In the figure, AC = 10cm, PC = 15cm, PQ = 12 cm, find PB.
9 cm
In ΔABC and ΔPQC,∠PQC=∠ABC=90∘∠C=∠C (common angle)Therefore, ΔABC∼ΔPQC by AA similarityACPC=ABPQ=BCQC⇒1015=AB12=BCQC⇒AB=8cmIn ΔABC,AB2+BC2=AC2⇒102=82+BC2⇒BC=6cmPB=PC−BC=15−6=9cmHence, PB=9cm.