# If a right triangle’s side lengths are a, b, and c, and b−a=2, c−b=2, what is the area minus the perimeter?

Oct 29, 2017

${\text{Area"_triangle - "Perimeter}}_{\triangle} = 0$

#### Explanation:

For a right triangle with sides $a , b , \text{ and } c$
color(white)("xx"){:("If",b-a=2,rarr,b=a+2,,), ("If",c-b=2,rarr,c=b+2,rarr,c=a+4) :}

Since $c > b > a$, the hypotenuse must be of length $c$

${a}^{2} + {b}^{2} = {c}^{2}$

${a}^{2} + {\left(a + 2\right)}^{2} = {\left(a + 4\right)}^{2}$

${a}^{2} + {a}^{2} + 4 a + 4 = {a}^{2} + 8 a + 16$

$2 {a}^{2} + 4 a + 4 = {a}^{2} + 8 a + 16$

${a}^{2} - 4 a - 12 = 0$

$\left(a - 6\right) \left(a + 2\right) = 0$

{: (rarr,(a-6)=0," or ",(a+2)=0), (,rarr a=6,,rarr a=-2), (,,,"impossible") :}

Therefore
$\textcolor{w h i t e}{\text{XXX}} a = 6$,
$\textcolor{w h i t e}{\text{XXX}} b = 8$, and
$\textcolor{w h i t e}{\text{XXX}} c = 12$

${\text{Perimeter}}_{\triangle} = 6 + 8 + 12 = 24$

${\text{Area}}_{\triangle} = \frac{6 \times 8}{2} = 24$

${\text{Area"_triangle - "Perimeter}}_{\triangle} = 24 - 24 = 0$