In the figure (i) given below, PB and QA are perpendiculars to line segment AB. If PO = 6 cm, OQ = 9 cm and the area of △POB = 120 cm², find the area of △QOA.

Considering △QOA and △POB,

∠ QOA = ∠ POB (Vertically opposite angles are equal)
∠ QAO = ∠ PBO (Both are equal to 90°)

Hence, by AA axiom △QOA ~ △POB.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

Let the area of △QOA be x cm².

$\therefore \dfrac{\text{Area of △QOA}}{\text{Area of △POB}} = \dfrac{QO^2}{PO^2} \\[1em] \Rightarrow \dfrac{x}{120} = \dfrac{9^2}{6^2} \\[1em] \Rightarrow \dfrac{x}{120} = \dfrac{81}{36} \\[1em] \Rightarrow x = \dfrac{120 \times 81}{36} \\[1em] \Rightarrow x = \dfrac{9720}{36} \\[1em] \Rightarrow x = 270.$

Hence, the area of △QOA is 270 cm²