2008 AMC 8 Problems/Problem 23
Contents
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$}<cmath> = </cmath>\frac{\frac{$ (Error compiling LaTeX. Unknown error_msg)5x^2$}{2}}{$ (Error compiling LaTeX. Unknown error_msg)9x^2$}<cmath> = </cmath>\frac{$ (Error compiling LaTeX. Unknown error_msg)5x^2$}{$ (Error compiling LaTeX. Unknown error_msg)9x^2$}<cmath> = \boxed{\textbf{(C)}\ </cmath>\frac{5}{18}}$ (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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