1. Given sides of the triangle

   The three side lengths are:

   $$$$

5r, 5r^2$$

Since side lengths must be positive, we need: $r>0$

2. Use triangle inequality

   For three positive numbers to form a triangle, the sum of any two sides must be greater than the third.

   So we need: $5+5r>5r^2$ $5+5r^2>5r$ $5r+5r^2>5$

   Divide each inequality by $5$: $1+r>r^2 \quad ...(1)$ $1+r^2>r \quad ...(2)$ $r+r^2>1 \quad ...(3)$

3. Simplify each condition

   From (1):

   $r^2-r-1<0$ The roots of $r^2-r-1=0$ are: $r=\frac{1\pm\sqrt{5}}{2}$ Since the parabola opens upward, $\frac{1-\sqrt{5}}{2}<r<\frac{1+\sqrt{5}}{2}$

   Because $r>0$, this becomes: $0<r<\frac{1+\sqrt{5}}{2}$

   From (2):

   $r^2-r+1>0$ Its discriminant is: $(-1)^2-4(1)(1)=1-4=-3<0$ So this is always true for all real $r$.

   From (3):

   $r^2+r-1>0$ The roots of $r^2+r-1=0$ are: $r=\frac{-1\pm\sqrt{5}}{2}$ Since the parabola opens upward, $r<\frac{-1-\sqrt{5}}{2} \quad \text{or} \quad r>\frac{-1+\sqrt{5}}{2}$ As $r>0$, this gives: $r>\frac{\sqrt{5}-1}{2}$

4. Combine all conditions

   Therefore, $\frac{\sqrt{5}-1}{2}<r<\frac{\sqrt{5}+1}{2}$

   Numerically, $0.618<r<1.618$

5. Check the options

   • A: $\frac{7}{4}=1.75$ which is greater than $1.618$  cannot form a triangle.
   • B: $\frac{5}{4}=1.25$ lies in the interval  possible.
   • C: $\frac{3}{4}=0.75$ lies in the interval  possible.
   • D: $\frac{3}{2}=1.5$ lies in the interval  possible.

6. Conclusion

   Hence, $r$ cannot be equal to: $\boxed{\frac{7}{4}}$

   So the correct option is A.