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3 February, 01:24

For what value of x, the triangle having the sides x, x+7, x+8 is a right angled triangle

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  1. 3 February, 03:42
    0
    x = 5

    Step-by-step explanation:

    Using Pythagoras' identity.

    The square on the hypotenuse is equal to the sum of the squares on the other 2 sides.

    Here the longest side, the hypotenuse is x + 8, thus

    x² + (x + 7) ² = (x + 8) ² ← expand factors using FOIL

    x² + x² + 14x + 49 = x² + 16x + 64

    2x² + 14x + 49 = x² + 16x + 64 ← subtract terms on right side from both sides

    x² - 2x - 15 = 0 ← in standard form

    (x - 5) (x + 3) = 0 ← in factored form

    Equate each factor to zero and solve for x

    x - 5 = 0 ⇒ x = 5

    x + 3 = 0 ⇒ x = - 3

    However x > 0, thus x = 5

    The sides of the right triangle are 5, 12, 13
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