Mathematics
khekking
14

2. The height of a triangle is 5 m less than its base. The area of the triangle is 42 m2. Find the length of the base. (Points : 1)

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(1) Answers
gonzalez1999

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

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