Ratio Of Two Triangles | AMC-10A, 2004 | Problem 20

Points \(E\) and \(F\) are located on square \(ABCD\) so that \(\triangle BEF\) is equilateral. What is the ratio of the area of \(\triangle DEF\) to that of \(\triangle ABE\)

We have to find out the ratio of the areas of two Triangles \(\triangle DEF\) and \(\triangle ABE\).Let us take the side length of \(AD\)=\(1\) & \(DE=x\),therefore \(AE=1-x\)

Now in the \(\triangle ABE\) & \(\triangle BCF\) ,

\(AB=BC\) and \(BE=BF\).using Pythagoras theorm we may say that \(AE=FC\).Therefore \(\triangle ABE \cong \triangle CEF\).So \(AE=FC\) \(\Rightarrow DE=DF\).Therefore the \(\triangle DEF\) is  an isosceles right triangle. Can you find out the area of isosceles right triangle \(\triangle DEF\)

Can you now finish the problem ..........

Length of \(DE=DF=x\).Then the the side length of \(EF=X \sqrt 2\)

Therefore the area of \(\triangle DEF= \frac{1}{2} \times x \times x=\frac{x^2}{2}\) and area of \(\triangle ABE\)=\(\frac{1}{2} \times 1 \times (1-x) = \frac{1-x}{2}\).Now from the Pythagoras theorm \((1-x)^2 +1 =2x^2 \Rightarrow x^2=2-2x=2(1-x)\)

can you finish the problem........

The ratio of the area of \(\triangle DEF\) to that of \(\triangle ABE\) is \(\frac{\frac{x^2}{2}}{\frac{(1-x)}{2}}\)=\(\frac{x^2}{1-x}\)=\(\frac {2(1-x)}{(1-x)}=2\)

AMC - AIME Boot Camp... for brilliant students.

Use our exclusive one-on-one plus group class system to prepare for Math Olympiad

Learn More

Subscribe to Cheenta at Youtube