Triangle + Square = Problem!

By Heron's Formula, the area(A) of the triangle is given by A = $\sqrt{[s(s - a)(s - b)(s - c)]}$ , where $s = \frac{(a + b + c)}{2}$ is the semi-perimeter of the triangle.

Then $s = \frac{(10 + 17 + 21)}{2} = 24$ , and $A = 84$ .

Now drop a perpendicular of length $h$ onto the side of length 21.

We also have $A =\frac{1}{2} × base × height$ .

Hence $A = \frac{21h}{2} = 84$ , from which $h = 8$ .

The triangle above the square is similar to the whole triangle.

$\Rightarrow$ Let the square have side of length $d$ .

Considering the ratio of altitude to base in each triangle, we have $\frac{8}{21} = \frac{(8 - d)}{d}$

Therefore the length of the side of the square is $\frac{168}{29}$ .

The triangle is build up of two right angled triangles. 10, 8, 6 + 15, 8, 17. 6+15=21.
So to the altitude on 21 is 8. If X is the side of the square, and X=r 21, also X= (1-r) 8. Therefore r=8/29. X=8/29*21=168/29. a+b=168+29=197.