please look at the attachment

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- Jun 24th 2007, 10:32 PMsanee66could somebody please help me with this
please look at the attachment

- Jun 24th 2007, 10:51 PMJhevon
I attached the diagram below. By the way, the triangle is AMN not ABC

Note that the altitude of an equilateral triangle bisects one of it's angles and divides the base it hits in two equal parts. This means that $\displaystyle MH = \frac {1}{2} (7m - 1)$

Also, since $\displaystyle \triangle AMN$ is equilateral, $\displaystyle AN = AM$

We are told that the sum of AM and MH gives 11m - 4

$\displaystyle \Rightarrow AM + MH = 11m - 4$

$\displaystyle \Rightarrow 7m - 1 + \frac {7m - 1}{2} = 11m - 4$

$\displaystyle \Rightarrow 7m + \frac {7m}{2} - 11m = -4 + \frac {3}{2}$

$\displaystyle \Rightarrow - \frac {1}{2} m = - \frac {5}{2}$

$\displaystyle \Rightarrow m = 5$

The side length of the triangle is equal to the side-length of AN

So, the side-length is 7m - 1 = 7(5) - 1 = 34