Difference between revisions of "2005 AMC 10A Problems/Problem 8"
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<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}</math>. | ||
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+ | ==See Also== | ||
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+ | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:Area Problems]] |
Revision as of 21:35, 11 April 2013
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
.