In ΔABC, D is a point on side AB such that BD = 2 cm and DA = 3 cm. E is a point on BC such that DE || AC, and AC = 4 cm. Then (Area of ΔBDE) : (Area of trapezium ACED) is?
Show answer & explanation
Correct answer: B
Explanation
The correct answer is option 2 i.e. 4 : 21
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Concept |
Understanding |
Application |
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If all three angles of 2 triangles are equal, then these two triangles are similar triangles. And (Area of triangle1/ Area of triangle2) = (Side12/Side22) |
Given BD = 2 cm DA = 3 cm. DE || AC, and AC = 4 cm. Since DE || AC, all three angles of triangle BDE and BAC are equal. Hence, triangles BDE and BAC are similar triangles. |
Area of ∆BDE/Area of ∆BAC = (BD/BA)2 ⇒ Area of ∆BDE/Area of ∆BAC = [2/(2 + 3)]2 = 4 : 25
In the figure: Area of trapezium ACED = Area of ∆BAC – Area of ∆BDE (Area of ∆BDE) : (Area of trapezium ACED) = 4 : (25 - 4) = 4 : 21 |