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Difference between revisions of "2008 AMC 8 Problems/Problem 23"

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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>
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==Solution 2==
 
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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".
 
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Let the side of the square be <math>3x</math>.
 
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Which means <math>AF</math>=<math>2x</math>=<math>CD</math> and <math>EF</math>=<math>x</math>=<math>ED</math>.
 
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Therefore the ratio of the area of <math>\triangle BFD</math> to the area of <math>ABCE</math> is
 
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<math></math>\frac{<math>9x^2</math>-<math>3x^2</math>-<math>3x^2</math>-\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></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 05:32, 15 November 2019

Problem

In square $ABCE$, $AF=2FE$ and $CD=2DE$. What is the ratio of the area of $\triangle BFD$ to the area of square $ABCE$? [asy] size((100)); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((3,0)--(9,9)--(0,3)--cycle); dot((3,0)); dot((0,3)); dot((9,9)); dot((0,0)); dot((9,0)); dot((0,9)); label("$A$", (0,9), NW); label("$B$", (9,9), NE); label("$C$", (9,0), SE); label("$D$", (3,0), S); label("$E$", (0,0), SW); label("$F$", (0,3), W); [/asy] $\textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{2}{9}\qquad\textbf{(C)}\ \frac{5}{18}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{7}{20}$

Solution 1

The area of $\triangle BFD$ is the area of square $ABCE$ subtracted by the the area of the three triangles around it. Arbitrarily assign the side length of the square to be $6$.

[asy] size((100)); pair A=(0,9), B=(9,9), C=(9,0), D=(3,0), E=(0,0), F=(0,3); pair[] ps={A,B,C,D,E,F}; dot(ps); draw(A--B--C--E--cycle); draw(B--F--D--cycle); label("$A$",A, NW); label("$B$",B, NE); label("$C$",C, SE); label("$D$",D, S); label("$E$",E, SW); label("$F$",F, W); label("$6$",A--B,N); label("$6$",(10,4.5),E); label("$4$",D--C,S); label("$2$",E--D,S); label("$2$",E--F,W); label("$4$",F--A,W); [/asy]

The ratio of the area of $\triangle BFD$ to the area of $ABCE$ is

\[\frac{36-12-12-2}{36} = \frac{10}{36} = \boxed{\textbf{(C)}\ \frac{5}{18}}\]

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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