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Math Help - Solving a quadratic with Geometry.

  1. #1
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    Solving a quadratic with Geometry.

    Can someone please tell me if I managed to do this correctly?

    "The length of one leg of a right triangle is 2 ft longer than the other leg. If the hypotenuse is 10 ft, find the perimeter of the triangle."

    My answer was

    P (perimeter) = 12+2y

    If this is wrong someone please tell me. I don't understand how to do these types of problems at all.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Solving a quadratic with Geometry.

    I would let x be the length of the shorter leg, then x + 2 is the length of the longer leg. By Pythagoras we then have:

    x^2+(x+2)^2=10^2

    x^2+x^2+4x+4=100

    Now collect like terms, simplify, and factor to find x.

    Then the perimeter P will be:

    P=x+x+2+10=2(x+6)
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  3. #3
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    e^(i*pi)'s Avatar
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    Re: Solving a quadratic with Geometry.

    Quote Originally Posted by WannaBSmart View Post
    Can someone please tell me if I managed to do this correctly?

    "The length of one leg of a right triangle is 2 ft longer than the other leg. If the hypotenuse is 10 ft, find the perimeter of the triangle."

    My answer was

    P (perimeter) = 12+2y

    If this is wrong someone please tell me. I don't understand how to do these types of problems at all.
    Let x be the shorter leg. Therefore the longer leg has length x+2.

    The perimeter is given by P=2(l+w) where l and w are length and width respectively. In your case l = x and w = x+2 and so P = 2(x+x+2) = 2(2x+2) = 4x+4

    To find the length x use Pythagoras since you are given the hypotenuse: x^2 + (x+2)^2 = 10^2 \ \ ,\ \ x > 0
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  4. #4
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    Re: Solving a quadratic with Geometry.

    Hi,

    Let the length of one side be 'x'. We now have a triangle such that we have

    one side has length 'x'
    the other side is 'x+2'
    and the hypotenuse is '10'

    so your perimeter = 10+(x+2)+x which simplifies too 12+2x

    Now in order to solve that you need to use r^2= a^2+b^2

    r = 10
    a = x+2
    b = x

    Solve for x and plug into the perimeter equation and you will have an answer



    I agree with MarkFL2.
    e^(i*pi), i think you solved using a rectangle instead of a triangle :P

    Also how do i use math symbols n stuff like you guys did? :S
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  5. #5
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    Re: Solving a quadratic with Geometry.

    Quote Originally Posted by e^(i*pi) View Post
    Let x be the shorter leg. Therefore the longer leg has length x+2.

    The perimeter is given by P=2(l+w) where l and w are length and width respectively. In your case l = x and w = x+2 and so P = 2(x+x+2) = 2(2x+2) = 4x+4

    To find the length x use Pythagoras since you are given the hypotenuse: x^2 + (x+2)^2 = 10^2 \ \ ,\ \ x > 0
    Hmm, I don't think I understand. I don't think that is the formula for perimeter of a right triangle.

    Since I have X on the shorter leg, X+2 on the longer leg, and 10ft as the hypotenuse, shouldn't I just group common variables and get

    P= 2x+12 OR P= 2(x+6) as my answer?
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  6. #6
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    Re: Solving a quadratic with Geometry.

    Quote Originally Posted by WannaBSmart View Post
    Hmm, I don't think I understand. I don't think that is the formula for perimeter of a right triangle.

    Since I have X on the shorter leg, X+2 on the longer leg, and 10ft as the hypotenuse, shouldn't I just group common variables and get

    P= 2x+12 OR P= 2(x+6) as my answer?
    That is right, but is not the complete answer. Solve for 'x' using pythagorean theorem!
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  7. #7
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    Re: Solving a quadratic with Geometry.

    Quote Originally Posted by Crzyrio View Post
    That is right, but is not the complete answer. Solve for 'x' using pythagorean theorem!


    Hmm, I'm not following. I thought the purpose of Pythagorean theorem is to find the hypotenuse? I don't quite understand how I can use Pythagorean theorem to solve for 'x'.
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  8. #8
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    Re: Solving a quadratic with Geometry.

    Quote Originally Posted by WannaBSmart View Post
    Hmm, I'm not following. I thought the purpose of Pythagorean theorem is to find the hypotenuse? I don't quite understand how I can use Pythagorean theorem to solve for 'x'.
    Ohhhh haha. Never mind. As soon as I wrote this I realized what you were telling me.

    Thank you very much for all the help!
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