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AMC 10 Math Olympiad USA Math Olympiad

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

Try this beautiful problem from Geometry: Area of the inner square AMC-10A, 2005, Problem-8. You may use sequential hints to solve the problem.

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 $EFGH$?

Area of the Inner Square - Problem Figure
  • \(25\)
  • \(32\)
  • \(36\)
  • \(42\)
  • \(40\)

Key Concepts


Geometry

Square

similarity

Check the Answer


Answer: \(36\)

AMC-10A (2005) Problem 8

Pre College Mathematics

Try with Hints


Area of the Inner Square - Shaded Figure

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…..?

Explanatory Shading of the figure

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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