First, using Pythagoras's Theorem, we establish that these integer Heronian triangles are indeed proper: {{begin-eqn}} {{eqn | l = 6^2 + 25^2       | r = 661       | c =  }} {{eqn | o = \ne       | r = 29^2       | c = so not right-angled }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 7^2 + 15^2       | r = 274       | c =  }} {{eqn | o = \ne       | r = 20^2       | c = so not right-angled }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 9^2 + 10^2       | r = 181       | c =  }} {{eqn | o = \ne       | r = 17^2       | c = so not right-angled }} {{end-eqn}} Now we show they have area equal to perimeter. We use Heron's Formula throughout: :$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$ where: :$\AA$ denotes the area of the triangle :$a$, $b$ and $c$ denote the lengths of the sides of the triangle :$s = \dfrac {a + b + c} 2$ denotes the semiperimeter of the triangle. Thus we take the $3$ triangles in turn: {{begin-eqn}} {{eqn | n = 6, 25, 29       | l = s       | r = \frac {6 + 25 + 29} 2       | c =  }} {{eqn | r = 30       | c =  }} {{eqn | ll= \leadsto       | l = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }       | r = \sqrt {30 \paren {30 - 6} \paren {30 - 25} \paren {30 - 29} }       | c =  }} {{eqn | r = \sqrt {30 \times 24 \times 5 \times 1}       | c =  }} {{eqn | r = \sqrt {3600}       | c =  }} {{eqn | r = 60       | c =  }} {{eqn | r = 6 + 25 + 29       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | n = 7, 15, 20       | l = s       | r = \frac {7 + 15 + 20} 2       | c =  }} {{eqn | r = 21       | c =  }} {{eqn | ll= \leadsto       | l = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }       | r = \sqrt {21 \paren {21 - 7} \paren {21 - 15} \paren {21 - 20} }       | c =  }} {{eqn | r = \sqrt {21 \times 14 \times 6 \times 1}       | c =  }} {{eqn | r = \sqrt {1764}       | c =  }} {{eqn | r = 42       | c =  }} {{eqn | r = 7 + 15 + 20       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | n = 9, 10, 17       | l = s       | r = \frac {9 + 10 + 17} 2       | c =  }} {{eqn | r = 18       | c =  }} {{eqn | ll= \leadsto       | l = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }       | r = \sqrt {18 \paren {18 - 9} \paren {18 - 10} \paren {18 - 17} }       | c =  }} {{eqn | r = \sqrt {18 \times 9 \times 8 \times 1}       | c =  }} {{eqn | r = \sqrt {1296}       | c =  }} {{eqn | r = 36       | c =  }} {{eqn | r = 9 + 10 + 17       | c =  }} {{end-eqn}} It remains to be demonstrated that these are indeed the only such proper integer Heronian triangles which match the criterion. Let $\tuple {a, b, c}$ be the sides of such a triangle. Using Heron's Formula, we have: {{begin-eqn}} {{eqn | l = 2 s       | r = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} } }} {{eqn | ll= \leadsto        | l = 4 s^2       | r = s \paren {s - a} \paren {s - b} \paren {s - c} }} {{eqn | ll= \leadsto       | l = 4 s       | r = \paren {s - a} \paren {s - b} \paren {s - c} }} {{end-eqn}} Note that: :$\paren {s - a} + \paren {s - b} + \paren {s - c} = 3 s - a - b - c = s$ Hence by substituting $x = s - a$, $y = s - b$, $z = s - c$: :$4 \paren {x + y + z} = x y z$ By Semiperimeter of Integer Heronian Triangle is Composite, $s$ is an integer. Hence $s, x, y, z \in \N_{>0}$. By Triple with Product Quadruple the Sum, our equation has solutions: :$\tuple {1, 5, 24}, \tuple {1, 6, 14}, \tuple {1, 8, 9}, \tuple {2, 3, 10}, \tuple {2, 4, 6}$ Using: :$a = s - x = x + y + z - x = y + z$ :$b = s - y = x + z$ :$c = s - z = x + y$ the possible sets of side lengths are: :$\tuple {29, 25, 6}, \tuple {20, 15, 7}, \tuple {17, 10, 9}, \tuple {13, 12, 5}, \tuple {10, 8, 6}$ of which the final $2$ are Pythagorean Triples, so they are not proper Heronian triangles. Hence the result. {{qed}}