Equable Isosceles Triangle

Let the length of the congruent sides of the isosceles triangle be $a$ and that of the base be $b$ . Then the perimeter is $2a+b$ and the area is $\frac 12b^2$ . Then we have:

$\begin{aligned} 2a + b & = \frac {b^2}2 & \small \color{#3D99F6} \text{By Pythagorean theorem }a = \sqrt{b^2 + \frac {b^2}4} = \frac {\sqrt 5}2b \\ (\sqrt 5 + 1)b & = \frac {b^2}2 \\ \implies b & = 2(\sqrt 5 + 1) \end{aligned}$

Therefore, the perimeter $(\sqrt 5 + 1)b = 2(\sqrt 5+1)^2 = 12+4\sqrt 5$ . $\implies p+q+r = 12+4+5 = \boxed{21}$ .

Let the equal sides be $a$ each and the base (and hence the height) be $b$ . Then $a=$ $\dfrac{b√5}{2}$ and $b^2=4a+2b$ . Solving this two, we get $b=2√5+2$ and $a=5+√5$ . Hence the perimeter is $2a+b=12+4√5$ , and $p=12$ , $q=4$ and $r=5$ . So $p+q+r=21$ . There is a typo. $5$ should be replaced by $r$