Question

the length of a rectangle is 1 yd more than double the width, and the area of the rectangle is 45 yd². find the dimensions of the rectangle. length: yd width: yd

Explanation:

Step1: Define variables

Let the width of the rectangle be $w$ yd. Then the length $l = 2w + 1$ yd.

Step2: Set up the area - formula equation

The area of a rectangle is $A=l\times w$. Given $A = 45$ yd², we substitute $l$ and $A$ into the formula: $(2w + 1)\times w=45$.

Step3: Expand the equation

Expand $(2w + 1)w$ to get $2w^{2}+w = 45$, which can be rewritten as $2w^{2}+w - 45=0$.

Step4: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 2$, $b = 1$, $c=-45$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ or factor. Factoring $2w^{2}+w - 45=(2w - 9)(w + 5)=0$.
Setting each factor equal to zero gives $2w-9 = 0$ or $w + 5=0$.
From $2w-9 = 0$, we get $w=\frac{9}{2}=4.5$ yd. From $w + 5=0$, we get $w=-5$ yd. Since the width cannot be negative, we discard $w=-5$.

Step5: Find the length

Substitute $w = 4.5$ into the length formula $l=2w + 1$. Then $l=2\times4.5+1=9 + 1=10$ yd.

Answer:

Length: 10 yd
Width: 4.5 yd