In the given figure, QR is parallel to AB and DR is parallel to QB. If PD = 4 cm and PA = 9 cm, then PQ = ____.
6 cm
In ΔPQB & ΔPDR
∠P=∠P (common)
∠PDR=∠PQB (corresponding angles)
ΔPDR ∼ ΔPQB (AA similarity)
PDPR=PQPB (corresponding sides)
PDPQ=PRPB ---- (1)
In ΔPQR & ΔPAB
∠P=∠P (common)
∠PQR=∠PAB (corresponding angles)
ΔPQR ∼ ΔPAB (AA similarity)
PQPR=PAPB (Corresponding sides)
PQPA=PRPB ---- (2)
From (1) & (2), we get
PDPQ=PQPAPQ2=PD×PAPQ2=4×9
⟹PQ=6 cm