Difference between revisions of "1950 AHSME Problems/Problem 35"
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| − | 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 | + | ==Problem== |
| + | In triangle <math>ABC</math>, <math>AC=24</math> inches, <math>BC=10</math> inches, <math>AB=26</math> inches. The radius of the inscribed circle is: | ||
| + | |||
| + | <math>\textbf{(A)}\ 26\text{ in} \qquad | ||
| + | \textbf{(B)}\ 4\text{ in} \qquad | ||
| + | \textbf{(C)}\ 13\text{ in} \qquad | ||
| + | \textbf{(D)}\ 8\text{ in} \qquad | ||
| + | \textbf{(E)}\ \text{None of these}</math> | ||
| + | |||
| + | ==Solution== | ||
| + | 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>. Therefore the inradius is <math>\boxed{\textbf{(B)}\ 4}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 50p box|year=1950|num-b=34|num-a=36}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
Revision as of 07:38, 29 April 2012
Problem
In triangle 
, 
inches, 
inches, 
inches. The radius of the inscribed circle is:
Solution
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 
.
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 34 |
Followed by Problem 36 | |
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| All AHSME Problems and Solutions | ||
