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Math Help - Finding perimeter and area using polynomials

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    Senior Member vaironxxrd's Avatar
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    Finding perimeter and area using polynomials

    One side of an isoceles right triangle is x + 2 and its hypotenuse is x +3. Find the perimeter and the area. p = 2l * 2w

    How exactly could I solve this?

    I would say:

    (x+3)^2 = (x+2)^2 * h is this right just for a start?

    If so does it mean (x+3)-(x+2) = b?

    I believe the above idea is disappointing to you guys.
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    Re: Finding perimeter and area using polynomials

    I see P = (x+2)+(x+2)+(x+3) and A = \frac{1}{2}\times (x+2)\times (x+2)
    Last edited by pickslides; March 5th 2012 at 06:56 PM.
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    Senior Member vaironxxrd's Avatar
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    Re: Finding perimeter and area using polynomials

    Quote Originally Posted by pickslides View Post
    I see P = (x+2)+(x+2)+(x+3) and A = \frac{1}{2}\times (x+2)\times (x+2)
    So both the length and the width are the same size?
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    Re: Finding perimeter and area using polynomials

    I would call them base and height, but yes at you are told it is an 'isoceles' right triangle.

    Draw one, it'll make more sense.
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    Senior Member vaironxxrd's Avatar
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    Re: Finding perimeter and area using polynomials

    Quote Originally Posted by pickslides View Post
    I would call them base and height, but yes at you are told it is an 'isoceles' right triangle.

    Draw one, it'll make more sense.
    Couldn't you also say the perimeter is 2(x+2)2(x+2)? [Sorry for all of the (bad) questions]

    P =
    Code:
    x^3+7x^2+16x+12
    a = (1/2)(x^2+4x+4)
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    Re: Finding perimeter and area using polynomials

    P = (x+2)+(x+2)+(x+3) = 2(x+2)+(x+3) = 2x+4+x+3 group like terms.

    A = \frac{1}{2}\times (x+2)\times (x+2)= \frac{1}{2}\times (x+2)^2
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    Re: Finding perimeter and area using polynomials

    Quote Originally Posted by vaironxxrd View Post
    One side of an isoceles right triangle is x + 2 and its hypotenuse is x +3.
    Find the perimeter and the area.
    p = 2l * 2w
    That triangle is NOT isosceles: isosceles means 2 equal sides, so 2 sides each equal to x +2; then:
    hypotenuse = SQRT[(x + 2)^2 + (x + 2)^2] : NOT x + 3

    What you have is a right triangle with one leg = x + 2 and the hypotenuse = x + 3; so other leg
    = SQRT[(x + 3)^2 - (x + 2)^2]
    = SQRT(2x + 5)

    So triangle has sides SQRT(2x + 5), x +2 and x + 3

    Please use google to find out about perimeter and area of a right triangle.
    You assumed: p = 2l * 2w
    That's perimeter of a rectangle!
    You are quite unaware of the basics. Are you attending math classes?
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    Re: Finding perimeter and area using polynomials

    Quote Originally Posted by Wilmer View Post
    That triangle is NOT isosceles: isosceles means 2 equal sides, so 2 sides each equal to x +2; then:
    hypotenuse = SQRT[(x + 2)^2 + (x + 2)^2] : NOT x + 3

    What you have is a right triangle with one leg = x + 2 and the hypotenuse = x + 3; so other leg
    = SQRT[(x + 3)^2 - (x + 2)^2]
    = SQRT(2x + 5)

    So triangle has sides SQRT(2x + 5), x +2 and x + 3

    Please use google to find out about perimeter and area of a right triangle.
    You assumed: p = 2l * 2w
    That's perimeter of a rectangle!
    You are quite unaware of the basics. Are you attending math classes?
    Yes I am sorry for the errors.
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    Re: Finding perimeter and area using polynomials

    Quote Originally Posted by vaironxxrd View Post
    Yes I am sorry for the errors.
    Nobody's asking you to be "sorry"; just be careful, and make sure what you post makes sense...
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