# A rectangle is 5cm longer than it is wide. If its length and its width are both increased by 3cm its area is increased by 60cm, how do you find the dimension of the original rectangle?

Nov 23, 2016

The original width was 6cm and the length was 11cm

#### Explanation:

You are working with two different rectangles. The original one and the larger one.

Write what you know about each of them:

Original rectangle:$\textcolor{w h i t e}{\times \times \times \times \times}$ larger rectangle

$w = x \textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times} w = \left(x + 3\right)$
$l = x + 5 \textcolor{w h i t e}{\times \times \times \times \times x . \times \times x} l = \left(x + 8\right)$

$A = l \times b \textcolor{w h i t e}{\times \times \times \times . \times \times . \times x} A = l \times b$
$A = x \left(x + 5\right) \textcolor{w h i t e}{\times \times \times \times \times . \times x} A = \left(x + 3\right) \left(x + 8\right)$

The new area is 60 cm larger than the original area.

$x \left(x + 5\right) + 60 = \left(x + 3\right) \left(x + 8\right)$

${x}^{2} + 5 x + 60 = {x}^{2} + 11 x + 24$

#60-24 = 11x-5x

$36 = 6 x$

$x = 6$

The original width was 6cm and the length was 11cm