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Difference between revisions of "2015 AMC 8 Problems/Problem 11"

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\textbf{(E) } \frac{1}{1,050}
 
\textbf{(E) } \frac{1}{1,050}
 
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</math>
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The area of <math>\triangle ABC</math> is equal to half the product of its base and height.  By the Pythagorean Theorem, we find its height is <math>\sqrt{1^2+2^2}=\sqrt{5}</math>, and its base is <math>\sqrt{2^2+4^2}=\sqrt{20}</math>.  We multiply these and divide by 2 to find the of the triangle is <math>\frac{\sqrt{5 \cdot 20}}2=\frac{\sqrt{100}}2=\frac{10}2=5</math>.  Since the grid has an area of <math>30</math>, the fraction of the grid covered by the triangle is <math>\frac 5{30}=\boxed{\textbf{(A) }\frac{1}{6}}</math>.
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==See Also==
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{{AMC8 box|year=2015|num-b=10|num-a=12}}
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{{MAA Notice}}

Revision as of 16:23, 25 November 2015

In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the 21 non-vowels, and the fourth must be a digit (0 through 9). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read "AMC8"?

$\textbf{(A) } \frac{1}{22,050} \qquad \textbf{(B) } \frac{1}{21,000}\qquad \textbf{(C) } \frac{1}{10,500}\qquad \textbf{(D) } \frac{1}{2,100} \qquad \textbf{(E) } \frac{1}{1,050}$

The area of $\triangle ABC$ is equal to half the product of its base and height. By the Pythagorean Theorem, we find its height is $\sqrt{1^2+2^2}=\sqrt{5}$, and its base is $\sqrt{2^2+4^2}=\sqrt{20}$. We multiply these and divide by 2 to find the of the triangle is $\frac{\sqrt{5 \cdot 20}}2=\frac{\sqrt{100}}2=\frac{10}2=5$. Since the grid has an area of $30$, the fraction of the grid covered by the triangle is $\frac 5{30}=\boxed{\textbf{(A) }\frac{1}{6}}$.

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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

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