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...............

 A: We start as you did. Use Heron's Formula to find the area of the triangle. It is $(12)(7)$. 
From that we find, as you did, that the triangle has height $8$. 
Drop a perpendicular from the top to the bottom, meeting the bottom at $W$. This divides the big triangle into two Pythagorean triangles with bases $6$ and $15$. (The numbers are "nice." But even if not so nice, we could have found them by using the Pythagorean Theorem.)
Let $x$ be the side length of the square. The point $W$ divides the bottom of the square into two parts.  Let $s$ be the length of the left part, and $t$ the length of the right part. Then $s+t=x$.
By similar triangles, 
$$\frac{8}{6}=\frac{x}{6-s}=\frac{s+t}{6-s}.$$
By similar triangles,
$$\frac{8}{15}=\frac{x}{15-t}=\frac{s+t}{15-t}.$$
We end up with two linear equations in $s$ and $t$. Solve. Then $x=s+t$, so we can find the perimeter of the square.
A: Let $AB=x$. Then the area of trapezoid $QABR=\dfrac{x(21+x)}2$. The area of triangle $ABP=\dfrac{x(8-x)}2$. The sum of these is the area of the whole triangle, which you have already calculated. Solve for x.
EDIT: More simply, note that since the little triangle ABP is similar to the whole triangle, it's base, $AB=\dfrac{21(8-x)}8$. So set this equal to x and solve.
A: 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 $ABCD= 4x = 4\cdot5.8= 23.2.$
