Difference between revisions of "2018 AMC 8 Problems/Problem 22"
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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}} | {{MAA Notice}} | ||
Revision as of 15:32, 21 November 2018
Problem 22
Point 
is the midpoint of side 
in square 
and 
meets diagonal 
at 
The area of quadrilateral 
is 
What is the area of 
![[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]](http://latex.artofproblemsolving.com/7/c/1/7c13c3d233bfe4e1b9a7ecd2e6343fe5b8aa2e3a.png)
| 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 
. Notice that 
and 
are similar with ratio 
. Than the altitude from 
to
has length of 
. Then ![$[ABF]=\frac{2x^2}{6}=\frac{x^2}{3}$](http://latex.artofproblemsolving.com/b/9/7/b97a7937be1346f6459aa87e61900642b5564c49.png)
. Also ![$[BEC]=\frac{\frac{x}{2}\cdot x}{2}=\frac{x^2}{4}$](http://latex.artofproblemsolving.com/c/0/a/c0a583566dbe5d2bb17ef0d1b44bf2ab13c9db14.png)
. That means ![$[AFED]=x^2-\frac{x^2}{3}-\frac{x^2}{4}=\frac{5}{12}x$](http://latex.artofproblemsolving.com/6/5/a/65a443d718da2bd4e04ecf1ed78cf08615634422.png)
. So ![$[ABCD]=45\cdot\frac{12}{5}=108$](http://latex.artofproblemsolving.com/1/f/4/1f46e5611583b79938abb60f27003cf040ff3fd7.png)
or B
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
