Difference between revisions of "1953 AHSME Problems/Problem 37"

(Created page with "==Problem== The base of an isosceles triangle is <math>6</math> inches and one of the equal sides is <math>12</math> inches. The radius of the circle through the vertices of...")
 
Line 14: Line 14:
 
==See Also==
 
==See Also==
  
 +
{{AHSME 50p box|year=1953|num-b=36|num-a=38}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:40, 24 January 2020

Problem

The base of an isosceles triangle is $6$ inches and one of the equal sides is $12$ inches. The radius of the circle through the vertices of the triangle is:

$\textbf{(A)}\ \frac{7\sqrt{15}}{5} \qquad \textbf{(B)}\ 4\sqrt{3} \qquad \textbf{(C)}\ 3\sqrt{5} \qquad \textbf{(D)}\ 6\sqrt{3}\qquad \textbf{(E)}\ \text{none of these}$

Solution

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 36
Followed by
Problem 38
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 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png