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Math Help - [SOLVED] Triangle Question!

  1. #1
    Junior Member
    Joined
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    [SOLVED] Triangle Question!

    Hi, I'm a little wary regarding this problem. I'm pretty sure I'm doing it right but would appreciate it if someone verified my solution. The question is:

    The perimeter of a right triangle is 60 cm. The length of the hypotenuse is 6 cm more than twice the length of one of the other sides. Find the lengths of all three sides.

    So here' what i did.
    Let one side of the triangle be x, let the other be y
    60 - x - (2x+6) = y
    (54-3x) = y

    a^2 + b^2 = c^2
    So X^2 + (54-3x)^2 = (2x+6)^2
    After expanding i got
    6x^2 - 348x + 2880 = 0
    Using the quadratic formula, I got 10, 48.

    Choosing 10 as the best answer for x and solving for y. I got the following at my answers:
    10, 24, and 26 as the sides of the triangle. Is my approach correct?

    THANKS!
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by demo1 View Post
    Hi, I'm a little wary regarding this problem. I'm pretty sure I'm doing it right but would appreciate it if someone verified my solution. The question is:

    The perimeter of a right triangle is 60 cm. The length of the hypotenuse is 6 cm more than twice the length of one of the other sides. Find the lengths of all three sides.

    So here' what i did.
    Let one side of the triangle be x, let the other be y
    60 - x - (2x+6) = y
    (54-3x) = y

    a^2 + b^2 = c^2
    So X^2 + (54-3x)^2 = (2x+6)^2
    After expanding i got
    6x^2 - 348x + 2880 = 0
    Using the quadratic formula, I got 10, 48.

    Choosing 10 as the best answer for x and solving for y. I got the following at my answers:
    10, 24, and 26 as the sides of the triangle. Is my approach correct?

    THANKS!
    Hi demo1,

    You did good!
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  3. #3
    Junior Member
    Joined
    Feb 2009
    Posts
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    I'm so glad! Thanks for confirming my answer.
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