Difference between revisions of "1962 AHSME Problems/Problem 39"
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Simplifying and squaring both sides: | Simplifying and squaring both sides: | ||
<cmath>1215 = (9-m)(9+m)(m+3)(m-3)</cmath> | <cmath>1215 = (9-m)(9+m)(m+3)(m-3)</cmath> | ||
| − | Now, we can just plug in the answer choices. | + | Now, we can just plug in the answer choices and find that <math>\boxed{3\sqrt6}</math> works. |
Revision as of 11:31, 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:
![\[A = \frac43\sqrt{s(s-m_1)(s-m_2)(s-m_3)}\]](http://latex.artofproblemsolving.com/4/f/8/4f8ec5eece7a34c0e17a97eca41becab71b193e7.png)
Where 
.
Plugging in the numbers:
![\[3\sqrt{15} = \frac43\sqrt{\frac{9+m}2\cdot\frac{9-m}2\cdot\frac{m+3}2\cdot\frac{m-3}2}\]](http://latex.artofproblemsolving.com/a/8/8/a8882a05dbc2f0a1830dd083b9d829f3857df33e.png)
Simplifying and squaring both sides:
![\[1215 = (9-m)(9+m)(m+3)(m-3)\]](http://latex.artofproblemsolving.com/5/7/0/57020569f58996d0508b52058477832099dbd111.png)
Now, we can just plug in the answer choices and find that 
works.
