perimeter of square inscribed in the triagle

Question

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

Answer 1

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.

Answer 2

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.