Difference between revisions of "2002 AMC 12P Problems/Problem 19"
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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>. | 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>. | ||
| − | 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 | + | 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>. |
== 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 14:30, 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 
.
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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