Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

enter image description here

share|improve this question
    
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

2 Answers 2

up vote 2 down vote accepted

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

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

share|improve this answer
    
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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.