Question

the area of a rectangle is 66 m², and the length of the rectangle is 1 m more than double the width. find the dimensions of the rectangle.

Explanation:

Step1: Define variables

Let the width of the rectangle be $x$ meters. Then the length is $(2x + 1)$ meters.

Step2: Set up the area - equation

The area formula of a rectangle is $A=\text{length}\times\text{width}$. Given $A = 66$, we have the equation $x(2x + 1)=66$.
Expand it to get $2x^{2}+x - 66=0$.

Step3: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 2$, $b = 1$, $c=-66$), we use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(1)^{2}-4\times2\times(-66)=1 + 528 = 529$.
Then $x=\frac{-1\pm\sqrt{529}}{2\times2}=\frac{-1\pm23}{4}$.
We get two solutions for $x$: $x_1=\frac{-1 + 23}{4}=\frac{22}{4}=5.5$ and $x_2=\frac{-1-23}{4}=\frac{-24}{4}=-6$.
Since the width cannot be negative, we take $x = 5.5$.

Step4: Find the length

The length is $2x+1$. Substitute $x = 5.5$ into it, we get $2\times5.5+1=11 + 1=12$.

Answer:

The width of the rectangle is $5.5$ m and the length is $12$ m.