Difference between revisions of "1950 AHSME Problems/Problem 35"
(Created page with "The inradius is equal to the area divided by semiperimeter. The area is <math>(10)(24)/2 = 120</math> because it's a right triangle. The semiperimeter is <math>30</math>. Theref...") |
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Revision as of 23:08, 25 November 2011
The inradius is equal to the area divided by semiperimeter. The area is 
because it's a right triangle. The semiperimeter is 
. Therefore the inradius is 
(B)
