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

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.

• How about showing some of your attempts so that someone might point out where you’re going wrong? – amd Jan 17 at 1:38
• What does "the other leg" mean? – lulu Jan 17 at 1:38
• By other leg, do you mean the "other legs?" – Mohammad Zuhair Khan Jan 17 at 1:40
• Suppose that we call the length of the hypotenuse $h$ and the other legs $x$ and $y$, and say that $h$ is 2 units longer than the side with length $x$. i. What does the fact that it's a right triangle tell you about $h, x, y$? ii. What does the fact that the hypotenuse is 2 units longer than one leg tell you about $h, x, y$? iii. What does the fact that the perimeter is $84$ tell you about $h, x, y$? You can write these three equations into your question by clicking "edit" just below the question; then you can show us how you tried to solve them, too. – John Hughes Jan 17 at 1:46
• @amd I edited the question with more information. – William Coulter Jan 18 at 4:39

Let the sides of the right triangle be $$x,y,x+2$$.
$$2x+y=82 \tag{1}$$
$$x^2 + y^2 = (x+2)^2 \tag{2}$$ $$\implies x^2 + y^2 = x^2 +4x+4$$ $$\implies y^2 = 4x+4$$
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.
• It can be simplified to $x = 41 - \frac{y}{2}$. Try that, and further edit your question if you get stuck again. Good work so far! – John Hughes Jan 18 at 12:14