Question
In triangle ABC, DE is parallel to BC, where D lies on
AB and E lies on AC. If AD:AB = 3:5 and the area of triangle ABC is 200 sq cm, then the area of trapezium DBCE is:Solution
ATQ, Since DE || BC, triangle ADE is similar to triangle ABC. So, Area of triangle ADE / Area of triangle ABC = (AD/AB)^2 = (3/5)^2 = 9/25 Area of triangle ADE = 200 x 9/25 = 72 sq cm Area of trapezium DBCE = 200 - 72 = 128 sq cm
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