smc

We know that triangle $RCQ$ is similar to triangle $ABC$. We draw a line to point $D$ on hypotenuse $AB$ such that $\angle QDR$ is $90^{\circ }$ and that $RDQC$ is a rectangle. Since triangle $RCQ$ is similar to triangle $ABC$, let $RC$ be $4x$ and $RD/CQ$ be $3x$. Now we have line segment $AQ$ = $8-3x$, and line segment $RB$ = $6-4x$. Since $BD+DA=AB$, we use simple algebra and Pythagorean Theorem to get $\sqrt{(3x{)}^2+(6-4x{)}^2}$ + $\sqrt{(4x{)}^2+(8-3x{)}^2}$ = $10$. Expanding and simplifying gives us $\sqrt{25x^2-48x+36}$ + $\sqrt{25x^2-48x+64}$ = $10$. Squaring both sides is not an option since it is both messy and time consuming, so we should proceed to subtracting both sides by $\sqrt{25x^2-48x+36}$. Now, we can square both sides and simplify to get $0=72-20\sqrt{25x^2-48x+36}$. Dividing both sides by $4$, we get $18-5\sqrt{25x^2-48x+36}$ = $0$. We then add $5\sqrt{25x^2-48x+36}$ to both sides to get $18=5\sqrt{25x^2-48x+36}$. Since this is very messy, let $25x^2-48x=y$. Squaring both sides, we get $324=25y+900,25y=-576$. Solving for $y$, we have $y=-23.04$. Plugging in $y$ as $25x^2-48x$, we have $25x^2-48x+23.04=0$. Using the quadratic equation, we get $\frac{48+0}{50}$. Therefore, $x=\frac{48}{50}$. Remember that our desired answer is the hypotenuse of the triangle $3x-4x-5x$. Since $5x$ is the hypotenuse, our answer is $\boxed{4.8}$