# 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?

Jan 25, 2017

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

#### Explanation:

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

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

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