Difference between revisions of "1962 AHSME Problems/Problem 39"

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==Problem==
 
==Problem==
Two medians of a triangle with unequal sides are <math>3</math> inches and <math>6</math> inches. Its area is <math>3 \sqrt{15}</math> square inches. The length of the third median in inches, is:  
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Two medians of a triangle with unequal sides are <math>3</math> inches and <math>6</math> inches. Its area is <math>3 \sqrt{15}</math> square inches. The length of the third median in inches, is:
 
 
 
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{6}\qquad\textbf{(D)}\ 6\sqrt{3}\qquad\textbf{(E)}\ 6\sqrt{6} </math>
 
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{6}\qquad\textbf{(D)}\ 6\sqrt{3}\qquad\textbf{(E)}\ 6\sqrt{6} </math>
 
 
==Solution==
 
==Solution==
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By the area formula:
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<cmath>A = \frac43\sqrt{s(s-m_1)(s-m_2)(s-m_3)}</cmath>
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Where <math>s = \frac{m_1+m_2+m_3}{2}</math>.
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Plugging in the numbers:
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<cmath>3\sqrt{15} = \frac43\sqrt{\frac{9+m}2\cdot\frac{9-m}2\cdot\frac{m+3}2\cdot\frac{m-3}2}</cmath>
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Simplifying and squaring both sides:
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<cmath>1215 = (9-m)(9+m)(m+3)(m-3)</cmath>
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Now, we can just plug in the answer choices.
 
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Revision as of 12:29, 23 February 2020

Problem

Two medians of a triangle with unequal sides are $3$ inches and $6$ inches. Its area is $3 \sqrt{15}$ square inches. The length of the third median in inches, is: $\textbf{(A)}\ 4\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{6}\qquad\textbf{(D)}\ 6\sqrt{3}\qquad\textbf{(E)}\ 6\sqrt{6}$

Solution

By the area formula: \[A = \frac43\sqrt{s(s-m_1)(s-m_2)(s-m_3)}\] Where $s = \frac{m_1+m_2+m_3}{2}$. Plugging in the numbers: \[3\sqrt{15} = \frac43\sqrt{\frac{9+m}2\cdot\frac{9-m}2\cdot\frac{m+3}2\cdot\frac{m-3}2}\] Simplifying and squaring both sides: \[1215 = (9-m)(9+m)(m+3)(m-3)\] Now, we can just plug in the answer choices. This problem needs a solution. If you have a solution for it, please help us out by adding it.