Suppose that, in the triangle ABC, the hypothenuse is AC. Let's call its length [tex ] \overline{AC} = x [/tex].
The hypothenuse is 2 ft longer than one of the legs (say AB, for example). This means that
![\overline{AB} = x-2](https://img.qammunity.org/2019/formulas/mathematics/high-school/xa1ov5ftpryy17gkmxwbhyhtgpy7040hlw.png)
Finally, we can find the other leg with the pythagorean theorem:
![\overline{BC} = \sqrt{\overline{AC}^2 - \overline{AB}^2} = √(x^2 - (x-2)^2) = √(x^2 - x^2 + 4x - 4) = √(4x-4) = √(4(x-1)) = 2√(x-1)](https://img.qammunity.org/2019/formulas/mathematics/high-school/j9tfwzm53g0gtpmh5c1pbxn58oxugl7f7g.png)
So, the perimeter (i.e. the sum of the sides) is given by
![x + (x-2) + 2√(x-1) = 364 \iff 2x - 2 + 2√(x-1) = 364](https://img.qammunity.org/2019/formulas/mathematics/high-school/brywud462a1wv7o8nsggzcb6pkpkx1rlz6.png)
Isolate the square root to get
![2√(x-1) = 2 - 2x + 364 = 366 - 2x](https://img.qammunity.org/2019/formulas/mathematics/high-school/v1jf7uv040h9xpqpmpa8qpl2589epgbz0z.png)
Divide all sides by 2:
![√(x-1) = 183 - x](https://img.qammunity.org/2019/formulas/mathematics/high-school/likgntwrnis8hcpazqxkwxq6m5wgr7xtiy.png)
Square both sides:
![x-1 = x^2 - 366 x + 33489 \iff x^2 - 367x + 33490 = 0](https://img.qammunity.org/2019/formulas/mathematics/high-school/gzqe17qnomz41666dygz8urerll3zwec5i.png)
This equation has solutions
or
. These solutions lead to, in the first case,
![\overline{AC} = x = 170,\quad \overline{AB} = x-2 = 168,\quad \overline{BC} = 2√(x-1) = 2√(169) = 2\cdot 13 = 26](https://img.qammunity.org/2019/formulas/mathematics/high-school/vsdnbcympxagrrhih5c0o8a1yx8qtdtyuh.png)
In the second case, you have
![\overline{AC} = x = 197,\quad \overline{AB} = x-2 = 195,\quad \overline{BC} = 2√(x-1) = 2√(196) = 2\cdot 14 = 28](https://img.qammunity.org/2019/formulas/mathematics/high-school/sum8di95vfk1xd4qfbte561vkeo6xwil4c.png)