Difference between revisions of "2003 AMC 10A Problems/Problem 12"
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== Problem == | == Problem == | ||
| − | A point <math>(x,y)</math> is randomly picked from inside the rectangle with vertices <math>(0,0)</math>, <math>(4,0)</math>, <math>(4,1)</math>, and <math>(0,1)</math>. What is the probability that <math>x | + | A point <math>(x,y)</math> is randomly picked from inside the rectangle with vertices <math>(0,0)</math>, <math>(4,0)</math>, <math>(4,1)</math>, and <math>(0,1)</math>. What is the probability that <math>x<y</math>? |
<math> \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} </math> | <math> \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} </math> | ||
Revision as of 22:32, 21 August 2009
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 
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 | ||
