Difference between revisions of "2018 AMC 10A Problems/Problem 24"

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By angle bisector theorem, <math>BG=\frac{5a}{6}</math>. By similar triangles, <math>DF=\frac{5a}{12}</math>, and the height of this trapezoid is <math>\frac{h}{2}</math>, where <math>h</math> is the length of the altitude to <math>BC</math>. Then <math>\frac{ah}{2}=120</math> and we wish to compute <math>\frac{5a}{8}\cdot\frac{h}{2}=\boxed{75}</math>. (trumpeter)
 
By angle bisector theorem, <math>BG=\frac{5a}{6}</math>. By similar triangles, <math>DF=\frac{5a}{12}</math>, and the height of this trapezoid is <math>\frac{h}{2}</math>, where <math>h</math> is the length of the altitude to <math>BC</math>. Then <math>\frac{ah}{2}=120</math> and we wish to compute <math>\frac{5a}{8}\cdot\frac{h}{2}=\boxed{75}</math>. (trumpeter)
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== (More basic) Solution 2 (bad) ==
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<math>\overline{DE}</math> is midway from <math>A</math> to <math>\overline{BC}</math>, and <math>DE = \frac{BC}{2}</math>. Therefore, <math>DE</math> is a quarter of the area, which is <math>30</math>. Using the angle bisector theorem in the same fashion as the previous problem, we get that one segment is <math>5</math> times the other. We want the larger piece, as described by the problem. Because the heights are identical, one area is <math>5</math> times the other, and <math>\frac{5}{6} \cdot 90 = \boxed{75}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 19:56, 8 February 2018

Problem

Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$. Let $D$ be the midpoint of $\overline{AB}$, and let $E$ be the midpoint of $\overline{AC}$. The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$, respectively. What is the area of quadrilateral $FDBG$?

$\textbf{(A) }60 \qquad \textbf{(B) }65 \qquad \textbf{(C) }70 \qquad \textbf{(D) }75 \qquad \textbf{(E) }80 \qquad$

Solution

By angle bisector theorem, $BG=\frac{5a}{6}$. By similar triangles, $DF=\frac{5a}{12}$, and the height of this trapezoid is $\frac{h}{2}$, where $h$ is the length of the altitude to $BC$. Then $\frac{ah}{2}=120$ and we wish to compute $\frac{5a}{8}\cdot\frac{h}{2}=\boxed{75}$. (trumpeter)

(More basic) Solution 2 (bad)

$\overline{DE}$ is midway from $A$ to $\overline{BC}$, and $DE = \frac{BC}{2}$. Therefore, $DE$ is a quarter of the area, which is $30$. Using the angle bisector theorem in the same fashion as the previous problem, we get that one segment is $5$ times the other. We want the larger piece, as described by the problem. Because the heights are identical, one area is $5$ times the other, and $\frac{5}{6} \cdot 90 = \boxed{75}$.

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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 AMC 10 Problems and Solutions
2018 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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 AMC 12 Problems and Solutions

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