Parallelogram inside Triangle!

Area of triangle ABC = √((16.5)(6.5)(5.5)(4.5) = A, x = [ABC] – [FBD] – [EDC]. [FBD] ,[EDC] are both similar to [ABC]. Let the length of FB = z. Then by similarity
[FBD]=(z/12)2A, and [EDC]=((12 – z)/12)2A, so that z =A - (z/12)2A - ((12 – z)/12)2A = A(- z2/72 + 1/6 z), The stationary value of z (max.of X), dx/dz = A (-z/36 +1/6) = 0, then z=6. So xmax = ½ A = 25.7606. So└ 100x ┘ = 2576.

This is basically Ruslan's solution LaTeX-ified. I used a slightly different approach using vectors. (Although it is essentially the same thing.)

$\Delta AFDE = \Delta ABC - \Delta FBD - \Delta EDC$ .

Also, $FBD \sim ABC$ and $EDC \sim ABC$

If $FB=x \Rightarrow \Delta FBD = \left(\frac{x}{c} \right)^2 A$

$AF=c-x \Rightarrow DE=c-x \Rightarrow \Delta EDC = \left(\frac{c-x}{c} \right)^2 A$

Finally, $\Delta AFDE = \Delta A \left( 1 - \left(\frac{x}{c}\right)^2 - \left(\frac{c-x}{c}\right)^2 \right)$

Differentiating wrt $x$ we find that the expression attains maxima at $x=\frac{c}{2}$ .

Substituting in the equation gives the answer $\Delta AFDE =\frac{1}{2}\Delta A$

$\Delta A = \sqrt{s(s-a)(s-b)(s-c)}$ and on substiting values comes out to be $55.1212$

Therefore, maximum $\Delta AFDE = 27.5606$