1967 AHSME Problems/Problem 5: Difference between revisions
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== Solution == | == Solution == | ||
<math>\ | The area <math>K</math> of the triangle can be expressed in terms of its inradius <math>r</math> and its semiperimeter <math>s</math> as: | ||
<cmath> K = r \times s = r \times \frac{P}{2}</cmath> | |||
So, <math>\frac{P}{K} = \boxed{\textbf{(C) } \frac{2}{r}}</math>. | |||
~ proloto | |||
== See also == | == See also == | ||
Latest revision as of 18:33, 28 September 2023
Problem
A triangle is circumscribed about a circle of radius 
inches. If the perimeter of the triangle is 
inches and the area is 
square inches, then 
is:
Solution
The area of the triangle can be expressed in terms of its inradius and its semiperimeter 
as:
![\[K = r \times s = r \times \frac{P}{2}\]](http://latex.artofproblemsolving.com/6/b/2/6b2fab40f5c4b4a8f4917ac19fb6f7a29a92a0b7.png)
So, 
.
~ proloto
See also
| 1967 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
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