Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $4,4,$ and $6$, while those of $T'$ have lengths $a,a,$ and $b$. What is the value of $b$?
$T$ has perimeter $14$ and (by the area formula $\frac b4\sqrt{4a^2-b^2}$) area $\frac64\sqrt{64-36}=3\sqrt7$.
From $2a+b=14$ we get $b=14-2a$, and the height to the base $b$ is $\sqrt{a^2-(7-a)^2}=\sqrt{7(2a-7)}$, so the area is $(7-a)\sqrt{7(2a-7)}=3\sqrt7$, i.e. $(7-a)^2(2a-7)=9$. With $t=7-a$ this is $2t^3-7t^2+9=(t-3)(2t-3)(t+1)=0$. The root $t=3$ gives $T$ itself ($a=4$), $t=-1$ gives $b<0$, and $t=\frac32$ gives $a=\frac{11}2$, $b=3$.
