Question 337447
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The area of a triangle is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ \frac{bh}{2}]


So for your triangle


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{bh}{2}\ =\ 60]


But you are also given that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h\ =\ 3b\ +\ 2]


Substitute for *[tex \Large h] and then solve for *[tex \Large b].


Hint: You will end up with a factorable quadratic.  Discard the extraneous negative root.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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