2008 AMC 8 Problems/Problem 23
Problem
In square , and . What is the ratio of the area of to the area of square ?
Solution
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~Mr.BigBrain_AoPS
Say that has length , and that from there we can infer that . We also know that , and that . The area of triangle is the square's area subtracted from the area of the excess triangles, which is simply these equations: \begin{align*} 9x^2 - (3x^2 + \dfrac{x}{2}^2 + 3x^2) \\ 9x^2 - 6.5x^2\\ 2.5x^2 \end{align*} Thus, the area of the triangle is . We can now put the ratio of triangle 's area to the area of the square as a fraction. We have: \begin{align*} \dfrac{2.5x^2}{9x^2} \\ \dfrac{2.5\cancel{x^2}}{9\cancel{x^2}} \\ \dfrac{2.5}{9} \\ \dfrac{5}{18} \end{align*} Thus, our answer is , .
See Also
2008 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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