Try this beautiful problem from Geometry: Area of the inner square

## Area of Inner Square – AMC-10A, 2005- Problem 8

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 $EFGH$?

- \(25\)
- \(32\)
- \(36\)
- \(42\)
- \(40\)

**Key Concepts**

Geometry

Square

similarity

## Check the Answer

But try the problem first…

Answer: \(36\)

AMC-10A (2005) Problem 8

Pre College Mathematics

## Try with Hints

First hint

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…..?

Second Hint

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) =\({6}^2=36\)

## Other useful links

- https://www.cheenta.com/circumscribed-circle-amc-10a-2003-problem-17/
- https://www.youtube.com/watch?v=fRj9NuPGrLU&t=282s

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