In triangle ABC, a line DE is drawn parallel to BC, intersecting AB at D and AC at E. If AD : DB = 3 : 2 and the area of triangle ADE is 45 cm², find the area of quadrilateral BDEC.
80 cm²
Since \(DE \parallel BC\), triangle \(ADE\) is similar to triangle \(ABC\).
Given \(AD:DB = 3:2\), so \(AD:AB = 3:5\).
The ratio of areas of similar triangles equals the square of the ratio of corresponding sides: \(\frac{\text{Area}(ADE)}{\text{Area}(ABC)} = \left(\frac{3}{5}\right)^2 = \frac{9}{25}\).
So, \(\text{Area}(ABC) = 45 \times \frac{25}{9} = 125 \text{ cm}^2\).
Area of quadrilateral \(BDEC = \text{Area}(ABC) - \text{Area}(ADE) = 125 - 45 = 80 \text{ cm}^2\).
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