If we call the base b, and the other sides a and c, such that a>c (a can’t be equal to c as the mirror line would be parallel with the base). We want a and c to be as close as possible, so that the angle of the mirror line is as shallow as possible. Since they are integers the smallest difference is 1, so let’s say a-1=c. With that being the case the length that we want to be 45 will be bc (in general it would be bc/(a-c)). To minimise the perimeter we want b to be roughly double c. If we had said we wanted the dashed line to be 50 instead of 45, we could achieve that exactly with b=10, c=5 and a=6.

Here the closest is b=9, c=5, a=6, with a perimeter of 20, and area of 10*sqrt(2).