Let’s call the a+3 length ‘c’ and the base ‘B’.

The area of the triangle is half base times height, and that can be done in two different ways, so ac = 4B.

Using Pythagoras, c^2 + a^2 = B^2

Let’s expand (c-a)^2:

(c-a)^2 = c^2 + a^2 – 2ac

But we know that (c-a) is equal to 3, and we know the values of c^2+a^2 and ac in terms of B, so:

9 = B^2 – 8B

Solving this quadratic we get B is 9 or -1. As it’s a length it must be positive, so the base B is equal to 9.

If you’re interested, the value of a is 3(sqrt(17)-1)/2, which is about 4.68.