What's the measure of the side of the rombus in the figure below? For rereence:In a rhombus the sum of the measures of its diagonals is $70$ cm
and the radius of the inscribed circle is $12$ cm. Calculate the measure of the rhombus side,
My progress:
$CO +BO = \frac{70}{2} = 35$
$\frac{1}{h^2}=\frac{1}{a^2}+\frac{1}{c^2}\\
\frac{1}{144} = \frac{1}{(35-CO)^2}+\frac{1}{CO^2}(CO = x)\implies\\
\frac{1}{144} = \frac{2x^2-70x+1225}{x^2(x-35)^2}$
therefore (by Wolfran) $CO = 15 \implies OB = 20$
or
$CO = 20 \implies OB = 15\\
\therefore BC^2 = 15^2+20^2 \implies \boxed{BC = 25}$
Would there be a way to get a simpler equation?


$\mathsf{
AO(x)+OB(y) = \frac{70}{2}=35\\
(x+y)^2  = 35^2\implies x^2+y^2+2xy = 1225\\
Por~ métrica: x^2+y^2 = AB(a)^2\\
 a.h = x.y\implies 12a = xy\\
\therefore : a^2+24a-1225 = 0\\
\therefore \boxed{\color{red}a =AB =  25} }$
