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

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{{AMC8 box|year=2016|num-b=21|num-a=23}}
 
{{AMC8 box|year=2016|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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==Solution 2==
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Let <math>E=(0,0)</math>, <math>F=(3,0)</math>
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Now, we easily discover that line <math>CF</math> has lattice coordinates at <math>(1,4)</math> and <math>(3,0)</math>. Hence, the slope of line <math>CF=-2x</math>
 +
 +
Plugging in the rest of the coordinate points, we find that <math>CF=-2x+6</math>
 +
 +
Doing the same process to line <math>BE</math>, we find that line <math>BE=2x</math>.
 +
 +
Hence, setting them equal to find the intersection point...
 +
 +
<math>2x=-2x+6\implies 4x=6\implies x=\frac{3}{2}</math>.
 +
 +
Hence, we find that the intersection point is <math>(\frac{3}{2},3)</math>. Call it Z.
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Now, we can see that
 +
 +
<math>E=(0,0)</math>
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 +
<math>Z=(\frac{3}{2},3)</math>
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 +
<math>C=(1,4)</math>.
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 +
Shoelace!
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We find that the area of one of those small shaded triangles is <math>\frac{3}{2}</math>
 +
 +
Now because there are two of them, we multiple that area by <math>2</math> to get <math>\boxed{\textbf{(C) }3}</math>

Revision as of 11:13, 13 August 2017

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, 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 is $\boxed{\textbf{(C) }3}$.

2016 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Solution 2

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

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

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

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

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

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

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!

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 $\boxed{\textbf{(C) }3}$

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