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

Try this beautiful problem from AMC 10A, 2004 based on Geometry: Ratio Of Two Triangles

## 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 $\triangle ABE$

• $\frac{4}{3}$
• $\frac{3}{2}$
• $\sqrt 3$
• $2$
• $1+\sqrt 3$

Square

Triangle

Geometry

## Check the Answer

Answer: $2$

AMC-10A (2002) Problem 20

Pre College Mathematics

## Try with Hints

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 $\frac{\frac{x^2}{2}}{\frac{(1-x)}{2}}$=$\frac{x^2}{1-x}$=$\frac {2(1-x)}{(1-x)}=2$

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