In \(\triangle ACE\), B and D are points on the sides AC and CE respectively such that \(BD\parallel AE\) and \(\displaystyle AE=\frac{9}{5}BD\). What is the ratio of the area of \(\triangle BCD\) to the area of \(\triangle ACE\)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 25 : 81
A line parallel to one side cuts off a similar triangle, and the areas of similar triangles are in the ratio of the squares of their corresponding sides.
\(BD\parallel AE\), so \(\triangle BCD\sim\triangle ACE\) (common \(\angle C\), corresponding angles equal)
\(\displaystyle AE=\frac{9}{5}BD\) gives \(BD:AE=5:9\)
Area ratio \(=5^{2}:9^{2}=25:81\)
