Try this beautiful problem from Geometry: Area of the inner square
In the figure, the length of side $AB$ of square $ABCD$ is $\sqrt{50}$ and $BE=1$. What is the area of the inner square $EFGH$?
Geometry
Square
similarity
Answer: \(36\)
AMC-10A (2005) Problem 8
Pre College Mathematics
We have to find out the area of the region \(EFGH\) Which is a square shape .so if we can find out one of it's side length then we can easily find out the area of \(EFGH\). Now given that \(BE=1\) i.e \(BE=CF=DG=AH=1\) and side length of the square \(ABCD=\sqrt {50}\).Therefore \((AB)^2=(\sqrt {50})^2=50\).so using this information can you find out the length of \(EH\)?
Can you find out the required area.....?
Since \(EFGH\) is a square,therefore \(ABH\) is a Right -angle Triangle.
Therefore,\((AH)^2+(BH)^2=(AB)^2\)
\(\Rightarrow (AH)^2+(HE+EB)^2=(AB)^2\)
\(\Rightarrow (1)^2+(HE+1)^2=50\)
\(\Rightarrow (HE+1)^2=49\)
\(\Rightarrow (HE+1)=7\)
\(\Rightarrow HE=6\)
Therefore area of the inner square (red shaded region) =\({6}^2=36\)
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