# If a rectangle width is increased by 3, the rectangle becomes a square and the area of the...

## Question:

If a rectangle width is increased by 3, the rectangle becomes a square and the area of the rectangle is increased by {eq}54u^2, {/eq} what is the area of the original rectangle?

## The Area of a Rectangle:

A rectangle is a plane figure that has four sides. The opposite sides of a rectangle are equal, and each of the interior angles is 90 degrees. The area of a rectangle is the amount of space covered by the rectangular shape, and it is computed as {eq}A = l\times w {/eq}.

The area of a rectangle is given by:

• {eq}A = l\times w {/eq}, where {eq}l {/eq} is the length and {eq}w {/eq} is the width of the rectangle.

Let the original width of the rectangle in our question be:

• {eq}w {/eq} units.

If the width is increased by 3 such that the formed figure is a square, then each side of the formed square will have a length of:

• {eq}l_1 = w + 3 {/eq}

Recall that the length of the original rectangle was not changed. Thus, the length of the formed square is equal to the length of the original square.

• {eq}l = l_1= w + 3 {/eq}
• {eq}l= w + 3 {/eq}

The area of the formed square will be equal to:

• {eq}A1 = l^2 = (w + 3)^2 {/eq}

And the area of the original rectangle is:

• {eq}A = l\times w {/eq}

If the area of the new figure is 54 square units more than the area of the original rectangle, then:

• {eq}A1 - A = 54 {/eq}
• {eq}(w + 3)^2 - l\times w = 54 {/eq}

But {eq}l = w + 3 {/eq}. Therefore:

• {eq}(w + 3)^2 - (w + 3)w = 54 {/eq}

Expanding and collecting the like terms:

• {eq}w^2 + 6w + 9 - w^2 - 3w = 54 {/eq}
• {eq}3w = 45 {/eq}

Solving for w:

• {eq}w = \dfrac{45}{3} {/eq}
• {eq}w = 15 {/eq}

Therefore:

• {eq}l = w + 3, \quad l = 15 + 3 = 18 {/eq}

The dimensions of the original rectangle are:

• {eq}w = 15, \quad l = 18 {/eq}

Therefore, the area of the original rectangle is:

• {eq}A = 18\times 15 {/eq}
• {eq}\boxed{A = 270\; \rm square\; units} {/eq}