The length of a rectangle is 2 cm more than its width. The perimeter of this rectangle is 40 cm. Find the area of the rectangle.

1. Problem Description

The length of a rectangle is 2 cm more than its width. The perimeter of this rectangle is 40 cm. Find the area of the rectangle.

2. Solution Steps

Let $w$ be the width of the rectangle and $l$ be the length of the rectangle. We are given that the length is 2 cm more than the width, so $l = w + 2$.

The perimeter of a rectangle is given by the formula:

$P = 2l + 2w$

We are given that the perimeter is 40 cm, so

$40 = 2l + 2w$

Substitute $l = w + 2$ into the perimeter equation:

$40 = 2(w+2) + 2w$

$40 = 2w + 4 + 2w$

$40 = 4w + 4$

$36 = 4w$

$w = \frac{36}{4}$

$w = 9$ cm

Now we can find the length:

$l = w + 2 = 9 + 2 = 11$ cm

The area of a rectangle is given by the formula:

$A = l \times w$

$A = 11 \times 9 = 99$

The area of the rectangle is 99 square centimeters.

3. Final Answer

99 cm$^2$