In the figure (ii) given below, AB || DC and AB = 2DC. If AD = 3 cm, BC = 4 cm and AD, BC produced meet at E, find

(i) ED

(ii) BE

(iii) area of △EDC : area of trapezium ABCD.

(i) Given, AB = 2DC or, $\dfrac{AB}{DC} = \dfrac{2}{1}.$

Considering △AEB and △EDC.

∠E = ∠E (Common angles)

∠EDC = ∠EAB (Corresponding angles are equal)

Hence, by AA axiom △AEB ~ △EDC.

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

$\therefore \dfrac{AE}{ED} = \dfrac{AB}{DC} \\[1em] \Rightarrow \dfrac{AD + ED}{ED} = \dfrac{2}{1} \\[1em] \Rightarrow \dfrac{3 + ED}{ED} = \dfrac{2}{1} \\[1em] \Rightarrow 3 + ED = 2ED \\[1em] \Rightarrow ED = 3.$

Hence, the length of ED = 3 cm.

(ii) Since, △AEB ~ △EDC. Hence the ratio of the corresponding sides will be equal

$\therefore \dfrac{BE}{EC} = \dfrac{AB}{DC} \\[1em] \Rightarrow \dfrac{BC + EC}{EC} = \dfrac{2}{1} \\[1em] \Rightarrow \dfrac{4 + EC}{EC} = \dfrac{2}{1} \\[1em] \Rightarrow 4 + EC = 2EC \\[1em] \Rightarrow EC = 4.$

BE = BC + EC = 4 + 4 = 8 cm.

Hence, the length of BE = 8 cm.

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

$\therefore \dfrac{\text{Area of △EDC}}{\text{Area of △AEB}} = \dfrac{DC^2}{AB^2} \\[1em] \Rightarrow \dfrac{\text{Area of △EDC}}{\text{Area of △AEB}} = \dfrac{1^2}{2^2} \\[1em] \Rightarrow \dfrac{\text{Area of △EDC}}{\text{Area of △AEB}} = \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{\text{Area of △EDC}}{\text{Area of △EDC + Area of ⏢ABCD}} = \dfrac{1}{4} \\[1em] \Rightarrow 4 \text{Area of △EDC} = \text{Area of △EDC + Area of ⏢ABCD} \\[1em] \Rightarrow 4 \text{Area of △EDC} - \text{Area of △EDC} = \text { Area of ⏢ABCD} \\[1em] \Rightarrow 3 \text{ Area of △EDC} = \text { Area of ⏢ABCD} \\[1em] \Rightarrow \dfrac{\text{ Area of △EDC}}{\text { Area of ⏢ABCD}} = \dfrac{1}{3}.$

Hence, the ratio of area of △EDC : area of trapezium ABCD = 1 : 3.