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