Difference between revisions of "2018 AMC 8 Problems/Problem 22"

(Problem 22)
(Solution)
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<math>\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144</math>
 
<math>\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144</math>
 
{{AMC8 box|year=2018|num-b=21|num-a=23}}
 
{{AMC8 box|year=2018|num-b=21|num-a=23}}
==Solution==
 
Let the sidelength of the square be x. Then <math>\overline{EC}=\frac{x}{2}</math>. Notice that <math>\triangle{ABF}</math> and <math>\triangle{CEF}</math> are similar with ratio <math>2:1</math>. Than the altitude from <math>F</math> to<math>\overline{AB}</math> has length of <math>\frac{2x}{3}</math>. Then <math>[ABF]=\frac{2x^2}{6}=\frac{x^2}{3}</math>. Also <math>[BEC]=\frac{\frac{x}{2}\cdot x}{2}=\frac{x^2}{4}</math>. That means <math>[AFED]=x^2-\frac{x^2}{3}-\frac{x^2}{4}=\frac{5}{12}x</math>. So <math>[ABCD]=45\cdot\frac{12}{5}=108</math> or B
 
{{MAA Notice}}
 

Revision as of 16:32, 21 November 2018

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 (ProblemsAnswer KeyResources)
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Problem 21
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Problem 23
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