Difference between revisions of "2003 AMC 10A Problems/Problem 12"
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The line <math>x=y</math> intersects the rectangle at <math>(0,0)</math> and <math>(1,1)</math>. | The line <math>x=y</math> intersects the rectangle at <math>(0,0)</math> and <math>(1,1)</math>. | ||
− | The area which <math>x | + | The area which <math>x<y</math> is the right isosceles triangle with side length <math>1</math> that has vertices at <math>(0,0)</math>, <math>(1,1)</math>, and <math>(0,1)</math>. |
The area of this triangle is <math>\frac{1}{2}\cdot1^{2}=\frac{1}{2}</math> | The area of this triangle is <math>\frac{1}{2}\cdot1^{2}=\frac{1}{2}</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> | 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 by WhyMath== | ||
+ | https://youtu.be/JueaMdaNRrs | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=WdpGXsAYxTQ ~David | ||
== See Also == | == See Also == |
Latest revision as of 20:58, 19 July 2023
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 by WhyMath
~savannahsolver
Video Solution
https://www.youtube.com/watch?v=WdpGXsAYxTQ ~David
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.