perimeter of square inscribed in the triagle In the figure given below, PQR is a triangle with sides PQ=10, PR=17, QR=21. ABCD is a square inscribed in the triangle. I want to find perimeter of square ABCD that is to find the length of side AB. But by using of basic high school geometry concepts, not by trigonometry.
I have drop perpendicular to side QR, and by using heron's formula i found its length 8. but i am confused what to do next. So, please help me. 
Any other solutions are expected with above limitation(to use basic high school concepts not by trignometry)
THANKS...............


Area of Triangle $PQR = 84$ by using semiperimeter formula. From this we get height of Triangle as 8. 
Say, side of square $= x$, hence $AB=BC=CD=AD=x$
Now take small triangle $APB$, in this height will be $8-x$. Area of Triangle = $\frac{x\cdot(8-x)}{2}$   -----(1)
Now, Area of Trapezoid $ABRQ = \frac{(21+x)\cdot x}{2}$            -------(2)
As we know, adding (1) & (2), we get area of Triangle $PQR$ i.e 84.
Hence, $$8x-x^2+21x+x^2 = 84\cdot2$$
$$29x = 168$$
 $$x  = 5.8$$
Hence Perimeter of Square $ABCD= 4x = 4\cdot5.8= 23.2.$
