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

Latest revision as of 10:22, 24 July 2024

Problem

Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA=1$ . The area of the "bat wings" (shaded area) is

$[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$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 2

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 two non-colored isosceles triangles are similar, and are in a $3:1$ ratio by AA similarity (alternate interior and vertical angles) so the height of the larger 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 $\boxed{\textbf{(C) }3}$

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https://youtu.be/oBzkBYeHFa8

~Education, the Study of Everything

Video Solutions

https://youtu.be/q3MAXwNBkcg ~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/FDgcLW4frg8?t=4448 ~ pi_is_3.14

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.

@@ Line 1: / Line 1: @@
-Rectangle <math>DEFA</math> below is a <math>3 \times 4</math> rectangle with <math>DC=CB=BA</math>. What is the area of the "bat wings" (shaded area)?
+== Problem ==
+Rectangle <math>DEFA</math> below is a <math>3 \times 4</math> rectangle with <math>DC=CB=BA=1</math>. The area of the "bat wings" (shaded area) is
 <asy>
 draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0));
@@ Line 19: / Line 22: @@
 <math>\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</math>
-==Solution==
-The area of trapezoid <math>CBFE</math> is <math>\frac{1+3}2\cdot 4=8</math>. Next, we find the height of each triangle to calculate their area. The triangles are similar, and are in a <math>3:1</math> ratio[SOMEBODY PROVE THIS], so the height of the larger one is <math>3,</math> while the height of the smaller one is <math>1.</math> Thus, their areas are <math>\frac12</math> and <math>\frac92</math>. Subtracting these areas from the trapezoid, we get <math>8-\frac12-\frac92 =\boxed3</math>. Therefore, the answer to this problem is <math>\boxed{\textbf{(C) }3}</math>.
-{{AMC8 box|year=2016|num-b=21|num-a=23}}
-{{MAA Notice}}
 ==Solution 2==
+The area of trapezoid <math>CBFE</math> is <math>\frac{1+3}2\cdot 4=8</math>. Next, we find the height of each triangle to calculate their area. The two non-colored isosceles triangles are similar, and are in a <math>3:1</math> ratio by AA similarity (alternate interior and vertical angles) so the height of the larger is <math>3,</math> while the height of the smaller one is <math>1.</math> Thus, their areas are <math>\frac12</math> and <math>\frac92</math>. Subtracting these areas from the trapezoid, we get <math>8-\frac12-\frac92 =\boxed3</math>. Therefore, the answer to this problem is <math>\boxed{\textbf{(C) }3}</math>
-Setting coordinates!
+==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
+https://youtu.be/oBzkBYeHFa8
-Let <math>E=(0,0)</math>, <math>F=(3,0)</math>
+~Education, the Study of Everything
-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 line <math>CF=-2x+6</math>
-Doing the same process to line <math>BE</math>, we find that line <math>BE=2x</math>.
+==Video Solutions==
-Hence, setting them equal to find the intersection point...
+*https://youtu.be/q3MAXwNBkcg ~savannahsolver
-<math>y=2x=-2x+6\implies 4x=6\implies x=\frac{3}{2}\implies y=3</math>.
+==Video Solution by OmegaLearn==
+https://youtu.be/FDgcLW4frg8?t=4448 ~ pi_is_3.14
-Hence, we find that the intersection point is <math>(\frac{3}{2},3)</math>. Call it Z.
+==See Also==
+{{AMC8 box|year=2016|num-b=21|num-a=23}}
-Now, we can see that
+{{MAA Notice}}
-<math>E=(0,0)</math>
-<math>Z=(\frac{3}{2},3)</math>
-<math>C=(1,4)</math>.
-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 <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>

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