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2018 AMC 8 Problems/Problem 22

Revision as of 09:35, 23 November 2018 by Frostfox (talk | contribs) (→‎Solution 1.5)

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 AAA 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 1.5

zooming into solution 1 We notice some similar triangles. This is good, but you want to know the ratios in order to begin using them. Say that 45 is the area of half the triangle subtracted by $a$, so then the area of the whole square would be $90+2a$. Similar triangles come into use $here$. $\triangle CEF \sim \triangle ABF$= $1/2$. From this we can deduce 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 $90+2x = 90 + 2 \cdot 9 = \boxed{\textbf{(B)} 108}$. -scrabbler94

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 $\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B)}108}$.

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

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