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Difference between revisions of "2003 AMC 10A Problems/Problem 12"

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Therefore, the probability that <math>x<y</math> is <math>\frac{\frac{1}{2}}{4}=\frac{1}{8} \Rightarrow A</math>
 
Therefore, the probability that <math>x<y</math> is <math>\frac{\frac{1}{2}}{4}=\frac{1}{8} \Rightarrow A</math>
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== See Also ==
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*[[2003 AMC 10A Problems]]
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*[[2003 AMC 10A Problems/Problem 11|Previous Problem]]
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*[[2003 AMC 10A Problems/Problem 13|Next Problem]]
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[[Category:Introductory Algebra Problems]]
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[[Category:Introductory Geometry Problems]]

Revision as of 19:15, 4 November 2006

Problem

A point $(x,y)$ is randomly picked from inside the rectangle with vertices $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$. What is the probability that $x<y$?

$\mathrm{(A) \ } \frac{1}{8}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{3}{8}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{3}{4}$

Solution

The rectangle has a width of $4$ and a height of $1$.

The area of this rectangle is $4\cdot1=4$.

The line $x=y$ intersects the rectangle at $(0,0)$ and $(1,1)$.

The area which $x>y$ is the right isosceles triangle with side length $1$ that has vertices at $(0,0)$, $(1,1)$, and $(0,1)$.

The area of this triangle is $\frac{1}{2}\cdot1^{2}=\frac{1}{2}$

Therefore, the probability that $x<y$ is $\frac{\frac{1}{2}}{4}=\frac{1}{8} \Rightarrow A$

See Also

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