In △ABC, DE ∥ AC, where D and E are the points on sides AB and BC, respectively. If BD = 5 cm and AD = 14 cm, then what is the ratio of the area of △BDE to that of the trapezium ADEC?
25 : 336
D lies on AB and E lies on BC with DE ∥ AC, so in △BDE and △BAC the corresponding angles are equal and the triangles are similar.
BD = 5 cm and AD = 14 cm, so BA = BD + AD = 5 + 14 = 19 cm.
For similar triangles the areas are in the ratio of the squares of corresponding sides: area(△BDE) : area(△BAC) = BD2 : BA2 = 52 : 192 = 25 : 361.
The trapezium ADEC is what is left of △BAC after △BDE is removed, so its share is 361 − 25 = 336.
Required ratio = 25 : 336.
Hence, the answer is 25 : 336.
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