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2002 AMC 12P Problems/Problem 19

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
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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