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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.

 May 14, 2020
 #1
avatar+111456 
+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

 

cool cool cool

 May 14, 2020
 #2
avatar+21953 
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!

 May 14, 2020

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