Try this beautiful problem from AMC 10A, 2004 based on Geometry: Ratio Of Two Triangles
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 \(\triangle ABE\)
Square
Triangle
Geometry
Answer: \(2\)
AMC-10A (2002) Problem 20
Pre College Mathematics
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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