In the figure (i) given below, ABCD is a trapezium in which DC is parallel to AB. If AB = 9 cm, DC = 6 cm and BD = 12 cm, find

(i) BP

(ii) the ratio of areas of △APB and △DPC.

(i) Considering △APB and △CPD.

∠APB = ∠CPD (Vertical opposite angles are equal)

∠PAB = ∠PCD (Alternate angles are equal)

Hence, by AA axiom △APB ~ △CPD.

Since triangles are similar, hence the ratio of the corresponding sides will be equal

$\therefore \dfrac{BP}{PD} = \dfrac{AB}{CD} \\[1em] \Rightarrow \dfrac{BP}{BD - BP} = \dfrac{9}{6} \\[1em] \Rightarrow \dfrac{BP}{12 - BP} = \dfrac{9}{6} \\[1em] \Rightarrow 6 BP = 9(12 - BP) \\[1em] \Rightarrow 6BP = 108 - 9BP \\[1em] \Rightarrow 15 BP = 108 \\[1em] \Rightarrow BP = \dfrac{108}{15} \\[1em] \Rightarrow BP = 7.2 \text{ cm}.$

Hence, the length of BP = 7.2 cm

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

$\therefore \dfrac{\text{Area of △APB}}{\text{Area of △DPC}} = \dfrac{AB^2}{CD^2} \\[1em] \Rightarrow \dfrac{\text{Area of △APB}}{\text{Area of △DPC}} = \dfrac{9^2}{6^2} \\[1em] \Rightarrow \dfrac{\text{Area of △APB}}{\text{Area of △DPC}} = \dfrac{81}{36} = \dfrac{9}{4}.$

Hence, area of △APB : area of △DPC = 9 : 4.