SOLUTION: The combined perimeters of the 2 squares is 52 in. When they are placed together, the third figure is formed. It has an outer perimeter of 42 in. Find the sides.
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Question 827182: The combined perimeters of the 2 squares is 52 in. When they are placed together, the third figure is formed. It has an outer perimeter of 42 in. Find the sides. Answer by LinnW(1048) (Show Source):
You can put this solution on YOUR website! S = side length of smaller
L = side length of larger
4S + 4L = 52
Now consider the figure where the two squares are set next to each other.
This figure will have 3 sides of length L, 3 sides of length S
and additional perimeter where the squares join.
The area where they join can be represented by L - S
So the equation for the new figure is
3L + 3S + L - ( L-S ) = 42 We are subtracting the covered portion.
3L + 3S + L - L + S = 42
3L + 4S = 42
Now we have two equations
4S + 4L = 52
4S + 3L = 42
subtract the bottom from the top
L = 10
Substitute L = 10 in 4S + 3L = 42
4S + 3(10) = 42
4S + 30 = 42
subtract 30 from each side
4S = 12
S = 4
So the smaller square is 4 x 4 and the larger 10 x 10
