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

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 $BF = CG = \frac{3\sqrt{3}}{2}$ .

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 $(1/2)(AE)(ED)(

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2002 AMC 12P Problems/Problem 19

Problem

Solution

See also