Fill the squares #2

Let the numbers be:

$\quad \quad \begin{matrix} A & B & C \\ D & E & F \\ G & H & I \end{matrix}$

Start with the middle column: $B+E-H =9$ . There are only 2 likely cases: $(B+E=3)\times (H=1)$ or $(B+E=9) \times (H=1)$ . In both cases $H = 1$ .

Case 1 is not acceptable because $E$ is too small for $D \times E - F = 29$ .

For case 2, $(B,E)$ cannot be $(8,1)$ , because $H=1$ . Let's start with $(7,2)$ ; $2$ is too small for $E$ for $D \times E - F = 29$ . Therefore, $B=2, E=7$ .

Now, consider $D \times 7 - F = 29$ . Then $D$ must be $5$ and then $F=6$ .

Now, $C+6-I=1$ . Since $H=1, B=2$ , then $C=3,4$ so that $I = 8,9$ . If $C=3$ , then $A=2=B$ , which is unacceptable. Therefore, $C=4$ , and $I = 9$ and $A = 3$ .

And it leaves $G=8$ . Since $3-5-8=-10$ and $8 \times 1 - 9 = -1$ are true, therefore, $G=8$ is correct.

The sum of the two diagonals $= A+E+I+C+E+G$

$= 3 +7+9 + 4 + 7 + 8 = \boxed{38}$ .