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1. In triangle XYZ, we have

2. A square is inscribed in a right triangle, as shown below. The side lengths of the triangle are 3, 4, and 5. Find the side length of the square.

3.In the diagram below, we have

 Feb 9, 2020

Best Answer 

 #1
avatar+21544 
+1

I'm assuming that you want help with problem #2.

 

Call the bottom-left vertex A, the bottom-right vertex B, and the top vertex C.

 

Label the points where the square hits the triangle P, Q, R, and S in this fashion:

-- along AB, let the points of the base be, in order, A, P, Q, and B

-- along BC, let the point of intersection be R, and

-- along AC, let the point of intersection be S.

 

Given:  AC = 3, CB = 4, and AB = 5

 

Triangle(APS) is similar to triangle(ACB), so we can call the length of AP = 3x, PS = 4x, and AD = 5x.

Triangle(BQR) is similar to triangle(BCA), so we can call the length of BQ = 4y, BR = 5y, and QR = 3y.

Triangle(CSR) is similar to triangle(CAB), so we can call the length of SR = 5z, CS = 3z, and CR = 4z.

 

For side(AC), we have AC = AS + SC  =  5x + 3z                   --->            5x + 3z = 3.        (1)

For side(AB), we have AB = AP + PQ + QB  =  3x + 5z + 4y   --->   3x + 4y + 5z  =  5.      (2)

For side(BC), we have BC = BR + RC  =  5y + 4z                   --->            5y + 4z  =  4.      (3)

 

Since we have a square:  SP  =  SR  =  QR     --->     4x  =  5z  =  3y

 

Using  5z  =  3y     --->     z  =  3y/5

Substituting this into equation 3:  5y + 4z  =  4   --->   5y + 4(3y/5)  =  4   --->   25y+ 12y  =  20

   --->      37y  =  20     --->     y  =  20/37.

Therefore, since the side of the square is  3y,  side  =  3(20/37)  =  60/37.

 Feb 10, 2020
 #1
avatar+21544 
+1
Best Answer

I'm assuming that you want help with problem #2.

 

Call the bottom-left vertex A, the bottom-right vertex B, and the top vertex C.

 

Label the points where the square hits the triangle P, Q, R, and S in this fashion:

-- along AB, let the points of the base be, in order, A, P, Q, and B

-- along BC, let the point of intersection be R, and

-- along AC, let the point of intersection be S.

 

Given:  AC = 3, CB = 4, and AB = 5

 

Triangle(APS) is similar to triangle(ACB), so we can call the length of AP = 3x, PS = 4x, and AD = 5x.

Triangle(BQR) is similar to triangle(BCA), so we can call the length of BQ = 4y, BR = 5y, and QR = 3y.

Triangle(CSR) is similar to triangle(CAB), so we can call the length of SR = 5z, CS = 3z, and CR = 4z.

 

For side(AC), we have AC = AS + SC  =  5x + 3z                   --->            5x + 3z = 3.        (1)

For side(AB), we have AB = AP + PQ + QB  =  3x + 5z + 4y   --->   3x + 4y + 5z  =  5.      (2)

For side(BC), we have BC = BR + RC  =  5y + 4z                   --->            5y + 4z  =  4.      (3)

 

Since we have a square:  SP  =  SR  =  QR     --->     4x  =  5z  =  3y

 

Using  5z  =  3y     --->     z  =  3y/5

Substituting this into equation 3:  5y + 4z  =  4   --->   5y + 4(3y/5)  =  4   --->   25y+ 12y  =  20

   --->      37y  =  20     --->     y  =  20/37.

Therefore, since the side of the square is  3y,  side  =  3(20/37)  =  60/37.

geno3141 Feb 10, 2020

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