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Math Help - Find Perimeter of a Right Traingle w/ Hypotenuse & Area

  1. #1
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    Find Perimeter of a Right Traingle w/ Hypotenuse & Area

    Hi everyone!

    I've been struggling with this problem recently, and don't really understand how I'm supposed to solve it:

    A hypotenuse of a right triangle has length 8. The area of the triangle is also 8 units squared. What is the perimeter of the triangle?

    Find Perimeter of a Right Traingle w/ Hypotenuse & Area-img_0458.jpg

    Any help would be appreciated!
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    Re: Find Perimeter of a Right Traingle w/ Hypotenuse & Area

    You have $\displaystyle \begin{align*} \frac{1}{2}\,a\,b = 8 \end{align*}$ and $\displaystyle \begin{align*} a^2 + b^2 = 64 \end{align*}$. Rearranging the first equation we have $\displaystyle \begin{align*} b = \frac{16}{a} \end{align*}$, so substituting into the second equation gives $\displaystyle \begin{align*} a^2 + \left( \frac{16}{a} \right) ^2 = 64 \end{align*}$. Expand and multiply both sides by $\displaystyle \begin{align*} a^2 \end{align*}$, giving you a quadratic equation you should be able to solve for $\displaystyle \begin{align*} a^2 \end{align*}$ and thus solve for $\displaystyle \begin{align*} a \end{align*}$.
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    Re: Find Perimeter of a Right Traingle w/ Hypotenuse & Area

    Alright, so I've tried 1/2ab=8^2 and solved for b, getting b = 128/a. Then I tried plugging that into the equation a^2 + b^2 = 64, like so:

    a^2 + (128/a)^2 = 64

    a^ 2 + (16384/a^2) = 64

    16384/a^2 = 64 - a^2

    16384 = 64a^2 -a^4

    -a^4 + 64a^2 - 16384 = 0

    -1(a^4 - 64a^2 + 16384) = 0

    a^4 - 64a^2 + 16384 = 0



    Then, when I tried to solve for that with the quadratic formula:

    -(-64) +/- sqrt[(-64)^2 - 4(1)(16384)]/2(1)


    I got an imaginary answer, since 4096 - 65536 = a negative number, which you can't find a real square root of.

    I'm still stuck, so can someone shed some light on what I did wrong?
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    Re: Find Perimeter of a Right Traingle w/ Hypotenuse & Area

    Quote Originally Posted by Lexielai View Post
    Alright, so I've tried 1/2ab=8^2 and solved for b, getting b = 128/a. Then I tried plugging that into the equation a^2 + b^2 = 64, like so:

    a^2 + (128/a)^2 = 64

    a^ 2 + (16384/a^2) = 64

    16384/a^2 = 64 - a^2

    16384 = 64a^2 -a^4

    -a^4 + 64a^2 - 16384 = 0

    -1(a^4 - 64a^2 + 16384) = 0

    a^4 - 64a^2 + 16384 = 0



    Then, when I tried to solve for that with the quadratic formula:

    -(-64) +/- sqrt[(-64)^2 - 4(1)(16384)]/2(1)


    I got an imaginary answer, since 4096 - 65536 = a negative number, which you can't find a real square root of.

    I'm still stuck, so can someone shed some light on what I did wrong?
    The question is flawed.Max area of a rt triangle with hypothenuse of 8 = 16 sq units
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    Re: Find Perimeter of a Right Traingle w/ Hypotenuse & Area

    Quote Originally Posted by Lexielai View Post
    Alright, so I've tried 1/2ab=8^2
    NO! You have been told the area of the triangle is $\displaystyle \begin{align*} 8\,\textrm{units}^2 \end{align*}$, NOT 64. Read my previous post properly!
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    Re: Find Perimeter of a Right Traingle w/ Hypotenuse & Area

    Oh, that makes much more sense now! I was totally thinking about this the wrong way. Thanks.
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    Re: Find Perimeter of a Right Traingle w/ Hypotenuse & Area

    Well, for practice, try this one:

    right triangle ABC with sides a,b,c (a<b, c = hypotenuse).

    area of ABC = 78a = 30c

    Calculate a,b,c.
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    Re: Find Perimeter of a Right Traingle w/ Hypotenuse & Area

    I misinterpreted area as 64 sq units.For an area of 8 sq units
    Draw a semicircle radius 4 Mark diameter ends A and C. On the arc where altitude is 2 mark B on arc.D is the base of alt from B to AC.
    ABC is a right angle. 1/2*180. Area triangle ABC = 1/2 * 2*8
    AD=x DC = 8-x
    ABD is similar to BDC Equal angles
    x/2 =2/8-x
    x^2-8x+4=0
    x=0.54
    AB^2= 2^2 +(0.54)^2
    AB = 2.07 short leg
    BC = 7.73 long leg
    AC = 8.00 hypothenuse
    perimeter = 17.8 units
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