Question

In the given figure, QR is parallel to AB and DR is parallel to QB. If PD = 4 cm and PA = 9 cm, then find PQ.

Answer 1

In $\Delta PQB\&\Delta PDR$

$\angle P=\angle P$ (common)

$\angle PDR=\angle PQB$ (corresponding angles)

$\Delta PDR\sim \Delta PQB$ (AA similarity)

$\frac{PD}{PR}=\frac{PQ}{PB}$ (corresponding sides)

$\frac{PD}{PQ}=\frac{PR}{PB}$ ---- (1)

In $\Delta PQR\&\Delta PAB$

$\angle P=\angle P$ (common)

$\angle PQR=\angle PAB$ (corresponding angles)

$\Delta PQR\sim \Delta PAB$ (AA similarity)

$\frac{PQ}{PR}=\frac{PA}{PB}$ (Corresponding sides)

$\frac{PQ}{PA}=\frac{PR}{PB}$ ---- (2)

From (1) & (2), we get

$\frac{PD}{PQ}=\frac{PQ}{PA}P{Q}^2=PD\times PAP{Q}^2=4\times 9$

$⟹PQ=6\text{cm}$