Difference between revisions of "2002 AMC 10P Problems/Problem 14"
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| + | == Problem 14 == | ||
| + | |||
| + | The vertex <math>E</math> of a square <math>EFGH</math> is at the center of square <math>ABCD.</math> The length of a side of <math>ABCD</math> is <math>1</math> and the length of a side of <math>EFGH</math> is <math>2.</math> Side <math>EF</math> intersects <math>CD</math> at <math>I</math> and <math>EH</math> intersects <math>AD</math> at <math>J.</math> If angle <math>EID=60^{\circ},</math> the area of quadrilateral <math>EIDJ</math> is | ||
| + | |||
| + | <math> | ||
| + | \text{(A) }\frac{1}{4} | ||
| + | \qquad | ||
| + | \text{(B) }\frac{\sqrt{3}}{6} | ||
| + | \qquad | ||
| + | \text{(C) }\frac{1}{3} | ||
| + | \qquad | ||
| + | \text{(D) }\frac{\sqrt{2}}{4} | ||
| + | \qquad | ||
| + | \text{(E) }\frac{\sqrt{3}}{2} | ||
| + | </math> | ||
| + | |||
== Solution 1== | == Solution 1== | ||
Latest revision as of 18:49, 14 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
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 | ||
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