1955 AHSME Problems/Problem 9
Revision as of 23:01, 22 July 2020 by Duck master (talk | contribs) (created page w/ solution & categorization)
A circle is inscribed in a triangle with sides , and . The radius of the circle is:
Solution
We know that , where is the triangle's area, its semiperimeter, and its inradius. Since this particular triangle is a right triangle (which we can verify by the Pythagorean theorem), the area is half of , and the semiperimeter is half of . Therefore, the inradius is , so our answer is and we are done.
See also
1955 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.