Difference between revisions of "1957 AHSME Problems/Problem 7"
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− | We can see that the radius of the circle is <math>4\sqrt{3}</math>. We know that the radius is <math>\frac{1}{3}</math> of each median line of the triangle; each median line is therefore <math>12\sqrt{3}</math>. Since the median line completes a <math>30</math>-<math>60</math>-<math>90</math> triangle, we can conclude that one of the sides of the triangle is <math>24</math>. Triple the side length and we get our answer, <math>\boxed{72}</math>. | + | We can see that the radius of the circle is <math>4\sqrt{3}</math>. We know that the radius is <math>\frac{1}{3}</math> of each median line of the triangle; each median line is therefore <math>12\sqrt{3}</math>. Since the median line completes a <math>30</math>-<math>60</math>-<math>90</math> triangle, we can conclude that one of the sides of the triangle is <math>24</math>. Triple the side length and we get our answer, <math>\boxed{\textbf{(E)}72}</math>. |
==See Also== | ==See Also== | ||
+ | {{AHSME 50p box|year=1957|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 09:07, 25 July 2024
Problem 7
The area of a circle inscribed in an equilateral triangle is . The perimeter of this triangle is:
Solution
We can see that the radius of the circle is
. We know that the radius is
of each median line of the triangle; each median line is therefore
. Since the median line completes a
-
-
triangle, we can conclude that one of the sides of the triangle is
. Triple the side length and we get our answer,
.
See Also
1957 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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All AHSME Problems and Solutions |
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