In △ABC, DE ∥ AC, where D and E are the points on sides AB and BC, respectively. If BD = 17 cm and AD = 14 cm, then what is the ratio of the area of △BDE to that of the trapezium ADEC?

Correct option is B

Given:

In triangle △ABC, DE∥AC,

D lies on AB and E on BC,

BD = 17 cm, AD = 14 cm.

Concept Used: 

By using the property of similar triangles, since DE∥AC,

△BDE∼△BAC

Thus, the ratio of the areas of similar triangles = square of the ratio of corresponding sides.

Also,

Area of trapezium ADEC = Area of △ABC - Area of △BDE

Solution:

​$$\frac{\text{Area of } \triangle BDE}{\text{Area of } \triangle ABC} = \left( \frac{BD}{AB} \right)^2$$​​

where , AB = AD + BD = 14 + 17 = 31

​$$\frac{\text{Area of } \triangle BDE}{\text{Area of } \triangle ABC} = \left( \frac{17}{31} \right)^2 = \frac{289}{961}$$​​

So,

​$$\frac{\text{Area of } \triangle BDE}{\text{Area of } \text{ADEC}} = \frac{289}{961 - 289} = \frac{289}{672}$$  = 289 : 672