# The length of a rectangle is more than twice its width, and the area of the rectangle is 20. How do you find the dimension?

May 20, 2016

Length is 10
Width is 2

#### Explanation:

Let length be $L$
Let width be $W$
Let area be $A$

Given that $L > 2 W$

Let $L = 2 W + x$

$A = L \times W$ .......................(1)

But $L = 2 W + x$ so by substituting for $L$ in equation (1)

$A = \left(2 W + x\right) \times W$

$A = 2 {W}^{2} + x W$...............(2)

But area is given as $A = 20$

Substitute for $A$ in equation (2)

$20 = 2 {W}^{2} + x W$

$\implies 2 {W}^{2} + x W - 20 = 0$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
as $x$ is a variable we can find it by default

Set $\textcolor{b r o w n}{\text{ "2W^2+xW-20" ")color(blue)( ->" } \left(2 W - 4\right) \left(W + 5\right)}$

Multiply out the brackets to determine the value of $x$

$\textcolor{b r o w n}{2 {W}^{2} + x W - 20} \textcolor{b l u e}{\to 2 {W}^{2} + 10 W - 4 W - 20}$

$\textcolor{b r o w n}{2 {W}^{2} + x W - 20} \textcolor{b l u e}{\to 2 {W}^{2} + 6 W - 20}$

color(green)("Thus "x=6" and "W=2" and "cancel(-5)

However, $W = - 5$ is not logical so discard it.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{g r e e n}{\text{Test } L = 2 W + x = 2 \left(2\right) + 6 = 10}$

$\textcolor{red}{\implies a r e a \to L \times W \to 10 \times 2 = 20 \text{ as given}}$