JAMB 1988 · UME · Q35

In the figure, $\angle STQ = \angle SRP$ , $PT = TQ = 6$ cm and $QS = 5$ cm. Find $SR$ .

Worked solution (try it first)

Triangles $QTS$ and $QRP$ share $\angle Q$ and have $\angle QTS = \angle QRP$ , so they are similar, with $T$ matching $R$ and $S$ matching $P$ .
So $\dfrac{QT}{QR} = \dfrac{QS}{QP}$ .
Here $QT = 6$ , $QS = 5$ and $QP = 6 + 6 = 12$ , so $\dfrac{6}{QR} = \dfrac{5}{12}$ .
Cross-multiply: $5 \times QR = 72$ , so $QR = \frac{72}{5}$ .
Then $SR = QR - QS$
$= \frac{72}{5} - \frac{25}{5}$
$= \frac{47}{5}$ , option A.