Quantitative Aptitude :: Problems on Numbers - Discussion

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Explanation:
The line DE divides the triangle ABC into 2 parts - triangle ADE and trapezium DBCE.The question is to find the ratio of area(ADE) to area(DBCE).

Now triangle ADE is similar to triangle ABC.
=> DE/BC = AD/AB = AE/AC ....(1)
Drawing a perpendicular from A to BC meeting BC at G and meeting DE at F.
Now AF and AG are the heights of the triangles ADE and ABC resp.
tri(ADF)~tri(ABG) => AD/AB = AF/AG ....(2)
Area of a triangle = 1/2(base*height).
Area of tri(ADE) = 1/2(DE*AF)
Area of tri(ABC) = 1/2(BC*AG)
Area of trapezium(DBCE) = area(ABC)-area(ADE) = 1/2(BC*AG-DE*AF)
Ratio - area(ADE)/area(DBCE) = (DE*AF)/(BC*AG-DE*AF) = 1/[(BC*AG-DE*AF)/(DE*AF)]
(rearranging) = 1/[(BC/DE)*(AG/AF)-1]
Using (1) and (2),(BC/DE)*(AG/AF) = (BC/DE)^2
(using given DE/BC=3/5) => BC/DE=5/3 => (BC/DE)^2 = 25/9
The required ratio = 1/(25/9-1) =1/(16/9) = 9/16