2002 AMC 12P Problems/Problem 19

Revision as of 14:30, 10 March 2024 by The 76923th (talk | contribs) (Solution)

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

Draw $AE$ parallel to $BC$ and draw $BF$ and $CG$ perpendicular to $AE$, 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 $BF = CG = \frac{3\sqrt{3}}{2}$.

See also

2002 AMC 12P (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AMC 12 Problems and Solutions

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