Find the length of the rectangle and its perimeter if the width is 10 dm and the area is 120 dm2.

Let’s designate everything that is given to us: S = 120 dm ^ 2, b (width) = 10 dm
We need to find: a (length) and P (perimeter)
Solution: S = ab, substitute what we know 120 = 10a, a = 120/10 = 12 dm;
P = (a + b) × 2 and we know everything, so we substitute: P = (12 + 10) × 2 = 22 × 2 = 44 dm
Answer: length = 12 dm, perimeter = 44 dm.