Question

A rectangle is 4 cm longer than it is wide. If the length and width are both

decreased by 2 cm, the area is decreased by 24 cm². Find the dimensions
of the original rectangle.

Answer 1

The original rectangle has a width of 5 cm and a length of 9 cm, with the width being 5 cm and the length being calculated by adding 4 cm to the width.

Step-by-step explanation:

Let's denote the width of the original rectangle as w cm. Therefore, the length would be w + 4 cm. We are told that when both the length and width are decreased by 2 cm, the new dimensions are (w - 2) cm and (w + 2) cm, respectively. The area of the original rectangle would be w(w + 4) cm², and the area of the modified rectangle is (w - 2)(w + 2) cm². The change in area is 24 cm², which gives us the equation:

w(w + 4) - (w - 2)(w + 2) = 24

Expanding and simplifying the equation:

w² + 4w - ((w - 2)(w + 2)) = 24

w² + 4w - (w² - 4) = 24

4w + 4 = 24

4w = 20

w = 5

Thus, the original width is 5 cm, and the original length is 9 cm (5 cm + 4 cm).

Answer 2

Let's use algebra to solve the problem.

Let x be the width of the rectangle in cm. Then, according to the problem, the length is 4 cm longer, so it is x + 4 cm.

The area of the original rectangle is:

A = length x width
A = (x + 4) x (x)
A = x^2 + 4x

When both the length and the width are decreased by 2 cm, the new area is:

A - 24 = (length - 2) x (width - 2)
x^2 + 4x - 24 = (x + 2) x (x - 2)
x^2 + 4x - 24 = x^2 - 4

Simplifying this equation, we get:

8x = 28
x = 3.5

Therefore, the width of the original rectangle is 3.5 cm, and the length is 4 cm longer, or 7.5 cm.

So the original rectangle has dimensions of 7.5 cm by 3.5 cm.