If c is the measure of the hypotenuse of a right triangle, how do you find each missing measure given a=x-32, b=x-1, c=x?

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If c is the measure of the hypotenuse of a right triangle, how do you find each missing measure given a=x-32, b=x-1, c=x?

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Answer 1 · 2017-01-25T16:09:24

$a = 9, b = 40, c = 41$ $a=9, b=40, c=41$

Explanation:

$c^{2} = a^{2} + b^{2}$ $c^2=a^2+b^2$
$x^{2} = {(x - 32)}^{2} + {(x - 1)}^{2}$ $x^2=(x-32)^2+(x-1)^2$
$x^{2} = x^{2} - 64 x + 1024 + x^{2} - 2 x + 1$ $x^2=x^2-64x+1024+x^2-2x+1$
$x^{2} = 2 x^{2} - 66 x + 1025$ $x^2=2x^2-66x+1025$
$0 = x^{2} - 66 x + 1025$ $0=x^2-66x+1025$
$0 = (x - 25) (x - 41)$ $0=(x-25)(x-41)$

$x = 25 and 41$ $x=25 and 41$
Since the unit of length cannot be -ve, then $x = 41$ $x=41$ only
e.g, $a = x - 32$ $a=x-32$ , if we take $x = 25, a = - 7$ $x=25, a=-7$ (-ve value).

Therefore,
$a = 41 - 32 = 9, b = 41 - 1 = 40, c = 41$ $a=41-32=9, b=41-1=40, c=41$

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If c is the measure of the hypotenuse of a right triangle, how do you find each missing measure given a=x-32, b=x-1, c=x?

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