Difference between revisions of "2008 AMC 8 Problems/Problem 23"
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<math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{2}{9}\qquad\textbf{(C)}\ \frac{5}{18}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{7}{20} </math> | <math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{2}{9}\qquad\textbf{(C)}\ \frac{5}{18}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{7}{20} </math> | ||
− | ==Solution== | + | ==Solution 1== |
The area of <math>\triangle BFD</math> is the area of square <math>ABCE</math> subtracted by the the area of the three triangles around it. Arbitrarily assign the side length of the square to be <math>6</math>. | The area of <math>\triangle BFD</math> is the area of square <math>ABCE</math> subtracted by the the area of the three triangles around it. Arbitrarily assign the side length of the square to be <math>6</math>. | ||
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<cmath>\frac{36-12-12-2}{36} = \frac{10}{36} = \boxed{\textbf{(C)}\ \frac{5}{18}}</cmath> | <cmath>\frac{36-12-12-2}{36} = \frac{10}{36} = \boxed{\textbf{(C)}\ \frac{5}{18}}</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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> | ||
+ | <cmath>9x^2 - 6.5x^2</cmath> | ||
+ | <cmath>2.5x^2</cmath> | ||
+ | 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> | ||
+ | <cmath>\dfrac{2.5\cancel{x^2}}{9\cancel{x^2}} </cmath> | ||
+ | <cmath>\dfrac{2.5}{9}</cmath> | ||
+ | <cmath>\dfrac{5}{18} </cmath> | ||
+ | Thus, our answer is <math>\boxed{C}</math>, <math>\boxed{\dfrac{5}{18}}</math>. | ||
+ | |||
+ | ~Mr.BigBrain_AoPS | ||
+ | |||
+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/abSgjn4Qs34?t=528 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== |
Latest revision as of 08:39, 20 June 2024
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
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 , .
~Mr.BigBrain_AoPS
Video Solution by OmegaLearn
https://youtu.be/abSgjn4Qs34?t=528
~ pi_is_3.14
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.