Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3)
(I just added another solution =))
Line 34: Line 34:
 
==Solution 3==
 
==Solution 3==
 
Note that triangle <math>ABC</math> has half the area of the square and triangle <math>FEC</math> has <math>\dfrac1{12}</math>th. Thus the area of the quadrilateral is <math>1-1/2-1/12=5/12</math>th the area of the square. The area of the square is then <math>45\cdot\dfrac{12}{5}=\boxed{\textbf{(B.)}108}</math>.
 
Note that triangle <math>ABC</math> has half the area of the square and triangle <math>FEC</math> has <math>\dfrac1{12}</math>th. Thus the area of the quadrilateral is <math>1-1/2-1/12=5/12</math>th the area of the square. The area of the square is then <math>45\cdot\dfrac{12}{5}=\boxed{\textbf{(B.)}108}</math>.
 +
 +
==Solution 4==
 +
Extend <math>\overline{AD}</math> and <math>\overline{BE}</math> to meet at <math>X</math>. Drop an altitude from <math>F</math> to <math>\overline{CE}</math> and call it <math>h</math>. Also, call <math>\overline{CE}</math> <math>x</math>. As stated before, we have <math>\triangle ABF \sim \triangle CEF</math>, so the ratio of their heights is in a <math>1:2</math> ratio, making the altitude from <math>F</math> to <math>\overline{AB}</math> <math>2h</math>. Note that this means that the side of the square is <math>3h</math>. In addition, <math>\triangle XDE \sim \triangle XAB</math> by AA Similarity in a <math>1:2</math> ratio. This means that the side length of the square is <math>2x</math>, making <math>3h=2x</math>.
 +
 +
Now, note that <math>[ADEF]=[XAB]-[XDE]-[ABF]</math>. We have <math>[\triangle XAB]=(4x)(2x)/2=4x^2,</math> <math>[\triangle XDE]=(x)(2x)/2=x^2,</math> and <math>[\triangle ABF]=(2x)(2h)/2=(2x)(4x/3)/2=4x^2/3.</math> Subtracting makes <math>[ADEF]=4x^2-x^2-4x^2/3=5x^2/3.</math> We are given that <math>[ADEF]=45,</math> so <math>5x^2/3=45 \Rightarrow x^2=27.</math> Therefore, <math>x= 3 \sqrt{3},</math> so our answer is <math>(2x)^2=4x^2=4(27)=\boxed{\textbf{(B) }108}.</math> - moony_eyed
  
 
==Video Solution==
 
==Video Solution==

Revision as of 21:04, 25 July 2020

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]

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

Solution 1

Let the area of $\triangle CEF$ be $x$. Thus, the area of triangle $\triangle ACD$ is $45+x$ and the area of the square is $2(45+x) = 90+2x$.

By AA similarity, $\triangle CEF \sim \triangle ABF$ with a 1:2 ratio, so the area of triangle $\triangle ABF$ is $4x$. Now consider trapezoid $ABED$. Its area is $45+4x$, which is three-fourths the area of the square. We set up an equation in $x$:

\[45+4x = \frac{3}{4}\left(90+2x\right)\] Solving, we get $x = 9$. The area of square $ABCD$ is $90+2x = 90 + 2 \cdot 9 = \boxed{\textbf{(B)} 108}$.

Solution 2

We can use analytic geometry for this problem.

Let us start by giving $D$ the coordinate $(0,0)$, $A$ the coordinate $(0,1)$, and so forth. $\overline{AC}$ and $\overline{EB}$ can be represented by the equations $y=-x+1$ and $y=2x-1$, respectively. Solving for their intersection gives point $F$ coordinates $\left(\frac{2}{3},\frac{1}{3}\right)$.

Now, $\triangle$$EFC$’s area is simply $\frac{\frac{1}{2}\cdot\frac{1}{3}}{2}$ or $\frac{1}{12}$. This means that pentagon $ABCEF$’s area is $\frac{1}{2}+\frac{1}{12}=\frac{7}{12}$ of the entire square, and it follows that quadrilateral $AFED$’s area is $\frac{5}{12}$ of the square.

The area of the square is then $\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B)}108}$.

Solution 3

Note that triangle $ABC$ has half the area of the square and triangle $FEC$ has $\dfrac1{12}$th. Thus the area of the quadrilateral is $1-1/2-1/12=5/12$th the area of the square. The area of the square is then $45\cdot\dfrac{12}{5}=\boxed{\textbf{(B.)}108}$.

Solution 4

Extend $\overline{AD}$ and $\overline{BE}$ to meet at $X$. Drop an altitude from $F$ to $\overline{CE}$ and call it $h$. Also, call $\overline{CE}$ $x$. As stated before, we have $\triangle ABF \sim \triangle CEF$, so the ratio of their heights is in a $1:2$ ratio, making the altitude from $F$ to $\overline{AB}$ $2h$. Note that this means that the side of the square is $3h$. In addition, $\triangle XDE \sim \triangle XAB$ by AA Similarity in a $1:2$ ratio. This means that the side length of the square is $2x$, making $3h=2x$.

Now, note that $[ADEF]=[XAB]-[XDE]-[ABF]$. We have $[\triangle XAB]=(4x)(2x)/2=4x^2,$ $[\triangle XDE]=(x)(2x)/2=x^2,$ and $[\triangle ABF]=(2x)(2h)/2=(2x)(4x/3)/2=4x^2/3.$ Subtracting makes $[ADEF]=4x^2-x^2-4x^2/3=5x^2/3.$ We are given that $[ADEF]=45,$ so $5x^2/3=45 \Rightarrow x^2=27.$ Therefore, $x= 3 \sqrt{3},$ so our answer is $(2x)^2=4x^2=4(27)=\boxed{\textbf{(B) }108}.$ - moony_eyed

Video Solution

https://www.youtube.com/watch?v=c4_-h7DsZFg - Happytwin

See Also

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

Set s to be the bottom left triangle. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_22&oldid=129303"