Difference between revisions of "2002 AMC 12P Problems/Problem 19"
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== Solution == | == Solution == | ||
− | Draw <math>AE</math> parallel to <math>BC</math> and draw <math>BF</math> and <math>CG | + | Draw <math>AE</math> parallel to <math>BC</math> and draw <math>BF</math> and <math>CG \perp 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 <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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It remains to find the area of triangle <math>AED</math>, which is <math>(\frac{1}{2})(AE)(ED)(\sin 120^{\circ}) = (\frac{1}{2})(7)(2)(\frac{\sqrt{3}}{2}) = \frac{7\sqrt{3}}{2}</math>. | It remains to find the area of triangle <math>AED</math>, which is <math>(\frac{1}{2})(AE)(ED)(\sin 120^{\circ}) = (\frac{1}{2})(7)(2)(\frac{\sqrt{3}}{2}) = \frac{7\sqrt{3}}{2}</math>. | ||
− | Therefore, the total area of quadrilateral <math>ABCD</math> is <math>\frac{33\sqrt{3}}{4} + \frac{7\sqrt{3}}{2} = \frac{47\sqrt{3}}{4}</math> | + | Therefore, the total area of quadrilateral <math>ABCD</math> is <math>\frac{33\sqrt{3}}{4} + \frac{7\sqrt{3}}{2} = \boxed{\frac{47\sqrt{3}}{4} \text{(D) }}</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}} |
Latest revision as of 19:22, 11 March 2024
Problem
In quadrilateral , and Find the area of
Solution
Draw parallel to and draw and , 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 .
It remains to find the area of triangle , which is .
Therefore, the total area of quadrilateral 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 |
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