Answer: Each side of the original square paper was 42 inches long.
Explanation:
Let's start by using algebra to solve the problem.
Let x be the length of a side of the original square paper, in inches.
When Juan cut the paper vertically to make two rectangle pieces, he created two rectangles with different dimensions. However, we know that the perimeter of each rectangle is 63 inches.
The formula for the perimeter of a rectangle is:
Perimeter = 2 x Length + 2 x Width
We can use this formula to create two equations, one for each rectangle:
63 = 2L1 + 2W1
63 = 2L2 + 2W2
Since the original square paper was cut vertically, we know that the length and width of each rectangle must add up to x.
We can use this information to write two more equations:
L1 + W1 = x
L2 + W2 = x
Now we have four equations:
63 = 2L1 + 2W1
63 = 2L2 + 2W2
L1 + W1 = x
L2 + W2 = x
We have four equations and four unknowns (L1, W1, L2, W2), so we can solve for x.
First, let's simplify the equations by solving for one of the variables in terms of the others. We can solve the equations for L1 and L2:
L1 = x - W1
L2 = x - W2
Now we can substitute these expressions into the perimeter equations:
63 = 2(x - W1) + 2W1
63 = 2(x - W2) + 2W2
Simplifying, we get:
63 = 2x + 2W1
63 = 2x + 2W2
Subtracting 2x from both sides of each equation, we get:
2W1 = 63 - 2x
2W2 = 63 - 2x
Dividing both sides of each equation by 2, we get:
W1 = (63 - 2x)/2
W2 = (63 - 2x)/2
Now we can substitute these expressions into the equations for L1 and L2:
L1 = x - (63 - 2x)/2
L2 = x - (63 - 2x)/2
Simplifying, we get:
L1 = (3x - 63)/2
L2 = (3x - 63)/2
Now we can use the fact that the perimeter of each rectangle is 63 inches to create one more equation:
2L1 + 2W1 = 63
Substituting in the expressions for L1 and W1, we get:
2[(3x - 63)/2] + [(63 - 2x)/2] = 63
Simplifying, we get:
3x = 126
Dividing both sides by 3, we get:
x = 42
Therefore, each side of the original square paper was 42 inches long.