In the figure (ii) given below, AB || DC. AO = 10 cm, OC = 5 cm, AB = 6.5 cm and OD = 2.8 cm.

(i) Prove that △OAB ~ △OCD.

(ii) Find CD and OB.

(iii) Find the ratio of areas of △OAB and △OCD.

(i) Considering △OAB and △OCD,

∠ AOB = ∠ COD (Vertically opposite angles are equal)
∠ BAO = ∠ OCD (Alternate angles are equal)

Hence, by AA axiom △OAB ~ △OCD.

(ii) Since triangles are similar hence ratio of corresponding sides are equal,

$\therefore \dfrac{AO}{OC} = \dfrac{AB}{CD} \\[1em] \Rightarrow \dfrac{10}{5} = \dfrac{6.5}{CD} \\[1em] \Rightarrow CD = \dfrac{6.5 \times 5}{10} \\[1em] \Rightarrow CD = \dfrac{32.5}{10} \\[1em] \Rightarrow CD = 3.25 \text{ cm}.$

Similarly,

$\dfrac{AO}{OC} = \dfrac{OB}{OD} \\[1em] \Rightarrow \dfrac{10}{5} = \dfrac{OB}{2.8} \\[1em] \Rightarrow OB = \dfrac{2.8 \times 10}{5} \\[1em] \Rightarrow OB = \dfrac{28}{5} \\[1em] \Rightarrow OB = 5.6 \text{ cm.}$

Hence, the length of CD = 3.25 cm and OB = 5.6 cm.

(iii) In part (i) we have proved that △OAB ~ △OCD.

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

$\therefore \dfrac{\text{Area of △OAB}}{\text{Area of △OCD}} = \dfrac{AO^2}{OC^2} \\[1em] = \dfrac{10^2}{5^2} \\[1em] = \dfrac{100}{25} \\[1em] = \dfrac{4}{1} \\[1em] = 4 : 1.$

Hence, the ratio of area of △OAB and △OCD is 4 : 1.