Difference between revisions of "2002 AMC 10P Problems/Problem 14"
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== Solution 1== | == Solution 1== | ||
| − | Draw a diagram. | + | Draw a diagram. The perpendicular from point <math>E</math> intersects <math>AD</math> at <math>X</math>, while the perpendicular from point <math>E</math> intersects <math>CD</math> at <math>Y.</math> We know Since <math>E</math> is at the center of square <math>ABCD</math>, <math>EX=EY=\frac{1}{2}.</math> |
== See also == | == See also == | ||
{{AMC10 box|year=2002|ab=P|num-b=13|num-a=15}} | {{AMC10 box|year=2002|ab=P|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 07:27, 15 July 2024
Problem 14
The vertex 
of a square 
is at the center of square 
The length of a side of 
is 
and the length of a side of is 
Side 
intersects 
at 
and 
intersects 
at 
If angle 
the area of quadrilateral 
is
Solution 1
Draw a diagram. The perpendicular from point intersects at 
, while the perpendicular from point intersects at 
We know Since is at the center of square , 
See also
| 2002 AMC 10P (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
