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

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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: | 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: | ||

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

− | < | + | <cmath>1 + (HE+1)^2=50</cmath> |

− | < | + | <cmath>(HE+1)^2=49</cmath> |

− | < | + | <cmath>HE+1=7</cmath> |

− | < | + | <cmath>HE=6</cmath> |

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

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

## Revision as of 22:37, 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.