Difference between revisions of "1970 AHSME Problems/Problem 27"

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== Solution ==
 
== Solution ==
<math>\fbox{A}</math>
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One of the most common formulas involving the inradius of a triangle is <math>A = rs</math>, where <math>A</math> is the area of the triangle, <math>r</math> is the inradius, and <math>s</math> is the semiperimeter.
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The problem states that <math>A = p = 2s</math>.  This means <math>2s = rs</math>, or <math>r = 2</math>, which is option <math>\fbox{A}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 17:01, 15 July 2019

Problem

In a triangle, the area is numerically equal to the perimeter. What is the radius of the inscribed circle?

$\text{(A) } 2\quad \text{(B) } 3\quad \text{(C) } 4\quad \text{(D) } 5\quad \text{(E) } 6$

Solution

One of the most common formulas involving the inradius of a triangle is $A = rs$, where $A$ is the area of the triangle, $r$ is the inradius, and $s$ is the semiperimeter.

The problem states that $A = p = 2s$. This means $2s = rs$, or $r = 2$, which is option $\fbox{A}$.

See also

1970 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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