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Math Help - Quadratics problem solving

  1. #1
    Newbie
    Joined
    Jun 2009
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    14

    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
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  2. #2
    Super Member
    Joined
    May 2009
    Posts
    527
    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}<br />
A &= \frac{1}{2}bh \\<br />
30 &= \frac{1}{2}b(20 - b) \\<br />
30 &= 10b - \frac{1}{2}b^2 \\<br />
60 &= 20b - b^2 \\<br />
0 &= -b^2 + 20b - 60 \\<br />
0 &= b^2 - 20b + 60<br />
\end{aligned}

    Now solve by using the quadratic formula.

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


    01
    Last edited by yeongil; June 6th 2009 at 09:07 PM.
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