Question

The area of a rectangle is 80 cm2.The length is x-8, and the width is x + 8.Find the value of x, and the dimensions of a rectangle.A=LW

Answer 1

SOLUTION

The area of the rectangle is 80 squared-centimeters

Area is given as

$\begin{gathered} A=lw \\ \text{Now, the length }l=x-8 \\ \text{and the width w }=x+8 \\ So,\text{ } \\ A=(x-8)(x+8) \\ 80=(x-8)(x+8) \\ (x-8)(x+8)=80 \end{gathered}$

Expanding the equation, we have

$\begin{gathered} (x-8)(x+8)=80 \\ x^2+8x-8x-64=80 \\ x^2+0-64=80 \\ x^2=80+64 \\ x^2=144 \\ x=\pm\sqrt[]{144} \\ x=\pm12 \\ So,\text{ we go with the positive value } \\ x=12 \end{gathered}$

Hence, the answer is x = 12

The length becomes

$\begin{gathered} l=x-8 \\ l=12-8 \\ l=4 \end{gathered}$

Hence, the length is 4 cm

The width becomes

$\begin{gathered} w=x+8 \\ w=12+8 \\ w=20 \end{gathered}$

Hence, the length is 20 cm