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

m (Solution)
Line 5: Line 5:
 
fill((0,0)--(1,4)--(1.5,3)--cycle, black);
 
fill((0,0)--(1,4)--(1.5,3)--cycle, black);
 
fill((3,0)--(2,4)--(1.5,3)--cycle, black);
 
fill((3,0)--(2,4)--(1.5,3)--cycle, black);
label("A",(3.05,4.2));
+
label("$A$",(3.05,4.2));
label("B",(2,4.2));
+
label("$B$",(2,4.2));
label("C",(1,4.2));
+
label("$C$",(1,4.2));
label("D",(0,4.2));
+
label("$D$",(0,4.2));
label("E", (0,-0.2));
+
label("$E$", (0,-0.2));
label("F", (3,-0.2));
+
label("$F$", (3,-0.2));
 
</asy>
 
</asy>
  

Revision as of 11:39, 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); 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)); [/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 $8-\frac12-\frac92 =\boxed3$. Therefore, the answer is $\boxed{(C) 3}$.


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

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