Question

the length of a rectangle is 3 m more than double the width, and the area of the rectangle is 35 m². find the dimensions of the rectangle. length: m width: m

Explanation:

Step1: Define variables

Let the width of the rectangle be $w$ meters. Then the length $l = 2w + 3$ meters.

Step2: Set up area - formula equation

The area of a rectangle is $A=l\times w$. Given $A = 35$ m², we substitute $l$ and $A$ into the formula: $(2w + 3)\times w=35$.
Expand the left - hand side to get $2w^{2}+3w = 35$, or $2w^{2}+3w−35 = 0$.

Step3: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 2$, $b = 3$, $c=-35$), we can use the quadratic formula $w=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ or factor the equation.
Factor $2w^{2}+3w−35$: $2w^{2}+3w−35=(2w - 7)(w + 5)=0$.
Set each factor equal to zero:
If $2w−7 = 0$, then $2w=7$, and $w=\frac{7}{2}=3.5$.
If $w + 5 = 0$, then $w=-5$. But since the width cannot be negative, we discard $w=-5$.

Step4: Find the length

Substitute $w = 3.5$ into the length formula $l = 2w+3$. Then $l=2\times3.5 + 3=7 + 3=10$.

Answer:

Length: 10 m
Width: 3.5 m