Difference between revisions of "2003 AMC 10B Problems/Problem 20"
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Revision as of 13:10, 31 July 2018
- The following problem is from both the 2003 AMC 12B #14 and 2003 AMC 10B #20, so both problems redirect to this page.
Contents
[hide]Problem
In rectangle and . Points and are on so that and . Lines and intersect at . Find the area of .
Solution 1
because The ratio of to is since and from subtraction. If we let be the height of
The height is so the area of is .
Solution 2
We can look at this diagram as if it were a coordinate plane with point being . This means that the equation of the line is and the equation of the line is . From this we can set of the follow equation to find the coordinate of point :
We can plug this into one of our original equations to find that the coordinate is , meaning the area of is
Solution 3
At points and , segment is 5 units from segment . At points and , the segments are 2 units from each other. This means that collectively, the two lines closed the distance between them by 3 units over a height of 3 units. Therefore, to close the next two units of distance, they will have to travel a height of 2 units.
Then calculate the area of trapezoid and triangle separately and add them. The area of the trapezoid is and the area of the triangle is .
See Also
2003 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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 12 Problems and Solutions |
2003 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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.