In \(\triangle ABC\), D and E are points on the sides AB and AC respectively such that \(DE\parallel BC\) and AD : DB \(=2:3\). If the area of \(\triangle ABC\) is 75 cm2, what is the area of the quadrilateral DBCE?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 63 cm2
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.
\(DE\parallel BC\), so \(\triangle ADE\sim\triangle ABC\) (common \(\angle A\), corresponding angles equal)
AD : DB \(=2:3\) gives AD : AB \(=2:5\)
Area ratio \(=2^2:5^2=4:25\), so ar\((ADE)=\displaystyle\frac{4}{25}\times75=12\) cm2
ar\((DBCE)=75-12=63\) cm2
