Question

In the adjoining figure, ABCD is a trapezium in which DC is parallel to AB. If AB = 9 cm, DC = 6 cm and BD = 12 cm, find:

1. BP
2. the ratio of the areas of ΔAPB and ΔDPC

Answer 1

Given: In trapezium ABCD, DC || AB,

AB = 9 cm, DC = 6 cm, BD = 12 cm

Diagonals AC and BD intersect at P.

(i) Since AB || DC, the diagonals of a trapezium divide each other proportionally.

Therefore,

`(BP)/(PD) = (AB)/(DC) = 9/6 = 3/2`

So, BP : PD = 3 : 2

But, BD = BP + PD = 12 cm

Hence,

`BP = 3/(3 + 2) xx 12`

`BP = 3/5 xx 12`

`BP = 36/5`

∴ BP = 7.2 cm

(ii) In triangles △APB and △DPC,

• ∠APB = ∠DPC (vertically opposite angles)
• ∠ABP = ∠CDP (alternate interior angles, since AB || DC)

Therefore, △APB ∼ △DPC

Hence,

`(AB)/(DC) = (BP)/(PD) = (AP)/(PC) = 3/2`

Now, the ratio of areas of similar triangles is equal to the square of the ratio of their corresponding sides.

Therefore,

ar(△APB) : ar(△DPC) = `(3/2)^2 = 9/4`

∴ ar(△APB) : ar(△DPC) = 9 : 4