Question : In the given figure, QRTS is a cyclic quadrilateral. If PT = 5 cm, SQ = 4 cm, PS = 6 cm, and $\angle$PQR = $63^{\circ}$, then what is the value (in cm) of TR?
Option 1: 3
Option 2: 7
Option 3: 9
Option 4: 15
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Correct Answer: 7
Solution :
QSTR is a cyclic quadrilateral.
PT = 5 cm
SQ = 4 cm
PS = 6 cm
$\angle$PQR = $63^{\circ}$
The sum of the opposite angles = $180^{\circ}$
⇒ $\angle$SQR + $\angle$STR = $180^{\circ}$
⇒ $63^{\circ}$ + $\angle$STR = $180^{\circ}$
⇒ $\angle$STR = $180^{\circ}$ – $63^{\circ}$ = $117^{\circ}$
Also, $\angle$PTS + $\angle$STR = $180^{\circ}$
⇒ $117^{\circ}$ + $\angle$PTS = $180^{\circ}$
⇒ $\angle$PTS = $180^{\circ}$ – $117^{\circ}$ = $63^{\circ}$
⇒ $\angle$PTS = $\angle$PQR
$\angle$PST + $\angle$QST = $180^{\circ}$ -----------(i)
The sum of the opposite angles = $180^{\circ}$
⇒ $\angle$QRT + $\angle$QST = $180^{\circ}$ -----------(ii)
From (i) and (ii),
⇒ $\angle$PST = $\angle$PRQ
$\therefore$ $\triangle$PTS ~ $\triangle$PQR
⇒ $\frac{PT}{PQ}$ = $\frac{PS}{PR}$
⇒ $\frac{5}{4+6}$ = $\frac{6}{PR}$
⇒ PR = 2 × 6 = 12 cm
⇒ PT + TR = 12 cm
$\therefore$ TR = 12 – 5 = 7 cm
Hence, the correct answer is 7.
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