Difference between revisions of "2005 AMC 10A Problems/Problem 8"
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− | '''( | + | '''(C)''' 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> |
Revision as of 20:51, 30 January 2011
Problem
In the figure, the length of side of square
is
and
=1. What is the area of the inner square
?
Solution
(C) 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
.