Difference between revisions of "1953 AHSME Problems/Problem 37"
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==Solution== | ==Solution== | ||
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+ | <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 <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>. | ||
+ | |||
+ | 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== | ||
+ | {{AHSME 50p box|year=1953|num-b=36|num-a=38}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 00:13, 25 January 2020
Problem
The base of an isosceles triangle is inches and one of the equal sides is inches. The radius of the circle through the vertices of the triangle is:
Solution
Let be an isosceles triangle with and . Draw altitude . Since is the apex of the triangle, the altitude is also a median of the triangle. Therefore, . Using the Pythagorean Theorem, . The area of is .
If are the sides of a triangle, and is its area, the circumradius of that triangle is . Using this formula, we find the circumradius of to be . The answer is
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 36 |
Followed by Problem 38 | |
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All AHSME Problems and Solutions |
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