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: | + | 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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<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> | ||
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==Solution== | ==Solution== | ||
− | + | By the area formula: | |
+ | <cmath>A = \frac43\sqrt{s(s-m_1)(s-m_2)(s-m_3)}</cmath> | ||
+ | Where <math>s = \frac{m_1+m_2+m_3}{2}</math>. | ||
+ | Plugging in the numbers: | ||
+ | <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> | ||
+ | Simplifying and squaring both sides: | ||
+ | <cmath>1215 = (9-m)(9+m)(m+3)(m-3)</cmath> | ||
+ | Now, we can just plug in the answer choices and find that <math>\boxed{3\sqrt6}</math> works. |
Revision as of 12:31, 23 February 2020
Problem
Two medians of a triangle with unequal sides are inches and inches. Its area is square inches. The length of the third median in inches, is:
Solution
By the area formula: Where . Plugging in the numbers: Simplifying and squaring both sides: Now, we can just plug in the answer choices and find that works.