From the text: A shape was said to be equable in my last two articles if the value of its area is equal to the value of its perimeter and I was particularly interested in shapes that had integers for all their side lengths. In the previous article, however, we found that no equable isosceles triangle exists, although a nice example did exist with a base of $12$ and equal sides being $7\frac{1}{2}$. We will now consider the general case for a triangle which has a slightly different approach to that in the last article. However, we will begin with Heron’s formula again.