2016 AMC 8 Problems/Problem 22

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

~23orimy412uc3478

Solution 2

Plot the point $G$ where the two "wings" intersect. Now notice how $\triangle CBG\sim\triangle EFG$ . Since the length of $\overline {CB}$ is one third that of $\overline {EF}$ , then that means $\triangle EFG$ 's height is $3$ times bigger than $\triangle CBG$ . Since both of their heights ( $h$ ) add up to four, then we have the equation $3h+h=4 \implies h=1$ . Now that we now the height and length of both triangles, we can use complementary counting, $\text{Area}-\text{Unshaded Region}$ .

$\text {Total Area}=12$

$[\triangle CDE]=2$

$[\triangle ABF]=2$

$[\triangle CBG]=\frac1{2}$

$[\triangle EFG]=\frac{9}{2}$

$\text {Unshaded Region}=9\implies\text{"Bat Wings"}=\boxed{\textbf{(C) }3}$

~AM24

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

Video Solution only problem 22's by SpreadTheMathLove

https://www.youtube.com/watch?v=sOF1Okc0jMc

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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