Difference between revisions of "1968 AHSME Problems/Problem 12"

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== Solution ==
 
== Solution ==
<math>\fbox{C}</math>
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The triangle that goes through all the vertices of the triangle is the circumcircle of the triangle.
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<math>(7\frac{1}{2})^{2}+10^{2}=(12\frac{1}{2})^{2}</math>, so the triangle is a right triangle.The radius of a circumcircle
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of a right triangle is half the hypotenuse. <math>\frac{1}{2}\cdot \frac{25}{2}=\frac{25}{4}\implies</math> <math>\fbox{C}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 20:02, 12 October 2015

Problem

A circle passes through the vertices of a triangle with side-lengths $7\tfrac{1}{2},10,12\tfrac{1}{2}.$ The radius of the circle is:

$\text{(A) } \frac{15}{4}\quad \text{(B) } 5\quad \text{(C) } \frac{25}{4}\quad \text{(D) } \frac{35}{4}\quad \text{(E) } \frac{15\sqrt{2}}{2}$

Solution

The triangle that goes through all the vertices of the triangle is the circumcircle of the triangle. $(7\frac{1}{2})^{2}+10^{2}=(12\frac{1}{2})^{2}$, so the triangle is a right triangle.The radius of a circumcircle of a right triangle is half the hypotenuse. $\frac{1}{2}\cdot \frac{25}{2}=\frac{25}{4}\implies$ $\fbox{C}$

See also

1968 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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