How do you find the lengths of the sides of a right triangle given the legs x and x+1 and the hypotenuse 7?

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How do you find the lengths of the sides of a right triangle given the legs x and x+1 and the hypotenuse 7?

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Answer 1 · 2016-07-01T15:32:42

Legs of the triangle are 4.424 and 5.524

Explanation:

Write an equation using Pythagoras' Theorem.

$x^{2} + {(x + 1)}^{2} = 7^{2}$ $x^2 + (x+1)^2 = 7^2$
$x^{2} + x^{2} + 2 x + 1 - 49 = 0$ $x^2 + x^2 +2x+1 - 49 = 0$

$2 x^{2} + 2 x - 48 = 0 \div 2$ $2x^2 +2x -48 = 0 " " ÷2$

$x^{2} + x - 24 = 0$ $x^2 +x -24 =0$

Solve by Completing the square method:

$x^{2} + x + {(\frac{1}{2})}^{2} = 24 + {(\frac{1}{2})}^{2}$ $x^2 +x + (1/2)^2 =24 +(1/2)^2$
${(x + \frac{1}{2})}^{2} = 24 + \frac{1}{4}$ $(x + 1/2)^2 = 24+ 1/4$

$x + \frac{1}{2} = \pm \sqrt{\frac{97}{4}}$ $x + 1/2 = +-sqrt(97/4)$
$x = \frac{+ \sqrt{97}}{2} - \frac{1}{2} or x = \frac{- \sqrt{97}}{2} - \frac{1}{2}$ $x = (+sqrt97)/2 - 1/2" or "x = (-sqrt97)/2 - 1/2$

$x = 4.424 or x = - 5.424$ $x= 4.424 " or " x= -5.424$ (reject as a side)

Legs of the triangle are 4.424 and 5.524

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How do you find the lengths of the sides of a right triangle given the legs x and x+1 and the hypotenuse 7?

1 Answer

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