Difference between revisions of "2002 AMC 10P Problems/Problem 14"

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== Solution 1==
 
== Solution 1==
 
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Draw a diagram. Let the perpendicular from point <math>E</math> to <math>AD</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 08:25, 15 July 2024

Problem 14

The vertex $E$ of a square $EFGH$ is at the center of square $ABCD.$ The length of a side of $ABCD$ is $1$ and the length of a side of $EFGH$ is $2.$ Side $EF$ intersects $CD$ at $I$ and $EH$ intersects $AD$ at $J.$ If angle $EID=60^{\circ},$ the area of quadrilateral $EIDJ$ is

$\text{(A) }\frac{1}{4} \qquad \text{(B) }\frac{\sqrt{3}}{6} \qquad \text{(C) }\frac{1}{3} \qquad \text{(D) }\frac{\sqrt{2}}{4} \qquad \text{(E) }\frac{\sqrt{3}}{2}$

Solution 1

Draw a diagram. Let the perpendicular from point $E$ to $AD$

See also

2002 AMC 10P (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 10 Problems and Solutions

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