The perimeter of a triangle is 58 centimeters. The longest side is 2 centimeters less than the sum of the other two sides. Twice the shortest is 12 centimeters less than the longest side. Find the length of each side of the triangles.
Solution
Let x be the longest side.
Let y be the shortest side.
Let z be the last side.
The perimeter of a triangle is 58 centimeters.
The equation for the sentence above is the following:
x + y + z = 58 equation 1
The longest side is 2 centimeters less than the sum of the other two sides.
The equation for the sentence above is the following:
x = y + z - 2 equation 2
Twice the shortest is 12 centimeters less than the longest side.
The equation for the sentence above is the following:
2y = x - 12 equation 3
You now have a system to solve
x + y + z = 58 equation 1
x = y + z - 2 equation 2
2y = x - 12 equation 3
Equation 2 is the same as x - y - z = -2
The system becomes
x + y + z = 58 equation 1
x - y - z = - 2 equation 2
2y = x - 12 equation 3
equation 1 + equation 2 gives the following
2x = 56
x = 28
Use equation 3 to get y
2y = 28 - 12
2y = 16
y = 8
Use equation 1 to get z
28 + 8 + z = 58
36 + z = 58
z = 58 - 36
z = 22
The longest side of this triangle is 28, the shortest side is 8, and the third side is 22
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