# If c is the measure of the hypotenuse of a right triangle, how do you find each missing measure given a=x-47, b=x, c=x+2?

Feb 8, 2017

$a = 2 + 9 \sqrt{57}$
$b = 49 + 9 \sqrt{57}$
$c = 51 + 9 \sqrt{57}$

#### Explanation:

$a = x - 47$
$b = x$
$c = x + 2$

and we know that the pythagoras theorem is:
${a}^{2} + {b}^{2} = {c}^{2}$

so if we fill that in:
${\left(x - 47\right)}^{2} + {x}^{2} = {\left(x + 2\right)}^{2}$

from here we can see that:
${a}^{2} = {x}^{2} - 94 x - 2209$
${b}^{2} = {x}^{2}$
${c}^{2} = {x}^{2} + 4 x + 4$

$2 {x}^{2} - 94 x - 2209 = {x}^{2} = 4 x + 4$
$x 62 - 98 x - 2216 = 0$

The ABC Formula tells us that in order to calculate $x$ we'll use the following method:

x = (-b ± sqrt(b^2 - 4ac))/(2a)
with $a = 1$, $b = - 98$ and $c = - 2216$

when we fill that in we get:
$x = \frac{98 + \sqrt{- {98}^{2} - \left(4 \cdot - 2216\right)}}{2} = 49 + 9 \sqrt{57}$
or
$x = \frac{98 - \sqrt{- {98}^{2} - \left(4 \cdot - 2216\right)}}{2} = 49 - 9 \sqrt{57}$

$49 - 9 \sqrt{57}$ equals to a negative number, and since we know that x is the length of side b we know that $x$ cannot be negative.
therefore we know that $x = 49 + 9 \sqrt{57}$

we fill that in for the sides:

$a = x - 47$ $\to$ $a = 2 + 9 \sqrt{57}$
$b = x \to b = 49 + 9 \sqrt{57}$
$c = x + 2 \to c = 51 + 9 \sqrt{57}$