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

m (Solution: I don’t get this)
(Solution 2)
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==Solution 2==
 
==Solution 2==
 +
 +
Setting coordinates!
 +
 
Let <math>E=(0,0)</math>, <math>F=(3,0)</math>
 
Let <math>E=(0,0)</math>, <math>F=(3,0)</math>
  
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>  
+
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=-2</math>  
  
Plugging in the rest of the coordinate points, we find that <math>CF=-2x+6</math>
+
Plugging in the rest of the coordinate points, we find that line <math>CF=-2x+6</math>
  
 
Doing the same process to line <math>BE</math>, we find that line <math>BE=2x</math>.
 
Doing the same process to line <math>BE</math>, we find that line <math>BE=2x</math>.
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Hence, setting them equal to find the intersection point...
 
Hence, setting them equal to find the intersection point...
  
<math>2x=-2x+6\implies 4x=6\implies x=\frac{3}{2}</math>.
+
<math>y=2x=-2x+6\implies 4x=6\implies x=\frac{3}{2}\implies y=3</math>.
  
 
Hence, we find that the intersection point is <math>(\frac{3}{2},3)</math>. Call it Z.
 
Hence, we find that the intersection point is <math>(\frac{3}{2},3)</math>. Call it Z.
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Shoelace!
 
Shoelace!
  
We find that the area of one of those small shaded triangles is <math>\frac{3}{2}</math>
+
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 <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>
 
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 20:23, 8 January 2018

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

2016 AMC 8 (ProblemsAnswer KeyResources)
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

Setting coordinates!

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 $CF=-2$

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

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