Difference between revisions of "2000 AMC 8 Problems/Problem 25"
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==Solution 2== | ==Solution 2== | ||
− | The above answer is fast, but | + | The above answer is fast, but satisfying, and assumes that the area of <math>\triangle AMN</math> is independent of the dimensions of the rectangle. All in all, it's a very good answer though. However this is an alternative if you don't get the above answer. Label <math>AB = CD = l</math> and <math>BC = DA = h</math> |
Labelling <math>M</math> and <math>N</math> as the right and lower midpoints respectively, and redoing all the work above, we get: | Labelling <math>M</math> and <math>N</math> as the right and lower midpoints respectively, and redoing all the work above, we get: |
Revision as of 23:21, 1 November 2016
Contents
Problem
The area of rectangle is units squared. If point and the midpoints of and are joined to form a triangle, the area of that triangle is
Solution 1
To quickly solve this multiple choice problem, make the (not necessarily valid, but very convenient) assumption that can have any dimension. Give the rectangle dimensions of and , which is the easiest way to avoid fractions. Labelling the right midpoint as , and the bottom midpoint as , we know that , and .
, and the answer is
Solution 2
The above answer is fast, but satisfying, and assumes that the area of is independent of the dimensions of the rectangle. All in all, it's a very good answer though. However this is an alternative if you don't get the above answer. Label and
Labelling and as the right and lower midpoints respectively, and redoing all the work above, we get:
, and the answer is
See Also
2000 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.