Can we find other triangles with different integer sides which have equal integer area and perimeter using this method?

Question

A few days ago, I was reading this Wikipedia page and this part caught my eyes:

As W. A. Whitworth and D. Biddle proved in 1904, there are exactly three solutions, beyond the right triangles already listed, with sides $(6,25,29)$, $(7,15,20)$, and $(9,10,17)$.

I tried to construct these three triangles and I found I can construct them all by drawing a great right triangle which have a smaller right triangle inside. For example, to construct the triangle with sides $(9,10,17)$, All I need is a right triangle with sides $(8,15,17)$, which has a right triangle with sides $(6,8,10)$ inside.

Or to construct a triangle with sides $(7,15,20)$, one needs a triangle with sides $(12,16,20)$, with a triangle with sides $(9,12,15)$ inside.

And finally, to construct a triangle with sides $(6,25,29)$ we should have a triangle with sides $(20,21,29)$ and a triangle with sides $(15,20,25)$ inside.

My question: Can we construct other triangles with equal area and perimeters using this method? Is there a general formula or rule to do so?

Update: I now figured out if we call the sides of the bigger right triangle $(a,b,c)$ $a<b$, and the sides of the smaller one $(d,a,e)$ $d<a$, then we need the triangle of interest to have sides $(b-d,e,c)$.

Answer 1

The following code shows there are only 5 triangles with sides up to 1000 that have equal perimeter and area. You could try larger numbers but I think it would just waste computer time. Only 3 are not Pythagorean triples and you said W. A. Whitworth and D. Biddle proved in 1904, there are exactly three solutions... [that are not right triangles].

130 for a1 =  1 to 1000
140 for b1 = a1 to 1000
150 for c1 = b1 to 1000
160 s1 = a1+b1+c1
170 s2 = s1/2
180 r1 = sqr(s2*(s2-a1)*(s2-b1)*(s2-c1))
190 if r1 = s1
200     print a1,b1,c1
210 endif
220 next c1
230 next b1
240 next a1


>run
5   12  13 
6   8   10 
6   25  29 
7   15  20 
9   10  17 
>