Difference between revisions of "1950 AHSME Problems/Problem 35"
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{{AHSME 50p box|year=1950|num-b=34|num-a=36}} | {{AHSME 50p box|year=1950|num-b=34|num-a=36}} | ||
| − | + | We know that the formula of in radius is area of triangle by its semiperimeter , hence it is a right triangle => area is 1/2×24×10=120 and semiperimeter is 60/2=30 => 120/30=4 | |
| + | HK🗿 | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:50, 31 January 2024
Contents
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, as it's side length satisfies the Pythagorean Theorem. The semiperimeter is 
. Therefore the inradius is 
.
Solution 2
Since this is a right triangle, we have
![\[\frac{a+b-c}{2}=\boxed{4}\]](http://latex.artofproblemsolving.com/e/5/3/e53e2d600b5c1b592b671476025d743990210a2c.png)
- kante314
Solution 3
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 34 |
Followed by Problem 36 | |
| 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 | ||
We know that the formula of in radius is area of triangle by its semiperimeter , hence it is a right triangle => area is 1/2×24×10=120 and semiperimeter is 60/2=30 => 120/30=4
HK🗿
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
