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

Revision as of 15:32, 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 (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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 AJHSME/AMC 8 Problems and Solutions

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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>
 {{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}}

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

Revision as of 15:32, 21 November 2018

Problem 22