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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.

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2   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
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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   May 14, 2020
#2
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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