In \(\Delta\)ABC, DE \(\parallel\) AC, where D and E are the points on sides AB and BC, respectively. If BD = 10 cm and AD = 4 cm, then what is the ratio of the area of \(\Delta\) BDE to the trapezium ADEC?
25 : 24
Since DE \(\parallel\) AC, \(\Delta BDE \sim \Delta BAC\) (AA similarity: \(\angle B\) is common and corresponding angles are equal).
AB = BD + AD = 10 + 4 = 14 cm, so the ratio of corresponding sides is \(\frac{BD}{BA} = \frac{10}{14} = \frac{5}{7}\).
The ratio of areas of similar triangles equals the square of the side ratio: \(\frac{ar(BDE)}{ar(BAC)} = \frac{25}{49}\).
Area of trapezium ADEC = ar(BAC) − ar(BDE) = 49 − 25 = 24 parts.
Hence, ar(BDE) : ar(ADEC) = 25 : 24.
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