2002 AMC 10P Problems/Problem 14

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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. Split quadrilateral $EIDJ$ into $\triangle EIJ$ and $\triangle JDI.$ Let the perpendicular from point $E$ intersect $AD$ at $X$, and let the perpendicular from point $E$ intersect $CD$ at $Y.$ We know $\angle EJD=120^{\circ}$ because $\angle JDC=90^{\circ}$ since $ABCD$ is a square, $\angle DCE=60^{\circ}$ as given, and $\angle CEJ = 90^{\circ},$ so $\angle EJD = 360^{\circ}-120^{\circ}-90^{\circ}-90^{\circ}-60^{\circ}=120^{\circ}.$ Since $E$ is at the center of square $ABCD$, $EX=EY=\frac{1}{2}.$ By $30^{\circ}-60^{\circ}-90^{\circ},$ $ED=\frac{EX}{\sqrt{3}}=EC=\frac{EY}{\sqrt{3}}=\frac{1}{3}.$ Additionally, we know $JD=AD-AX-XJ,$ so $JD=1-\frac{1}{2}-\frac{1}{2 \sqrt{3}}=\frac{1}{2}-\frac{1}{2 \sqrt{3}}$ and we know $ID=DY+IY,$ so $ID=\frac{1}{2}+\frac{1}{2 \sqrt{3}}.$ From here, we can sum the areas of $\triangle EIJ$ and $\triangle JDI.$ to get the area of quadrilateral $EIDJ.$ Therefore,

[EIDJ]=[EIJ]+[JDI]=12(13)(13)+12(12123)(12+123)=12(13)+12(14112)=16+112=14

Thus, our answer is $\boxed{\textbf{(A) } \frac{1}{4}}.$

Solution 2

If we draw a diagram as explained by the prompt, $\angle EJD = 360^{\circ}-90^{\circ}-90^{\circ}-60^{\circ} = 120^{\circ}$ as $\angle IEJ$ and $\angle JDI$ are both $90^{\circ}$ because they are angles of the squares, and $\angle EID=60^{\circ}$ given by the question. Similar to solution 1, if we draw $EX$ and $EY$ perpendicular to $AD$ and $CD$ respectively, square $EXDY$ with a side length of $\frac{1}{2}$ will be formed. Looking at $\triangle JXE$, we will notice $\angle EJX = 180^{\circ}-120^{\circ}=60^{\circ}$. Therefore $\triangle EYI$ and $\triangle EXJ$ are congruent by $A.A.S.$ as they both have a $90^{\circ}$ and a $60^{\circ}$ angle and $EX=EY$. Thus, the area of quadrilateral $EIDJ$ is the same as the area of square $EXDY$, which equals $\frac{1}{2} \times \frac{1}{2} = \boxed{\textbf{(A) } \frac{1}{4}}.$

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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