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.

Explanation:

Step1: Define variables

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

Step2: Use area formula

The area formula of a rectangle is $A=l\times w$. Given that $A = 45$ square - yards. Substitute $l = 2w + 1$ into the area formula: $(2w + 1)\times w=45$.

Step3: Expand the equation

Expand $(2w + 1)\times w$ to get $2w^{2}+w = 45$. Rearrange it to the standard quadratic - form $2w^{2}+w−45 = 0$.

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 the equation. Factoring $2w^{2}+w - 45=(2w - 9)(w + 5)=0$.

Step5: Find the value of width

Set each factor equal to zero:

• If $2w−9 = 0$, then $2w=9$, and $w=\frac{9}{2}=4.5$ yards.
• If $w + 5=0$, then $w=-5$. Since the width cannot be negative, we discard $w=-5$.

Step6: Find the value of length

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

Answer:

The width of the rectangle is $4.5$ yards and the length is $10$ yards.