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2016 AMC 8 Problems/Problem 22

Revision as of 19:45, 11 November 2019 by Xxsc (talk | contribs) (→‎Solution 2)

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 1

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 by AA similarity (alternate interior and vertical angles) 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(scale(0.7)*"$A(3,4)$",(3.25,4.2)); label(scale(0.7)*"$B(2,4)$",(2.1,4.2)); label(scale(0.7)*"$C(1,4)$",(0.9,4.2)); label(scale(0.7)*"$D(0,4)$",(-0.3,4.2)); label(scale(0.7)*"$E(0,0)$", (0,-0.2)); label(scale(0.7)*"$Z(\frac{3}{2},3)$", (1.5,1.8)); label(scale(0.7)*"$F(3,0)$", (3,-0.2)); label(scale(0.7)*"$1$", (0.3, 4), N); label(scale(0.7)*"$1$", (1.5, 4), N); label(scale(0.7)*"$1$", (2.7, 4), N); label(scale(0.7)*"$4$", (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 [url=Shoelace Formula]https://en.m.wikipedia.org/wiki/Shoelace_formula[/url], 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 by
Problem 21 		Followed by
Problem 23
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All AJHSME/AMC 8 Problems and Solutions

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