2008 AMC 8 Problems/Problem 23
Problem
In square
,
and
. What is the ratio of the area of
to the area of square
?
Solution 1
The area of
is the area of square
subtracted by the the area of the three triangles around it. Arbitrarily assign the side length of the square to be
.
The ratio of the area of
to the area of
is
Solution 2
As stated in
"The area of
is the area of square
subtracted by the the area of the three triangles around it".
Let the side of the square be
.
Which means
=
=
and
=
=
.
Therefore the ratio of the area of
to the area of
is
$$ (Error compiling LaTeX. Unknown error_msg)\frac{
-
-$3x^2-\frac{$ (Error compiling LaTeX. Unknown error_msg)x^2$}{2}}{$ (Error compiling LaTeX. Unknown error_msg)9x^2$} = \frac{\frac{$ (Error compiling LaTeX. Unknown error_msg)5x^2$}{2}}{$ (Error compiling LaTeX. Unknown error_msg)9x^2$} = \frac{$ (Error compiling LaTeX. Unknown error_msg)5x^2$}{$ (Error compiling LaTeX. Unknown error_msg)9x^2$} = \boxed{\textbf{(C)}\ \frac{5}{18}}$ (Error compiling LaTeX. Unknown error_msg)$
See Also
| 2008 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
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