2018 AMC 8 Problems/Problem 22

Problem 22

Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$

$[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label("$A$",(0,6),NW); label("$B$",(6,6),NE); label("$C$",(6,0),SE); label("$D$",(0,0),SW); label("$E$",(3,0),S); label("$F$",(4,2),E); [/asy]$

$\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144$

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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All AJHSME/AMC 8 Problems and Solutions

Solution

Let the sidelength of the square be x. Then $\overline{EC}=\frac{x}{2}$ . Notice that $\triangle{ABF}$ and $\triangle{CEF}$ are similar with ratio $2:1$ . Than the altitude from $F$ to $\overline{AB}$ has length of $\frac{2x}{3}$ . Then $[ABF]=\frac{2x^2}{6}=\frac{x^2}{3}$ . Also $[BEC]=\frac{\frac{x}{2}\cdot x}{2}=\frac{x^2}{4}$ . That means $[AFED]=x^2-\frac{x^2}{3}-\frac{x^2}{4}=\frac{5}{12}x$ . So $[ABCD]=45\cdot\frac{12}{5}=108$ or B The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2018 AMC 8 Problems/Problem 22

Problem 22

Solution