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Math Help - Deriving hypotenuse from perimeter of a triangle

  1. #1
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    Deriving hypotenuse from perimeter of a triangle

    Please help me work through the below problem, thanks in advance!

    The perimeter of a certain isosceles right triangle is 16+16√2. What is the hypotenuse of the triangle?
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  2. #2
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    Right triangle with side lengths a, b, c (where c is hypotnuse)
    Isosceles means a=b

    perimeter = a + b + c = a + a + c
    Pythagorean theorem says: a^2 + b^2 = a^2 + a^2 = c^2

    You have 2 equations and 2 unknowns. Solve for c.
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  3. #3
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    first draw the triangle. If the hypotenuse is "b" and the 2 other equal sides are "a" then the perimeter is:

    b+2a=16+16\sqrt{2}

    You know the angles of an isosceles right triangle, there are 2 45 degree angles \frac{\pi}{4} and one angle \frac{\pi}{2}

    So the sine of pi over 4 is sin(\frac{\pi}{4})=\frac{a}{b} solve this for "a" and plug that into the original equation for the perimeter and use algebra to solve for "b".
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  4. #4
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    Snowtea's solution is easier, I didn't catch that...
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  5. #5
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    Hello, mtylerrose!

    \text{The perimeter of a certain isosceles right triangle is }16+16\sqrt{2}
    \text{What is the hypotenuse of the triangle?}

    You should know that the sides of an isosceles right triangle are: . x,\:x,\:x\sqrt{2}


    The perimeter is: . x + x + x\sqrt{2} \;=\;16 + 16\sqrt{2}

    . . . . . . . . . . . . . . . (2 + \sqrt{2})x \:=\:16(1+\sqrt{2})

    . . . . . . . . . . . . . . . . . . . . . x \;=\;\dfrac{16(1+\sqrt{2})}{2+\sqrt{2}}


    Rationalize: . \displaystyle x\;=\;\frac{16(1+\sqrt{2})}{2 + \sqrt{2}} \cdot\frac{2-\sqrt{2}}{2-\sqrt{2}} \;=\;\frac{16(2 - \sqrt{2} + 2\sqrt{2} - 2)}{4 - 2}

    . . . . . . . . . x \;=\;\dfrac{16\sqrt{2}}{2} \;=\;8\sqrt{2}


    The hypotenuse is: . x\sqrt{2} \;=\;8\sqrt{2}\cdot\sqrt{2} \;=\;16

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  6. #6
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    Thanks everyone! I had no idea where to start, but now I am starting to understand.

    Soroban,

    How did you know to rationalize the way you did?
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  7. #7
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    Quote Originally Posted by mtylerrose View Post
    Thanks everyone! I had no idea where to start, but now I am starting to understand.

    Soroban,

    How did you know to rationalize the way you did?
    That's a general technique he used that's taught in algebra courses. You always multiply by the conjugate of the denominator to rationalize an expression like that.
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