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 \boxed{\mathrm{(A)}\ \frac{1}{8}}</math> | Therefore, the probability that <math>x<y</math> is <math>\frac{\frac{1}{2}}{4}=\frac{1}{8} \Rightarrow \boxed{\mathrm{(A)}\ \frac{1}{8}}</math> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/JueaMdaNRrs | ||
| + | |||
| + | ~savannahsolver | ||
== See Also == | == See Also == | ||
Revision as of 15:21, 23 June 2021
Contents
Problem
A point 
is randomly picked from inside the rectangle with vertices 
, 
, 
, and 
. What is the probability that 
?
Solution
The rectangle has a width of 
and a height of 
.
The area of this rectangle is 
.
The line 
intersects the rectangle at and 
.
The area which is the right isosceles triangle with side length that has vertices at , , and .
The area of this triangle is 
Therefore, the probability that is 
Video Solution
~savannahsolver
See Also
| 2003 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
