Question

In the given figure, if $\angle TQR+\angle STQ={180}^0$, PT = 4 cm, QT = 3 cm and TS = 6 cm, then the length of QR is

• A. 10 cm
• B. 10.5 cm
• C. 7.5 cm
• D. 15 cm

Answer 1

The correct option is B 10.5 cm


$\angle TQR+\angle STQ={180}^0$ (Given)
$\Rightarrow TS||QR$ (Interior angle property)
Thus, $\angle PTS=\angle PQR$ (Corresponding angles)
Also, $\angle TPS=\angle QPR$ (Common)
$\therefore \Delta PTS\sim \Delta PQR$ (AA similarity)
$\Rightarrow \frac{PT}{PQ}=\frac{TS}{QR}$
$\Rightarrow \frac{4}{4+3}=\frac{6}{QR}$
$\Rightarrow QR=\frac{6\times 7}{4}=10.5cm$
Hence, the correct answer is option (2).