Difference between revisions of "1957 AHSME Problems/Problem 25"
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import geometry; | import geometry; | ||
− | point P = (3 | + | point P = (0,3); |
point Q = (4,0); | point Q = (4,0); | ||
point R = (5,4); | point R = (5,4); |
Revision as of 14:27, 25 July 2024
Problem
The vertices of have coordinates as follows: , where and are positive. The origin and point lie on opposite sides of . The area of may be found from the expression:
Solution
To solve this problem, we could use the distance formula to find the lengths of ths sides and then use Heron's Formula to find the area of the triangle, but that solution seems messy and prone to mistakes with lots of square roots and polynomial expansions. Therefore, we look for a simpler, easier solution. Suppose and . This makes half of a rectangle with side lengths and , so it has an area of . Plugging in and into all of the answer choices, the only one which returns is .
See Also
1957 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.