# Difference between revisions of "2016 AMC 8 Problems/Problem 22"

Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA$. What is the area of the "bat wings" (shaded area)? $[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("A",(3.05,4.2)); label("B",(2,4.2)); label("C",(1,4.2)); label("D",(0,4.2)); label("E", (0,-0.2)); label("F", (3,-0.2)); label("1", (0.5, 4), N); label("1", (1.5, 4), N); label("1", (2.5, 4), N); label("4", (3.2, 2), E); [/asy]$

$\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }5$

## Solution

The area of trapezoid $CBFE$ is $\frac{1+3}2\cdot 4=8$. Next, we find the height of each triangle to calculate their area. The triangles are similar, and are in a $3:1$ ratio[SOMEBODY PROVE THIS], so the height of the larger one is $3,$ while the height of the smaller one is $1.$ Thus, their areas are $\frac12$ and $\frac92$. Subtracting these areas from the trapezoid, we get $8-\frac12-\frac92 =\boxed3$. Therefore, the answer to this problem is $\boxed{\textbf{(C) }3}$

## Solution 2

Setting coordinates!

Let $E=(0,0)$, $F=(3,0)$

$[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("A(3,4)",(3.05,4.2)); label("B(2,4)",(2,4.2)); label("C(1,4)",(1,4.2)); label("D(0,4)",(0,4.2)); label("E(0,0)", (0,-0.2)); label("Z(\frac{3}{2},3)", (1.5,1.8)); label("F(3,0)", (3,-0.2)); label("1", (0.5, 4), N); label("1", (1.5, 4), N); label("1", (2.5, 4), N); label("4", (3.2, 2), E); [/asy]$

Now, we easily discover that line $CF$ has lattice coordinates at $(1,4)$ and $(3,0)$. Hence, the slope of line $CF=-2$

Plugging in the rest of the coordinate points, we find that line $CF=-2x+6$

Doing the same process to line $BE$, we find that line $BE=2x$.

Hence, setting them equal to find the intersection point...

$y=2x=-2x+6\implies 4x=6\implies x=\frac{3}{2}\implies y=3$.

Hence, we find that the intersection point is $(\frac{3}{2},3)$. Call it Z.

Now, we can see that

$E=(0,0)$

$Z=(\frac{3}{2},3)$

$C=(1,4)$.

Shoelace!

Using the well known Shoelace Formula(https://en.m.wikipedia.org/wiki/Shoelace_formula), we find that the area of one of those small shaded triangles is $\frac{3}{2}$.

Now because there are two of them, we multiple that area by $2$ to get $\boxed{\textbf{(C) }3}$

 2016 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 21 Followed byProblem 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