Let's start with what we know
Area:
\(42 = \frac{1}{2}bh\) where 42 is the area, b = base, and h = height
Height:
Since we know the height is 5 less than the base, we can write that as an equation.
\(h = b - 5\)
Now let's go and plug \(h = b - 5\) into \(42 = \frac{1}{2}bh\)
\(42 = \frac{b}{2}(b-5)\)
Let's distribute b over (b-5)
\(42 = \frac{ b^{2} - 5b }{2}\)
Let's move 42 over to the right side to make a quadratic formula
\(0 = \frac{1}{2} b^{2} - \frac{5}{2}b - 42\)
Let's plug that into the quadratic equation, which is:
\(\frac{-b +/- \sqrt{ b^{2} - 4ac } }{2a}\)
And we can now plug the pieces in to calculate b
\(\frac{- (-\frac{5}{2}) +/- \sqrt{ (-\frac{5}{2})^{2} - 4 (\frac{1}{2})(-42) } }{2 (\frac{1}{2}) }\)
\(\frac{\frac{5}{2} +/- \sqrt{ \frac{25}{4} +84 } }{1 }\)
\({\frac{5}{2} +/- \sqrt{ \frac{361}{4} } }\)
\({\frac{5}{2} +/- { \frac{19}{2} }\)
Since we can't have a negative value for b (a base can't be negative meters), let's add:
\({\frac{5}{2} + { \frac{19}{2} }\)
\({ \frac{24}{2} }\)
\(12 = b\)
So the base of the triangle is 12m
