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

m
Line 11: Line 11:
 
==Solution==
 
==Solution==
 
{{solution}}
 
{{solution}}
 +
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).
 +
  
 
{{AMC8 box|year=2016|num-b=21|num-a=23}}
 
{{AMC8 box|year=2016|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:04, 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

This problem needs a solution. If you have a solution for it, please help us out by adding it. 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 (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

Invalid username
Login to AoPS