math question?

Question

In triangle ABC, AB = 14, BC = 15, and AC = 13. A square is inscribed in triangle ABC with one side on AB. What is the length of the side of the square?

Answer 1

Let the sides of the square =z
Let x be the length of the side AC between C and the square. The rest of the side = 13-x
Let y be the length of the side BC between C and the square. The rest of the side = 15-y
By the rules of similar triangles,
x/13=y/15=z/14
Two small right triangles are formed in the triangle ABC that is not occupied by the square. Let a be the length between A and the square on AB and let b be the length between B and the square on AB.
Then a+b+z=14
Moreover, by Phytag thm.
(13-x)^2=a^2+z^2
(15-y)^2=b^2+z^2
We have enough equations to find z.

Answer 2

c = 14, a = 15, b = 13 => s = (14+15+13)/2 = 21
area ABC = √s(s--a)(s--b)(s--c) = √21(6)(8)(7) = 84
height of ABC from C = 2Δ/AB = 2*84/21 = 8
now if side of square is x,
ΔABC = x^2 + (AB -- x)x/2 + x(height -- x)/2
=> x^2 + (14 -- x)x/2 + x(8 -- x)/2 = 84
=> 11x = 84
=> x = 84/11 = 7 7/11
side of square is 7 7/11 or 84/11 or 7.6364

Answer 3

25

Answer 4

36

Answer 5

6.62