Difference between revisions of "2008 AMC 8 Problems/Problem 23"
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==Solution 2~Mr.BigBrain_AoPS== | ==Solution 2~Mr.BigBrain_AoPS== | ||
Say that <math>\overline{FE}</math> has length <math>x</math>, and that from there we can infer that <math>\overline{AF} = 2x</math>. We also know that <math>\overline{ED} = x</math>, and that <math>\overline{DC} = 2x</math>. The area of triangle <math>BFD</math> is the square's area subtracted from the area of the excess triangles, which is simply these equations: | Say that <math>\overline{FE}</math> has length <math>x</math>, and that from there we can infer that <math>\overline{AF} = 2x</math>. We also know that <math>\overline{ED} = x</math>, and that <math>\overline{DC} = 2x</math>. The area of triangle <math>BFD</math> is the square's area subtracted from the area of the excess triangles, which is simply these equations: | ||
− | + | <cmath>9x^2 - (3x^2 + \dfrac{x}{2}^2 + 3x^2)</cmath> | |
− | 9x^2 - (3x^2 + \dfrac{x}{2}^2 + 3x^2) | + | <cmath>9x^2 - 6.5x^2</cmath> |
− | 9x^2 - 6.5x^2 | + | <cmath>2.5x^2</cmath> |
− | 2.5x^2 | ||
− | |||
Thus, the area of the triangle is <math>2.5x^2</math>. We can now put the ratio of triangle <math>BFD</math>'s area to the area of the square <math>ABCE</math> as a fraction. We have: | Thus, the area of the triangle is <math>2.5x^2</math>. We can now put the ratio of triangle <math>BFD</math>'s area to the area of the square <math>ABCE</math> as a fraction. We have: | ||
− | + | <cmath>\dfrac{2.5x^2}{9x^2} </cmath> | |
− | \dfrac{2.5x^2}{9x^2} | + | <cmath>\dfrac{2.5\cancel{x^2}}{9\cancel{x^2}}</cmath> |
− | \dfrac{2.5\cancel{x^2}}{9\cancel{x^2}} | + | <cmath>\dfrac{2.5}{9} </cmath> |
− | \dfrac{2.5}{9} | + | <cmath>\dfrac{5}{18}</cmath> |
− | \dfrac{5}{18} | ||
Thus, our answer is <math>\boxed{C}</math>, <math>\boxed{\dfrac{5}{18}}</math>. | Thus, our answer is <math>\boxed{C}</math>, <math>\boxed{\dfrac{5}{18}}</math>. | ||
Revision as of 15:42, 20 November 2022
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: 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: Thus, our answer is , .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.