Question

the area of a rectangle is 77 m², and the length of the rectangle is 3 m more than twice the width. 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 $A = 77$ m², we have the equation $x(2x + 3)=77$.
Expand it: $2x^{2}+3x=77$, or $2x^{2}+3x - 77 = 0$.

Step3: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 2$, $b = 3$, $c=-77$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ or factor.
Factor $2x^{2}+3x - 77=(2x + 11)(x - 7)=0$.
Set each factor equal to zero:
$2x+11 = 0$ gives $x=-\frac{11}{2}$, but the width cannot be negative, so we discard this solution.
$x - 7=0$ gives $x = 7$.

Step4: Find the length and width

The width $x = 7$ m.
The length is $2x+3=2\times7 + 3=14 + 3=17$ m.

Answer:

Length: 17 m
Width: 7 m