Let sides are a, a, h So, 4a + 4h + 4a = 2(a2+ah+ah) 2a + 2h + 2a =a2+2ah ⇒a2−4a=2h(1−a) ⇒(a2−1)+1−4(a−1)−4=2h(1−a) ⇒(a−1)(a+1)−4(a−1)−3=2h(1−a) ⇒2h=
3
a−1
+4−(a+1) So a = 2 & h = 2 are the only integral solution (a & h are positive integers)