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1950 AHSME Problems/Problem 35

Revision as of 23:08, 25 November 2011 by Evanyao10 (talk | contribs) (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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The inradius is equal to the area divided by semiperimeter. The area is $(10)(24)/2 = 120$ because it's a right triangle. The semiperimeter is $30$. Therefore the inradius is $4$ (B)

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