Difference between revisions of "2016 AMC 8 Problems/Problem 22"
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| − | Rectangle <math>DEFA</math> below is a <math>3 \times 4</math> rectangle with <math>DC=CB=BA</math>. | + | == 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> | <asy> | ||
draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); | draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); | ||
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<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> | <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> | ||
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| + | ==Solution 1== | ||
| + | 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> | ||
| + | |||
| + | ==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 | ||
| + | ==See Also== | ||
{{AMC8 box|year=2016|num-b=21|num-a=23}} | {{AMC8 box|year=2016|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 15:22, 31 August 2024
Contents
Problem
Rectangle 
below is a 
rectangle with 
. 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]](http://latex.artofproblemsolving.com/a/0/7/a07a9e12af7228a9f783f34687384df0042b721e.png)
Solution 1
The area of trapezoid 
is 
. Next, we find the height of each triangle to calculate their area. The two non-colored isosceles triangles are similar, and are in a 
ratio by AA similarity (alternate interior and vertical angles) so the height of the larger is 
while the height of the smaller one is 
Thus, their areas are 
and 
. Subtracting these areas from the trapezoid, we get 
. Therefore, the answer to this problem is 
Video Solution (CREATIVE THINKING + ANALYSIS!!!)
~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
See Also
| 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. 
