2005 AMC 10A Problems/Problem 8
Problem
In the figure, the length of side of square
is
,
is between
and
, and
. What is the area of the inner square
?
Solution
We see that side , which we know is
, is also the shorter leg of one of the four right triangles (which are obviously congruent, using the symmetry of the diagram). So
, and hence
. Since
is one of the sides of the square whose area we want to find, we can now simply apply Pythagoras' theorem:
Thus the area of the square is .
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.