

$\Rightarrow$ ${60}^o+{90}^o+\angle PTQ={180}^o.$

$\Rightarrow$ $\angle PTQ={30}^o$

$\therefore$ $\angle PTQ={30}^o,\angle PQT={60}^o$ and $QPT={90}^o.$