The first side of a triangle is 8 meters longer than twice the second side. The third side is 3 times the second side. The perimeter is 128 meters. Find the lengths of the three sides.
Let x = second side
First side = 2x +8
Third side = 3x
x + (2x + 8) + 3x = 128
Solve for x, then the other sides.
I hope this helps.
Let b = 2nd side
2b + 8 = 1st side
3b = 3rd side
x + (2x + 8) + 3x = 128
6b + 8 = 128
*REMOVE 8 FROM BOTH SIDES*
6b = 120
*DIVIDE*
b = 20
To solve this problem, let's break it down step by step:
Step 1: Assign variables
Let's assume the second side of the triangle is represented by 'x'.
Therefore, the first side is 8 meters longer than twice the second side, which can be represented as 2x + 8.
Lastly, the third side is 3 times the second side, which can be represented as 3x.
Step 2: Set up the equation
The perimeter of a triangle is the sum of all its sides. In this case, the perimeter is given as 128 meters. So we can set up the equation:
x + (2x + 8) + 3x = 128
Step 3: Solve the equation
Combine like terms:
6x + 8 = 128
Subtract 8 from both sides:
6x = 120
Divide both sides by 6:
x = 20
Step 4: Find the lengths of the sides
Now that we know x is 20, we can substitute it back into our expressions for the sides:
First side = 2x + 8 = 2(20) + 8 = 40 + 8 = 48 meters
Second side = x = 20 meters
Third side = 3x = 3(20) = 60 meters
So, the lengths of the three sides of the triangle are 48 meters, 20 meters, and 60 meters, respectively.