# Difference between revisions of "2005 AMC 10A Problems/Problem 8"

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==Solution== | ==Solution== | ||

− | + | We see that side <math>BE</math>, which we know is 1, is also the shorter leg of one of the four right triangles (which are congruent, I'll not prove this). So, <math>AH = 1</math>. Then <math>HB = HE + BE = HE + 1</math>, and <math>HE</math> is one of the sides of the square whose area we want to find. So: | |

<math>1^2 + (HE+1)^2=\sqrt{50}^2</math> | <math>1^2 + (HE+1)^2=\sqrt{50}^2</math> | ||

Line 17: | Line 17: | ||

<math>HE+1=7</math> | <math>HE+1=7</math> | ||

− | <math>HE=6</math> So, the area of the square is <math>6^2=\boxed{36}</math>. | + | <math>HE=6</math> So, the area of the square is <math>6^2=\boxed{36} \rightarrow (C)</math>. |

==See Also== | ==See Also== |

## Revision as of 22:35, 3 January 2016

## Problem

In the figure, the length of side of square is and =1. What is the area of the inner square ?

## Solution

We see that side , which we know is 1, is also the shorter leg of one of the four right triangles (which are congruent, I'll not prove this). So, . Then , and is one of the sides of the square whose area we want to find. So:

So, the area of the square is .

## See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.