The triangle system

$\text{Let the sides be }L,B \text{ where }L>B . \\ \text{It is given That } LB = 154 \text{ and } L^2+B^2=317 \quad \rightarrow \text{Equation 1} \\ (L^2-B^2)^2=(L^2+B^2)^2-4\cdot L^2B^2 \\ (L^2-B^2)^2= 317^2-308^2 = 9\cdot 625 \\ L^2-B^2=75 \quad \rightarrow \text{Equation 2}\\ \text{After adding the Two Equations : } L^2=196 \\ \boxed{L=14}$

A triangle with hypotenuse is a right-angle triangle. Let lengths of the two legs be $a$ and $b$ . Then we have:

$\begin{cases} \frac 12 ab = 77 & \implies ab = 154 \\ \sqrt{a^2+b^2} = \sqrt{317} & \implies a^2+b^2 = 317 \end{cases}$

Consider:

$\begin{aligned} (a+b)^2 & = a^2 + b^2 + 2ab = 317 + 308 = 625 \\ \implies a + b & = \sqrt{625} = 25 \quad ...(1) \\ (a-b)^2 & = a^2 + b^2 - 2ab = 317 - 308 = 9 \\ \implies a - b & = \sqrt 9 = 3 \quad ...(2) \\ (1)+(2): \quad 2a & = 28 \\ \implies a & = 14 \\ b & = a-3 = 11 \end{aligned}$

The longest leg is $a= \boxed{14}$ .

$\large \frac { 1 }{ 2 } bh=77\\ \\ \sqrt { { b }^{ 2 }+{ h }^{ 2 } } =\sqrt { 317 } \\ { b }^{ 2 }+{ h }^{ 2 }=317\\ b=\sqrt { 317-{ h }^{ 2 } } \\ \\ now\quad plug\quad in\\ \frac { 1 }{ 2 } h\sqrt { 317-{ h }^{ 2 } } =77\\ h\sqrt { 317-{ h }^{ 2 } } =154\\ square\quad both\quad sides\\ { h }^{ 2 }\left( 317-{ h }^{ 2 } \right) =23716\quad \\ distribute\\ { h }^{ 2 }317-{ h }^{ 4 }=23716\\ subtract\quad 23716\quad from\quad both\quad sides\\ -{ h }^{ 4 }+317{ h }^{ 2 }-23716=0\\ factor\quad -1\\ { h }^{ 4 }-317{ h }^{ 2 }+23716=0\\ let\quad a={ h }^{ 2 }\\ { a }^{ 2 }-317a+23716=0\\ factor\quad (a-121)(a-196)=0\\ now\quad replace\quad a\quad with\quad { h }^{ 2 }\\ ({ h }^{ 2 }-121)({ h }^{ 2 }-196)=0\\ factor\quad again\\ (h-11)(h+11)(h-14)(h+14)=0\\ h=11,-11,14,-14\\ Since\quad we\quad are\quad looking\quad for\quad a\quad unit\quad of\quad length\quad we\quad only\quad take\quad the\quad positive.\\ Therefore\quad the\quad lengths\quad of\quad the\quad legs\quad are\quad the\quad triangle\quad are\quad 11,\quad 14.$