Question

the length of a rectangle is 3 m more than double the width, and the area of the rectangle is 54 m². find the dimensions of the rectangle. length : m width : m

Explanation:

Step1: Define variables

Let the width of the rectangle be $x$ meters. Then the length is $(2x + 3)$ meters.

Step2: Set up the area - equation

The area formula of a rectangle is $A=\text{length}\times\text{width}$. Given $A = 54$ m², we have the equation $x(2x + 3)=54$.
Expand it: $2x^{2}+3x=54$, or $2x^{2}+3x - 54 = 0$.

Step3: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 2$, $b = 3$, $c=-54$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(3)^{2}-4\times2\times(-54)=9 + 432 = 441$.
Then $x=\frac{-3\pm\sqrt{441}}{2\times2}=\frac{-3\pm21}{4}$.
We get two solutions for $x$:
$x_1=\frac{-3 + 21}{4}=\frac{18}{4}=4.5$ and $x_2=\frac{-3 - 21}{4}=\frac{-24}{4}=-6$.
Since the width cannot be negative, we take $x = 4.5$ m.

Step4: Find the length

The length is $2x+3$. Substitute $x = 4.5$ into it: $2\times4.5+3=9 + 3=12$ m.

Answer:

Length : 12 m
Width : 4.5 m