Square up to this Triangle

With Heron's formula we get $A=84$ .

Let $AD$ be the altitude from $A$ to $BC$ .

Then we get the altitude from the area $AD=12$ .

Let the length of the square be $x$ .

$\Delta ABC\sim \Delta ADE$ then $\frac {AE}{AC}=\frac {DE}{BC}=\frac {x}{14}$ .

$\Delta ADC\sim \Delta EFC$ then $\frac {CE}{AC}=\frac {EF}{AD}=\frac {x}{12}$ .

We know that $\frac {AE}{AC}+\frac {CE}{AC}=1$ .

Then $\frac {x}{14}+\frac {x}{12}=1$ .

Then $x=\frac{84}{13}$ .

Draw the altitude AI of the triangle which divides into two right angled triangles. By pythagorean triples, they are 15-12-9 and 13-12-5 sided triangles. So AI = 12 (Alternatively AI can be calculated after using Herons Formula to get area.)

Area of triangle ABC = 1/2 bh= 84.

Area of triangle ABC= Area of ADE + ( combined Area of BDG and EFC ) + x^2

84= 1/2x (12-x) + 1/2x(14-x) + x^2

x= 6.46