Difference between revisions of "2002 AMC 12P Problems/Problem 19"
The 76923th (talk | contribs) m (→Solution) |
The 76923th (talk | contribs) m (→Solution) |
||
| Line 15: | Line 15: | ||
== 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>. | ||
Revision as of 18:22, 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
