Question

Find the area of the rectangle that has a perimeter of 8x + 12 units and a length of 2x − 3 units.

Answer 1

We are given the following information.

Perimeter of rectangle = 8x + 12 units

Length of rectangle = 2x − 3 units

Recall that the perimeter of a rectangle is given by

`P=2(L+W)`

Where L is the length and W is the width of the rectangle.

Let us substitute the values of perimeter and length and solve for width.

`\begin{gathered} P=2(L+W) \\ 8x+12=2(2x-3+W) \\ 8x+12=4x-6+2W \\ 8x-4x+12+6=2W \\ 4x+18=2W \\ 2W=4x+18 \\ W=(4x+18)/(2) \\ W=2x+9 \end{gathered}`

Recall that the area of a rectangle is given by

`A=L\cdot W`

Substitute the values of length and width

`\begin{gathered} A=L\cdot W \\ A=(2x-3)\cdot(2x+9) \\ A=2x\cdot2x+2x\cdot9-3\cdot2x-3\cdot9 \\ A=4x^2+18x-6x-27 \\ A=4x^2+12x-27 \end{gathered}`

Therefore, the area of the rectangle is

`A=4x^2+12x-27`