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Difference between revisions of "2016 AMC 8 Problems/Problem 22"
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==Solution==
 
==Solution==
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The area of the trapezoid containing the shaded region and the two isosceles triangles 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, so the height of the bigger one is 3, while the height of the smaller one is 1. 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>. The answer is (C).
 
The area of the trapezoid containing the shaded region and the two isosceles triangles 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, so the height of the bigger one is 3, while the height of the smaller one is 1. 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>. The answer is (C).
  

Revision as of 11:35, 23 November 2016

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); [/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 the trapezoid containing the shaded region and the two isosceles triangles 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 bigger 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 $8-\frac12-\frac92 =\boxed3$. The answer is (C).


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

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