In ΔABC, DE || AC, where D and E are the points on sides AB and BC, respectively. If BD = 12 cm and AD = 11 cm, then what is the ratio of the area of ΔBDE to that of the trapezium ADEC?

Correct option is D

Given:

In ΔABC, DE∥AC, with D on AB and E on BC

BD = 12 cm

AD = 11 cm

Concept Used:

Basic Proportionality Theorem (Thales’ Theorem) and area ratio of similar triangles.

Formula Used:

Area ratio of two similar triangles = ( ratio of corresponding sides)²

Solution:

AB = AD + BD = 11 + 12 = 23 cm

​$$\frac{BD} { AB} = \frac{12} {23}$$​​

Since DE || AC, ΔBDE∼ΔBAC with ratio 12 : 23.

​$$\frac{Ar.(\Delta BDE)} { Ar.(\Delta BAC)} = \left( \frac{12}{23} \right)^2 = \frac{144}{529}$$​​

Area(trapezium ADEC) = Area(ΔBAC) – Area(ΔBDE) = 529 – 144 = 385 

Now,

Area(ΔBDE) : Area(ADEC) = 144 : 385