Question 200174
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By width, I presume you mean the measure of the base.  Let *[tex \Large x] represent the measure of the base, then the height must be *[tex \Large x + 4]


The area of a triangle, in terms of the measures of the base and the height is:


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


Substitute what you know:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 48 = \frac{x(x + 4)}{2}]


Simplify, put it into standard form, and solve the resulting quadratic for *[tex \Large x]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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