The length of a rectangle is 10 centimeters more than its width. If both the length and the width are doubled, the resulting area becomes 576 square centimeters. What are the dimensions of the original rectangle?Select one:a. length = 18 cm; width = 8 cmb. length = 16 cm; width = 6 cmc. length = 14 cm; width = 4 cmd. length = 12 cm; width = 8 cm

To solve this problem, we need to set up two equations based on the information given in the problem.

1. The length of the rectangle is 10 centimeters more than its width. If we let W represent the width of the rectangle, then the length of the rectangle can be represented as W + 10.

2. If both the length and the width are doubled, the resulting area becomes 576 square centimeters. The area of a rectangle is calculated by multiplying the length by the width. So, if we double the length and the width, the area of the rectangle becomes (2W) * (2(W + 10)) = 576.

Solving the second equation:

4W * (W + 10) = 576 4W^2 + 40W - 576 = 0

Divide the entire equation by 4 to simplify:

W^2 + 10W - 144 = 0

This is a quadratic equation in the form of ax^2 + bx + c = 0. We can solve for W by factoring:

(W - 8)(W + 18) = 0

Setting each factor equal to zero gives the solutions W = 8 and W = -18. Since width cannot be negative, W = 8 is the only valid solution.

Substituting W = 8 into the first equation gives the length of the rectangle as 8 + 10 = 18.

So, the dimensions of the original rectangle are length = 18 cm and width = 8 cm, which corresponds to option a.