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

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== Solution 3 ==
 
== Solution 3 ==
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The area of <math>\bigtriangleup ABG</math> to the area of <math>\bigtriangleup ACG</math> is <math>5:1</math> by Law of Sines. So the area of <math>\bigtriangleup ABG</math> is <math>100</math>. Since <math>\overline{DE}</math> is the midsegment of <math>\bigtriangleup ABC</math>, so <math>\overline{DF}</math> is the midsegment of <math>\bigtriangleup ABG</math> . So the area of <math>\bigtriangleup ACG</math>  to the area of <math>\bigtriangleup ABG</math> is <math>1:4</math> , so the area of <math>\bigtriangleup ACG</math> is <math>25</math>, by similar triangles. Therefore the area of quad <math>FDBG</math> is <math>100-25=\boxed{75}</math>
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The area of <math>\bigtriangleup ABG</math> to the area of <math>\bigtriangleup ACG</math> is <math>5:1</math> by Law of Sines. So the area of <math>\bigtriangleup ABG</math> is <math>100</math>. Since <math>\overline{DE}</math> is the midsegment of <math>\bigtriangleup ABC</math>, so <math>\overline{DF}</math> is the midsegment of <math>\bigtriangleup ABG</math> . So the area of <math>\bigtriangleup ADF</math>  to the area of <math>\bigtriangleup ABG</math> is <math>1:4</math> , so the area of <math>\bigtriangleup ACG</math> is <math>25</math>, by similar triangles. Therefore the area of quad <math>FDBG</math> is <math>100-25=\boxed{75}</math>
  
 
==Solution 4 ==
 
==Solution 4 ==

Revision as of 13:43, 31 July 2019

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 1

Let $BC = a$, $BG = x$, $GC = y$, and the length of the perpendicular to $BC$ through $A$ be $h$. By angle bisector theorem, we have that \[\frac{50}{x} = \frac{10}{y},\] where $y = -x+a$. Therefore substituting we have that $BG=\frac{5a}{6}$. By similar triangles, we have that $DF=\frac{5a}{12}$, and the height of this trapezoid is $\frac{h}{2}$. Then, we have that $\frac{ah}{2}=120$. We wish to compute $\frac{5a}{8}\cdot\frac{h}{2}$, and we have that it is $\boxed{75}$ by substituting.

Solution 2

For this problem, we have $\triangle{ADE}\sim\triangle{ABC}$ because of SAS and $DE = \frac{BC}{2}$. Therefore, $\bigtriangleup ADE$ is a quarter of the area of $\bigtriangleup ABC$, which is $30$. Subsequently, we can compute the area of quadrilateral $BDEC$ to be $120 - 30 = 90$. Using the angle bisector theorem in the same fashion as the previous problem, we get that $\overline{BG}$ is $5$ times the length of $\overline{GC}$. We want the larger piece, as described by the problem. Because the heights are identical, one area is $5$ times the other, and $\frac{5}{6} \cdot 90 = \boxed{75}$.

Solution 3

The area of $\bigtriangleup ABG$ to the area of $\bigtriangleup ACG$ is $5:1$ by Law of Sines. So the area of $\bigtriangleup ABG$ is $100$. Since $\overline{DE}$ is the midsegment of $\bigtriangleup ABC$, so $\overline{DF}$ is the midsegment of $\bigtriangleup ABG$ . So the area of $\bigtriangleup ADF$ to the area of $\bigtriangleup ABG$ is $1:4$ , so the area of $\bigtriangleup ACG$ is $25$, by similar triangles. Therefore the area of quad $FDBG$ is $100-25=\boxed{75}$

Solution 4

The area of quadrilateral $FDBG$ is the area of $\bigtriangleup ABG$ minus the area of $\bigtriangleup ADF$. Notice, $\overline{DE} || \overline{BC}$, so $\bigtriangleup ABG \sim \bigtriangleup ADF$, and since $\overline{AD}:\overline{AB}=1:2$, the area of $\bigtriangleup ADF:\bigtriangleup ABG=(1:2)^2=1:4$. Given that the area of $\bigtriangleup ABC$ is $120$, using $\frac{bh}{2}$ on side $AB$ yields $\frac{50h}{2}=120\implies h=\frac{240}{50}=\frac{24}{5}$. Using the Angle Bisector Theorem, $\overline{BG}:\overline{BC}=50:(10+50)=5:6$, so the height of $\bigtriangleup ABG: \bigtriangleup ACB=5:6$. Therefore our answer is $\big[ FDBG\big] = \big[ABG\big]-\big[ ADF\big] = \big[ ABG\big]\big(1-\frac{1}{4}\big)=\frac{3}{4}\cdot \frac{bh}{2}=\frac{3}{8}\cdot 50\cdot \frac{5}{6}\cdot \frac{24}{5}=\frac{3}{8}\cdot 200=\boxed{75}$

See Also

2018 AMC 10A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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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