Difference between revisions of "1970 AHSME Problems/Problem 27"
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== Solution == | == Solution == | ||
| − | <math>\fbox{A}</math> | + | |
| + | 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. | ||
| + | |||
| + | 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?
Solution
One of the most common formulas involving the inradius of a triangle is 
, where 
is the area of the triangle, 
is the inradius, and 
is the semiperimeter.
The problem states that 
. This means 
, or 
, which is option 
.
See also
| 1970 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 26 |
Followed by Problem 28 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
