2002 AMC 10P Problems/Problem 14

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. The perpendicular from point $E$ intersects $AD$ at $X$ , while the perpendicular from point $E$ intersects $CD$ at $Y.$ We know Since $E$ is at the center of square $ABCD$ , $EX=EY=\frac{1}{2}.$

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.

Page

Toolbox

Search

2002 AMC 10P Problems/Problem 14

Problem 14

Solution 1

See also