Question

the area of a rectangle is 35 yd², and the length of the rectangle is 3 yd more than double the width. find the dimensions of the rectangle. length: yd width: yd

Explanation:

Step1: Define variables

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

Step2: Set up area - formula equation

The area of a rectangle is $A=l\times w$. Given $A = 35$ yd², we substitute $l$ and $A$ into the formula: $(2w + 3)w=35$.

Step3: Expand and rearrange the equation

Expand the left - hand side: $2w^{2}+3w = 35$. Rearrange to get a quadratic equation: $2w^{2}+3w−35 = 0$.

Step4: 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 $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ or factor. Factoring gives $(2w - 7)(w + 5)=0$.

Step5: Find the value of $w$

Set each factor equal to zero:
If $2w−7 = 0$, then $2w=7$, and $w=\frac{7}{2}=3.5$ yd.
If $w + 5=0$, then $w=-5$. But the width cannot be negative, so we discard $w=-5$.

Step6: Find the value of $l$

Substitute $w = 3.5$ into the formula for $l$: $l=2w + 3=2\times3.5+3=7 + 3=10$ yd.

Answer:

Length: 10 yd
Width: 3.5 yd