Difference between revisions of "2008 AMC 8 Problems/Problem 4"

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</asy>
 
</asy>
 
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad  \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7</math>
 
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad  \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7</math>
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==Solution==
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The area outside the small triangle but inside the large triangle is <math>16-1=15</math>. This is equally distributed between the three trapezoids. Each trapezoid has an area of <math>15/3 = \boxed{\textbf{(C)}\ 5}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2008|num-b=3|num-a=5}}
 
{{AMC8 box|year=2008|num-b=3|num-a=5}}

Revision as of 21:58, 24 December 2012

Problem

In the figure, the outer equilateral triangle has area $16$, the inner equilateral triangle has area $1$, and the three trapezoids are congruent. What is the area of one of the trapezoids? [asy] size((70)); draw((0,0)--(7.5,13)--(15,0)--(0,0)); draw((1.88,3.25)--(9.45,3.25)); draw((11.2,0)--(7.5,6.5)); draw((9.4,9.7)--(5.6,3.25)); [/asy] $\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad  \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$

Solution

The area outside the small triangle but inside the large triangle is $16-1=15$. This is equally distributed between the three trapezoids. Each trapezoid has an area of $15/3 = \boxed{\textbf{(C)}\ 5}$.

See Also

2008 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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