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

Revision as of 11:36, 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 $8-\frac12-\frac92 =\boxed3$ . The answer is (C).

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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Difference between revisions of "2016 AMC 8 Problems/Problem 22"

Revision as of 11:36, 23 November 2016

Solution

@@ Line 5: / Line 5: @@
 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>