"the length of a rectangle is 10 m greater than twice its width". if the lengths were doubled and the widths were halved, the perimeter of the new rectangle would be 80 m more than the perimeter of the original rectangle. what are the dimensions of the original rectangle?

the original length of the rectangle is 50 m

the original width of the rectangle is 20 m

Step-by-step explanation:

Let L represent the original length of the rectangle

Let W represent the original width of the rectangle.

The length of a rectangle is 10 m greater than twice its width. This means that

L = 2W + 10

Perimeter = 2(L + W)

= 2(2W + 10 + W) = 6W + 20

If the lengths were doubled and the widths were halved, the perimeter of the new rectangle would be 80 m more than the perimeter of the original rectangle. This means that

2(2L + W/2) = 6W + 20 + 80

4L + W = 6W + 100 - - - - - - - 1

Substituting L = 2W + 10, it becomes

4(2W + 10) + W = 6W + 100

8W + 40 + W = 6W + 100

8W + W - 6W = 100 - 40

3W = 60

W = 60/3 = 20

Substituting W = 20 into L = 2W + 10, it becomes

L = 2 × 20 + 10 = 50

the answers c