jmc

Since the smaller triangle has hypotenuse 5, we guess that it is a 3-4-5 triangle. Sure enough, the area of a right triangle with legs of lengths 3 and 4 is $(3)(4)/2=6$, so this works. The area of the larger triangle is $150/6=25$ times the area of the smaller triangle, so its side lengths are $\sqrt{25}=5$ times as long as the side lengths of the smaller triangle. Therefore, the sum of the lengths of the legs of the larger triangle is $5(3+4)=\boxed{35}$.

Proof that the only possibility for the smaller triangle is that it is a 3-4-5 triangle: Let's call the legs of the smaller triangle $a$ and $b$ (with $b$ being the longer leg) and the hypotenuse of the smaller triangle $c$. Similarly, let's call the corresponding legs of the larger triangle $A$ and $B$ and the hypotenuse of the larger triangle $C$. Since the area of the smaller triangle is 6 square inches, we can say $$\frac{1}{2}ab=6.$$ Additionally, we are told that the hypotenuse of the smaller triangle is 5 inches, so $c=5$ and $$a^2+b^2=25.$$ Because $\frac{1}{2}ab=6$, we get $ab=12$ or $a=\frac{12}{b}$. We can now write the equation in terms of $b$. We get $$\begin{array}{rl}{a}^2+b^2& =25\\ {\left(\frac{12}{b}\right)}^2+b^2& =25\\ 12^2+b^4& =25b^2\\ b^4-25b^2+144& =0.\end{array}$$ Solving for $b$, we get $$b^4-25b^2+144=(b-4)(b+4)(b-3)(b+3)=0.$$ Since we said that $b$ is the longer of the two legs, $b=4$ and $a=3$. Therefore, the triangle must be a 3-4-5 right triangle.