The area of a triangle is 24cm² [squared]. The base is 8cm longer than the height. Use this information to set up a quadratic equation. Solve the equation to find the length of the base?

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Answer 1 · 2018-03-10T13:33:30

Let the length of the base is $x$ $x$ ,so height will be $x - 8$ $x-8$

so,area of the triangle is $\frac{1}{2} x (x - 8) = 24$ $1/2 x (x-8)=24$

or, $x^{2} - 8 x - 48 = 0$ $x^2 -8x-48=0$

or, $x^{2} - 12 x + 4 x - 48 = 0$ $x^2 -12x +4x-48=0$

or, $x (x - 12) + 4 (x - 12) = 0$ $x(x-12) +4(x-12)=0$

or, $(x - 12) (x + 4) = 0$ $(x-12)(x+4)=0$

so,either $x = 12$ $x=12$ or $x = - 4$ $x=-4$ but length of triangle can't be negative,so here length of the base is $12$ $12$ cm

Answer 2 · 2018-03-10T13:35:35

$12 c m$ $12 cm$

The area of a triangle is $\frac{base \times height}{2}$ $("base " xx " height" )/2$

Let the height be $x$ $x$ then if the base is 8 longer, then the base is $x + 8$ $x+8$

$\Rightarrow \frac{x \times (x + 8)}{2} = area$ $=> (x xx (x+8) )/2 = " area "$

$\Rightarrow \frac{x (x + 8)}{2} = 24$ $=> (x(x+8))/2 = 24$

$\Rightarrow x (x + 8) = 48$ $=> x(x+8) = 48$

Expanding and simplifying...

$\Rightarrow x^{2} + 8 x = 48$ $=> x^2 + 8x = 48$

$\Rightarrow x^{2} + 8 x - 48 = 0$ $=> x^2 +8x - 48 = 0$

$\Rightarrow (x - 4) (x + 12) = 0$ $=> (x-4)(x+12) = 0$

$\Rightarrow x = 4 and x = - 12$ $=> x = 4 " and " x = -12$

We know $x = - 12$ $x = -12$ can't be a solution as length can't be negative

Hence $x = 4$ $x = 4$

We know the base is $x + 8$ $x+8$

$\Rightarrow 4 + 8 = 12$ $=> 4+8 = 12$