In the figure given below, sides PB and QA are perpendiculars drawn to the line segment AB.
If PO = 6 cm, QO = 9 cm and area of ΔPOB=120 cm2, then the area of ΔQOA is
270 cm2
In ΔPOB and ΔQOA,
∠PBO=∠QAO=90∘
∠POB=∠QOA
(Since vertically opposite angles are equal)
ΔPOB ∼ ΔQOA
(by AA similarity criterion)
⇒ar(ΔPOB)ar(ΔQOA)=PO2QO2
(The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides).
⇒120ar(QOA)=6292
⇒ar(QOA)=120×8136=270 cm2