Let W = width of rectangle
Let W + 2 = length of rectangle = L
Perimeter = 2L + 2W
56 = 2L + 2W
since L = W + 2, we can substitute and solve
56 = 2(W + 2) + 2W
56 = 2W + 4 + 2W
56 = 4W + 4
52 = 4W
W = 13
the length 2 more than 13, which is 15
Area = L • W
= 15 • 13
= 195