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# 3.In the diagram below, we have Feb 9, 2020

#1
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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
+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.

geno3141 Feb 10, 2020