The height of a triangle is 5 units more than twice it's base. If the area of a triangles is 21 square units, what is its height?

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Answer 1 · 2018-04-19T12:56:17

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We know that Area of a triangle is given by $A = \frac{b h}{2}$ $A=(bh)/2$

where $h = 2 b + 5$ $h=2b+5$ and $A = 21$ $A=21$ . Then

$21 = b \frac{2 b + 5}{2}$ $21=b(2b+5)/2$

$42 = 2 b^{2} + 5 b$ $42=2b^2+5b$ Then $2 b^{2} + 5 b - 42 = 0$ $2b^2+5b-42=0$

$b = \frac{- 5 \pm \sqrt{25 + 4 \cdot 42 \cdot 2}}{4} = \frac{- 5 \pm 19}{4}$ $b=(-5+-sqrt(25+4·42·2))/4=(-5+-19)/4$

There are two possible values for b $b = \frac{7}{2}$ $b=7/2$ and $b = - 6$ $b=-6$ rejected

With this value, substitution in $h = 2 b + 5 = 2 \cdot \frac{7}{2} + 5 = 12$ $h=2b+5=2·7/2+5=12$

And both dimensions are now calculated.

$A = \frac{\frac{7}{2} \cdot 12}{2} = \frac{42}{2} = 21$ $A=(7/2·12)/2=42/2=21$