# The length of a rectangle exceeds its width by 3 cm. If the length and width each increased by 2...

## Question:

The length of a rectangle exceeds its width by 3 cm. If the length and width each increased by 2 cm, then the area of the new rectangle will be 70 {eq}cm^2 {/eq} more than that of the original rectangle. Find the length and width of the original rectangle.

## Area of a Rectangle:

A rectangle is a four-sided figure with 4 lines or sides. The opposite sides in a rectangle are parallel and equal, and all the interior angles are right angles. For example, a football pitch has a rectangular shape.

The area of a rectangle is given by:

{eq}A = l\times w{/eq}

Where l is the length and w is the width of the rectangle.

Let's start by solving for the area of the original rectangle.

Let the width of the original rectangle be {eq}x cm{/eq}. Thus, its length will be equal to {eq}x + 3{/eq}.

• {eq}w_1 = x\, cm{/eq}
• {eq}l_1(x + 3)\, cm{/eq}

Therefore, the area of the original rectangle is equal to:

• {eq}A_1 = l_1\times w_1 = x(x + 3)cm^2{/eq}

If we increase the length and the width by {eq}2 cm{/eq} each, the measurements of the new rectangle formed will be:

• {eq}w_2 = (x + 2) \,cm{/eq}
• {eq}l_2 = (x + 3 + 2) = (x + 5)\, cm.{/eq}

Thus, the area of the new rectangle formed is equal to:

• {eq}A_2 = l_2\times w_2 = (x + 2)(x + 5)cm^2{/eq}

We are told that the area of the new rectangle is {eq}70cm^2{/eq} more than the area of the original rectangle. Therefore:

• {eq}A_2 - A_1 = 70cm^2{/eq}
• {eq}(x + 2)(x + 5)cm^2 - x(x + 3)cm^2 = 70cm^2{/eq}

Expanding the LHS of the equation:

• {eq}x^2 + 7x + 10 - x^2 - 3x = 70 {/eq}

Collecting the like terms:

• {eq}4x = 70 - 10 {/eq}
• {eq}4x = 60 {/eq}
• {eq}x = \dfrac{60}{4} = 15\, cm {/eq}

Thus, the width of the original rectangle is equal to:

• {eq}w_1 = x = \boxed{15\, cm}{/eq}

And the length is equal to:

• {eq}l_1 = x + 3 = 15 + 3= \boxed{18\, cm}{/eq} 