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Difference between revisions of "2002 AMC 12P Problems/Problem 19"

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== Solution ==
 
== Solution ==
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Draw <math>AE</math> parallel to <math>BC</math> and draw <math>BF</math> and <math>CG</math> perpendicular to <math>AE</math>, where <math>F</math> and <math>G</math> are on <math>AE</math>.
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=== Solution 1 ===
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Draw <math>AE</math> parallel to <math>BC</math> and draw <math>BF</math> and <math>CG \perp AE</math>, where <math>F</math> and <math>G</math> are on <math>AE</math>.
  
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It is clear that triangles <math>AFB</math> and <math>EGC</math> are congruent 30-60-90 triangles. Therefore, <math>AF = EG = \frac{3}{2}</math> and <math>BF = CG = \frac{3\sqrt{3}}{2}</math>.
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It is clear that triangles <math>AFB</math> and <math>EGC</math> are congruent <math>30-60-90</math> triangles. Therefore, <math>AF = EG = \frac{3}{2}</math> and <math>BF = CG = \frac{3\sqrt{3}}{2}</math>.
  
 
Therefore, <math>AE = AF + FG + GC = 4 + (2)(\frac{3}{2}) = 7</math> and the area of trapezoid <math>ABCE</math> is <math>(\frac{1}{2})(4+7)(\frac{3\sqrt{3}}{2}) = \frac{33\sqrt{3}}{4}</math>.
 
Therefore, <math>AE = AF + FG + GC = 4 + (2)(\frac{3}{2}) = 7</math> and the area of trapezoid <math>ABCE</math> is <math>(\frac{1}{2})(4+7)(\frac{3\sqrt{3}}{2}) = \frac{33\sqrt{3}}{4}</math>.
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It remains to find the area of triangle <math>AED</math>, which is <math>(\frac{1}{2})(AE)(ED)(\sin 120^{\circ}) = (\frac{1}{2})(7)(2)(\frac{\sqrt{3}}{2}) = \frac{7\sqrt{3}}{2}</math>.
 
It remains to find the area of triangle <math>AED</math>, which is <math>(\frac{1}{2})(AE)(ED)(\sin 120^{\circ}) = (\frac{1}{2})(7)(2)(\frac{\sqrt{3}}{2}) = \frac{7\sqrt{3}}{2}</math>.
  
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Therefore, the total area of quadrilateral <math>ABCD</math> is <math>\frac{33\sqrt{3}}{4} + \frac{7\sqrt{3}}{2} = \boxed{\frac{47\sqrt{3}}{4} \text{(D) }}</math>.
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Therefore, the total area of quadrilateral <math>ABCD</math> is <math>\frac{33\sqrt{3}}{4} + \frac{7\sqrt{3}}{2} = \boxed{\textbf{(D) }\frac{47\sqrt{3}}{4}}</math>.
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=== Solution 2 ===
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Extend <math>AB</math> and <math>CD</math> to meet at <math>E</math>. Then triangle <math>BEC</math> is equilateral, as <math>\angle EBC = \angle ECB = 60^{\circ}</math>. Thus, <math>BC = CE = EB = 4</math>.
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Note <math>AE = AB + BE = 7</math> and <math>DE = DC + CE = 9</math> Then, the area of triangle <math>ADE</math> is <math>\frac {1} {2} \cdot AE \cdot DE \cdot \sin{60^{\circ}} = \frac {63\sqrt{3}} {4}</math>. As triangle <math>BEC</math> is equilateral, its area is <math>\frac {\sqrt{3}} {4} \cdot 4^2 = \frac {16\sqrt{3}} {4}</math>.
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The area of <math>ABCD</math> is the area of triangle <math>ADE</math> minus the area of triangle <math>BEC</math>, or <math>\boxed{\textbf{(D) }\frac{47\sqrt{3}}{4}}</math>.
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~Michw08
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=P|num-b=18|num-a=20}}
 
{{AMC12 box|year=2002|ab=P|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:11, 22 August 2024

Problem

In quadrilateral $ABCD$, $m\angle B = m \angle C = 120^{\circ}, AB=3, BC=4,$ and $CD=5.$ Find the area of $ABCD.$

$\text{(A) }15 \qquad \text{(B) }9 \sqrt{3} \qquad \text{(C) }\frac{45 \sqrt{3}}{4} \qquad \text{(D) }\frac{47 \sqrt{3}}{4} \qquad \text{(E) }15 \sqrt{3}$

Solution

Solution 1

Draw $AE$ parallel to $BC$ and draw $BF$ and $CG \perp AE$, where $F$ and $G$ are on $AE$.

It is clear that triangles $AFB$ and $EGC$ are congruent $30-60-90$ triangles. Therefore, $AF = EG = \frac{3}{2}$ and $BF = CG = \frac{3\sqrt{3}}{2}$.

Therefore, $AE = AF + FG + GC = 4 + (2)(\frac{3}{2}) = 7$ and the area of trapezoid $ABCE$ is $(\frac{1}{2})(4+7)(\frac{3\sqrt{3}}{2}) = \frac{33\sqrt{3}}{4}$.

It remains to find the area of triangle $AED$, which is $(\frac{1}{2})(AE)(ED)(\sin 120^{\circ}) = (\frac{1}{2})(7)(2)(\frac{\sqrt{3}}{2}) = \frac{7\sqrt{3}}{2}$.

Therefore, the total area of quadrilateral $ABCD$ is $\frac{33\sqrt{3}}{4} + \frac{7\sqrt{3}}{2} = \boxed{\textbf{(D) }\frac{47\sqrt{3}}{4}}$.

Solution 2

Extend $AB$ and $CD$ to meet at $E$. Then triangle $BEC$ is equilateral, as $\angle EBC = \angle ECB = 60^{\circ}$. Thus, $BC = CE = EB = 4$.

Note $AE = AB + BE = 7$ and $DE = DC + CE = 9$ Then, the area of triangle $ADE$ is $\frac {1} {2} \cdot AE \cdot DE \cdot \sin{60^{\circ}} = \frac {63\sqrt{3}} {4}$. As triangle $BEC$ is equilateral, its area is $\frac {\sqrt{3}} {4} \cdot 4^2 = \frac {16\sqrt{3}} {4}$.

The area of $ABCD$ is the area of triangle $ADE$ minus the area of triangle $BEC$, or $\boxed{\textbf{(D) }\frac{47\sqrt{3}}{4}}$.

~Michw08

See also

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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All AMC 12 Problems and Solutions

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