In a triangle, ABC, D, and E are two points on the sides AB and AC respectively so DE || BC and AD/BD = 4/5. The ratio of the area of ΔABC to the area of trapezium DECB is?
Show answer & explanation
Correct answer: B
Explanation
The correct answer is Option 2 i.e. 81 : 65.
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
DE ||BC, and AD/BD = 4/5
Since DE || BC, all three angles of triangle ADE and ABC are equal.
Hence, triangles ADE and ABC are similar triangles.
Area of ∆ADE/Area of ∆ABC = (AD/AB)2
⇒ Area of ∆ADE/Area of ∆ABC = [4/(4 + 5)]2 = 16 : 81
In the figure:
Area of trapezium DECB = Area of ∆ABC – Area of ∆ADE
(Area of ∆ABC) : (Area of trapezium ACED)
⇒ 81 : (81 - 16)
⇒ 81 : 65