# Math Help - Quadratics problem solving

1. ## Quadratics problem solving

I cannot figure this question out:

A straight length of wire is 20cm long. It is bent at right angles to form the two shorter sides of a right angled triangle. If the triangle's area is 30cm squared, find:

a). the length of the hypotenuse
b) the triangle's perimeter

My logic went like this:
a squared + b squared = c squared
therefore if 20 is a+b
20 squared = csquared, and
20 = c
That was incorrect and I have no idea how to do the problem

2. This is a problem about area, so you'll need a different formula:

$A = \frac{1}{2}bh$

where b is the base and h is the height.

The base and height are the two legs in a right triangle. You don't know the lengths exactly, but you do know that b + h = 20, so h = 20 - b. Substitute that into the formula:

\begin{aligned}
A &= \frac{1}{2}bh \\
30 &= \frac{1}{2}b(20 - b) \\
30 &= 10b - \frac{1}{2}b^2 \\
60 &= 20b - b^2 \\
0 &= -b^2 + 20b - 60 \\
0 &= b^2 - 20b + 60
\end{aligned}

Now solve by using the quadratic formula.

\begin{aligned}
b &= \frac{20 \pm \sqrt{(-20)^2 - 4(1)(60)}}{2(1)} \\
&= \frac{20 \pm \sqrt{400 - 240}}{2} \\
&= \frac{20 \pm \sqrt{160}}{2} \\
&= \frac{20 \pm 4\sqrt{10}}{2} \\
&= 10 \pm 2\sqrt{10}
\end{aligned}

If $b = 10 + 2\sqrt{10}$, then $h = 20 - b = 20 - (10 + 2\sqrt{10}) = 10 - 2\sqrt{10}$.

If $b = 10 - 2\sqrt{10}$, then $h = 20 - b = 20 - (10 - 2\sqrt{10}) = 10 + 2\sqrt{10}$.

Now use the Pythagorean theorem to find the hypotenuse. (It doesn't matter which b you use, because you'll get the same answer.) Then add the sides together to find the perimeter.

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