If $A (0, 2), B (2, 1), C (3, 0)$ ) are the vertices of a Triangle. $B D$ is the altitude. The line through “ $D$ " parallel to the side $A B$ intersects the side $B C$ at $E$ . The product of the area of Triangles $A B C$ and $B D E$ is $\frac{a}{b}$ sq units then $a + b =$

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Detailed Solution

Area of triangle $B D E = \frac{1}{3}$
Area of Triangle $A B C = \frac{3}{2}$
Products areas $= \frac{1}{3} \times \frac{3}{2} = \frac{1}{2} s s$
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