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Difference between revisions of "1953 AHSME Problems/Problem 37"
 
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==Solution==
 
==Solution==
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<asy>
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draw((0,0)--(3,3sqrt(15))--(6,0)--cycle);
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draw((3,3sqrt(15))--(3,0));
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label("$A$",(3,3sqrt(15)),N);
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label("$B$",(0,0),SW);
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label("$C$",(6,0),SE);
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label("$D$",(3,0),S);
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label("12",(1.5,5.8),WNW);
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label("12",(4.5,5.8),ENE);
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label("3",(1.5,0),N);
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</asy>
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Let <math>\triangle ABC</math> be an isosceles triangle with <math>AB=AC=12,</math> and <math>BC = 6</math>. Draw altitude <math>\overline{AD}</math>. Since <math>A</math> is the apex of the triangle, the altitude <math>\overline{AD}</math> is also a median of the triangle. Therefore, <math>BD=CD=3</math>. Using the [[Pythagorean Theorem]], <math>AD=\sqrt{12^2-3^2}=3\sqrt{15}</math>. The area of <math>\triangle ABC</math> is <math>\frac12\cdot 6\cdot 3\sqrt{15}=9\sqrt{15}</math>.
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If <math>a,b,c</math> are the sides of a triangle, and <math>A</math> is its area, the circumradius of that triangle is <math>\frac{abc}{4A}</math>. Using this formula, we find the circumradius of <math>\triangle ABC</math> to be <math>\frac{6\cdot 12\cdot 12}{4\cdot 9\sqrt{15}}=\frac{2\cdot 12}{\sqrt{15}}=\frac{24\sqrt{15}}{15}=\frac{8\sqrt{15}}{5}</math>. The answer is <math>\boxed{\textbf{(E)}\ \text{none of these}}</math>
  
 
==See Also==
 
==See Also==

Latest revision as of 01:13, 25 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

[asy] draw((0,0)--(3,3sqrt(15))--(6,0)--cycle); draw((3,3sqrt(15))--(3,0)); label("$A$",(3,3sqrt(15)),N); label("$B$",(0,0),SW); label("$C$",(6,0),SE); label("$D$",(3,0),S); label("12",(1.5,5.8),WNW); label("12",(4.5,5.8),ENE); label("3",(1.5,0),N); [/asy] Let $\triangle ABC$ be an isosceles triangle with $AB=AC=12,$ and $BC = 6$. Draw altitude $\overline{AD}$. Since $A$ is the apex of the triangle, the altitude $\overline{AD}$ is also a median of the triangle. Therefore, $BD=CD=3$. Using the Pythagorean Theorem, $AD=\sqrt{12^2-3^2}=3\sqrt{15}$. The area of $\triangle ABC$ is $\frac12\cdot 6\cdot 3\sqrt{15}=9\sqrt{15}$.

If $a,b,c$ are the sides of a triangle, and $A$ is its area, the circumradius of that triangle is $\frac{abc}{4A}$. Using this formula, we find the circumradius of $\triangle ABC$ to be $\frac{6\cdot 12\cdot 12}{4\cdot 9\sqrt{15}}=\frac{2\cdot 12}{\sqrt{15}}=\frac{24\sqrt{15}}{15}=\frac{8\sqrt{15}}{5}$. The answer is $\boxed{\textbf{(E)}\ \text{none of these}}$

See Also

1953 AHSC (Problems • Answer Key • Resources)
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

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