Difference between revisions of "2008 AMC 8 Problems/Problem 23"
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==Solution 2== | ==Solution 2== | ||
− | As stated in <math>solution 1</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". | + | As stated in <math>solution 1</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". |
Let the side of the square be <math>3x</math>. | Let the side of the square be <math>3x</math>. | ||
Line 56: | Line 56: | ||
Therefore the ratio of the area of <math>\triangle BFD</math> to the area of <math>ABCE</math> is | Therefore the ratio of the area of <math>\triangle BFD</math> to the area of <math>ABCE</math> is | ||
− | <math></math>\frac{<math>9x^2</math>-<math>3x^2</math>-<math>3x^2-\frac{</math>x^2<math>}{2}}{</math>9x^2<math>} = \frac{\frac{</math>5x^2<math>}{2}}{</math>9x^2<math>} = \frac{</math>5x^2<math>}{</math>9x^2<math>} = \boxed{\textbf{(C)}\ \frac{5}{18}}</math>$ | + | <math></math>\frac{<math>9x^2</math>-<math>3x^2</math>-<math>3x^2-\frac{</math>x^2<math>}{2}}{</math>9x^2<math>}<cmath> = </cmath>\frac{\frac{</math>5x^2<math>}{2}}{</math>9x^2<math>}<cmath> = </cmath>\frac{</math>5x^2<math>}{</math>9x^2<math>}<cmath> = \boxed{\textbf{(C)}\ </cmath>\frac{5}{18}}</math>$ |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2008|num-b=22|num-a=24}} | {{AMC8 box|year=2008|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 05:30, 15 November 2019
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.