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Math Help - Given Radius of incricle and Hypotenuse solve for the other 2 sides?

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    Given Radius of incricle and Hypotenuse solve for the other 2 sides?

    The right angle triangle has a hypotenuse of 403 and a incricle with a radius of 74 and i need to find the lengths of the other 2 sides....... i honestly have no idea how to do this everything i try does not work.
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  2. #2
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    Re: Given Radius of incricle and Hypotenuse solve for the other 2 sides?

    Sum of areas of triangles COB, COA, and AOB = area of triangle ABC

    \frac{1}{2}ra + \frac{1}{2}rb + \frac{1}{2}rc = \frac{1}{2}ab

    r(a+b+c) = ab

    you are given r and c ... and you know a = \sqrt{c^2 - b^2}

    finish.
    Attached Thumbnails Attached Thumbnails Given Radius of incricle and Hypotenuse solve for the other 2 sides?-rttriangleincircle.png  
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    Re: Given Radius of incricle and Hypotenuse solve for the other 2 sides?

    Hi,
    an other solution:
    The tangent points beetwen sides und incircle are D (on BC), E (on AC), F (on AB).
    DC=CE=r=74;
    BD=BF=x
    EA=AF=403-x
    =>
    a=x+74
    b=74+403-x= 477-x
    c=403
    You have to use the Pithagoras Formula and have a ecuation second grades in x
    Find the adequat solution for x and introduce in a and b
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