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Math Help - pythagoras

  1. #1
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    pythagoras

    the lengths, in centimetres, of the sides of a right-angled triangle are:
    x, x+2 and x+5

    i) use pythagoras' theorem to write down an equation in x. show that it simplifies to
    x^2-6x-21=0
    ii) solve the equation

    iii) write down the length of hypotenuse of the triangle.
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  2. #2
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    Hello, simonmann!

    The problem is straight-foward.
    Exactly where is your difficulty?


    The lengths of the sides (in cm) of a right triangle are: <br />
x,\:x+2,\:x+5

    a) Use Pythagoras' Theorem to write an equation in x.
    a^2 + b^2 \:=\:c^2 \quad\Rightarrow\quad x^2 + (x+2)^2 \:=\:(x+5)^2


    Show that it simplifies to : . x^2-6x-21\:=\:0
    We have: . x^2 + x^2 + 4x + 4 \;=\;x^2 + 10x + 25\quad\Rightarrow\quad x^2-6x-21 \:=\:0


    b) Solve the equation.
    Quadratic Formula: . x \;=\;\frac{-(\text{-}6) \pm\sqrt{(\text{-}6)^2 - 4(1)(\text{-}21)}}{2(1)} \;=\;\frac{6\pm\sqrt{120}}{2} \;=\;3 \pm\sqrt{30}

    And the positive root is: . x \:=\:3 + \sqrt{30} cm.



    c) Find the length of hypotenuse of the triangle.
    The hypotenuse is: . x + 5 \;=\;(3 + \sqrt{30}) + 5 \;=\;8 + \sqrt{30} cm.

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