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

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

Answer: $36$

AMC-10A (2005) Problem 8

Pre College Mathematics

## Try with Hints

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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