Difference between revisions of "1957 AHSME Problems/Problem 7"
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<math>\textbf{(A)}\ 72\sqrt{3} \qquad \textbf{(B)}\ 48\sqrt{3}\qquad \textbf{(C)}\ 36\qquad \textbf{(D)}\ 24\qquad \textbf{(E)}\ 72 </math> | <math>\textbf{(A)}\ 72\sqrt{3} \qquad \textbf{(B)}\ 48\sqrt{3}\qquad \textbf{(C)}\ 36\qquad \textbf{(D)}\ 24\qquad \textbf{(E)}\ 72 </math> | ||
==Solution== | ==Solution== | ||
+ | <asy> | ||
+ | draw((-3,-sqrt(3))--(3,-sqrt(3))--(0,2sqrt(3))--cycle); | ||
+ | draw(circle((0,0),sqrt(3))); | ||
+ | dot((0,0)); | ||
+ | draw((0,0)--(0,-sqrt(3))); | ||
+ | </asy> | ||
+ | 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 box|year=1957|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:20, 12 October 2020
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 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.