A square is inscribed in a right triangle, as shown below. The legs of the triangle are 2 and 3. Find the side length of the square.

Guest May 14, 2020

#1**+1 **

See the following image :

Using some trig

tan CAB = 3/2

tan CAB = S / AD where S = side of the square...let AD = x

3/2 = S/x

x = (2/3)S

And

tan CBA = 2/3

tan CBA = S/ (BC - (S + x)) BC = sqrt (13)

(2/3) = S / (sqrt (13) - (S + (2/3)S) sub in for x

(2/3) = S / ( sqrt (13) - (5/3)S)

(2/3) (sqrt (13) -(5/3)S) = S

(2/3)sqrt (13) - (10/9)S = S multiply through by 9

6sqrt (13) - 10S = 9S

6sqrt (13) = 19S

6sqrt (13) / 19 = S

CPhill May 14, 2020

#2**0 **

Getting the same answer as CPhill's but without trig.

By AA, all four triangles, the three smaller one and the full one are all similar.

Using CPhill's diagram, triangle(ABC) ~ triangle(FGC) ~ triangle(AFD).

Since triangle(ABC) is a right triangle, AB = sqrt(13).

Since the ratio of AC : CB : BA = 2 : 3 : sqrt(13),

FC : CG : GF = 2x : 3x : sqrt(13)·x (for some value of x, that we can find later)

and DA : DF : FA = 2y : 3y : sqrt(13)·y (for some value of , that we can find later)

Since DFGE is a square: FG = DF ---> sqrt(13)·x = 3y

---> y = (sqrt(13) / 3) · x

This makes AF = sqrt(13) · sqrt(13) / 3 · x = ( 13/3 )·x

AC = AF + FC ---> 2 = (13/3)x + 2x ---> 6 = 13x + 6x ---> 6 = 19x ---> x = 6/19

Since FG = sqrt(13)·x ---> FG = sqrt(13)·(6/9) = **6·sqrt(13)/19**

Same answer as CPhill's, so I feel confident!

geno3141 May 14, 2020