Difference between revisions of "2002 AMC 12P Problems/Problem 19"
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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>. | 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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| + | 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})</math>. | ||
== 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}} | ||
Revision as of 14:33, 10 March 2024
Problem
In quadrilateral 
, 
and 
Find the area of 
Solution
Draw 
parallel to 
and draw 
and 
perpendicular to , where 
and 
are on .
It is clear that triangles 
and 
are congruent 30-60-90 triangles. Therefore, 
and 
.
Therefore, 
and the area of trapezoid 
is 
.
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
