Determining the area of a right triangle, perimeter given, hypotenuse value given in terms of one of the legs.

Question

The problem states:

Right Triangle- perimeter of $84$, and the hypotenuse is $2$ greater than the other leg. Find the area of this triangle.

I have tried different methods of solving this problem using Pythagorean Theorem and systems of equations, but cannot find any of the side lengths or the area of the right triangle. I looked for similar problems on StackExchange and around the internet, but could not find anything.

Does anyone know anything that could help find the side lengths of the triangle and the area as well?
Method that I tried:

1. Made a system with the values given.
   \begin{align} a+b+c&=84 \\ c&=b+2 \end{align}
2. Substituted $c$ with $b+2$.
   \begin{align} a+b+b+2&=84 \\ a + 2b &= 82 & \text{subtracted $2$ from both sides}\\ a + a^2 - 4 &= 82 \end{align}
3. $c^2$ is $(b+2)(b+2)$, so I used Pythagorean Theorem to isolate one of the variables.
   \begin{align} a^2+b^2 &=c^2\\ a^2 + b^2 &=(b+2)(b+2)\\ a^2+b^2 &=b^2+2b+4\\ a^2&=2b+4 & \text{ (Subtracted $b^2$ from both sides) } \end{align} OR
   \begin{align} a^2-4&=2b \end{align}

I do not know what to do after this point.

Answer 1

Let the sides of the right triangle be $x,y,x+2$.

Given,

$2x+y=82 \tag{1}$

$x^2 + y^2 = (x+2)^2 \tag{2}$ $$\implies x^2 + y^2 = x^2 +4x+4 $$ $$\implies y^2 = 4x+4 $$

Now, substitute the value of $x$ from equation (1) in terms of $y,$ you will get a quadratic equation in $y$ whose roots can be easily found and hence, the sides and area.

Answer 2

$\\ \textbf{Finding triples, given perimeter using Euclid's formula}$ where $P=perimeter$

$$P=(m^2-n^2 )+2mn+(m^2+n^2 )=2m^2+2mn\implies n=\frac{P-2m^2}{2m}\quad where \quad \biggl\lceil\frac{\sqrt{P}}{2}\biggr\rceil\le m \le \biggl\lfloor\sqrt{\frac{P}{2}}\biggr\rfloor$$

Here, the lower limit ensures that $m>n$ and the upper limit insures that $n>0$.

Example: $$P=84\Rightarrow \biggl\lceil\frac{\sqrt{84}}{2} \biggr\rceil =5 \le m\le\biggl\lfloor\sqrt{\frac{84}{2}}\biggr\rfloor =6:\quad f(84,5)\notin\mathbb{N}\quad f(84,6)=1\Rightarrow F(6,1)=(35,12,37)$$

There are no other solutions for $P=84$.