# What is the area of triangle AFE?

If ED = 23 , and the value of the side of the square ABCD is a multiple of 11, what is the area of the red triangle AFE?! Find the very shortest way to solve this puzzle and use only basic geometry, trigonometry is not allowed.

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if no more condition is given can,can we assume that DF/DC is some rational number? –  dato datuashvili Sep 17 '11 at 13:03
@user3196,not 100% sure but I think that we can... –  pedja Sep 17 '11 at 13:13
so this problem says that value of AB can be above 22 or 33,44,55,if we know ration of DF/DC then we can find each other in case we know value of square side,so it means that as avik mentioned everything depend on x value –  dato datuashvili Sep 17 '11 at 13:15
@user3196,I see now,you are right –  pedja Sep 17 '11 at 13:21

Let $AB = 11x$. Triangles EDF and EAB are similar, so:

$\dfrac{ED}{EA} = \dfrac{DF}{AB}$

$\dfrac{23}{23 + 11x} = \dfrac{DF}{11x}$

$DF = \dfrac{253x}{23 + 11x}$

The area of $\triangle AFE$ is thus

$$\dfrac{1}{2} \cdot EA \cdot DF = \dfrac{1}{2} \cdot (23 + 11x) \cdot \dfrac{253x}{23 + 11x} = \dfrac{253}{2}x$$

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and what is a value of x? –  pedja Sep 17 '11 at 13:10
you can't calculate exact value of triangle if you dont have any more conditions,for example instead of x you can take 33,44,55, and so on. 11*x in general for x>2 –  dato datuashvili Sep 17 '11 at 13:12

let us consider one simple situation,suppose AB=33; you can check that 33 is multiple of 11,33/11=3,and also we know that DF/FC=1/2.if we denote DF by x,then FC=2*x so x+2*x=33, 3*x=33 x=11;(sorry in first coment instead of DC should be FC) so DF=11; length of AE=AD+DE or 33+23=56,so are of AEF=1/2*DF*AE`=1/2*56*11=28*11=308

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