Question

the area of a rectangle is 63 yd², and the length of the rectangle is 5 yd more than twice 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 $l = 2w + 5$ yards.

Step2: Set up the area - formula equation

The area formula of a rectangle is $A=l\times w$. Given $A = 63$ square - yards, we substitute $l$ and $A$ into the formula: $(2w + 5)\times w=63$.
Expanding this gives $2w^{2}+5w - 63 = 0$.

Step3: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 2$, $b = 5$, $c=-63$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ or factor the equation.
Factor $2w^{2}+5w - 63$:
$2w^{2}+5w - 63=2w^{2}+14w-9w - 63=2w(w + 7)-9(w + 7)=(2w - 9)(w + 7)=0$.
Setting each factor equal to zero gives:
$2w-9 = 0$ or $w + 7 = 0$.
From $2w-9 = 0$, we get $w=\frac{9}{2}=4.5$ yards.
From $w + 7 = 0$, we get $w=-7$. But since the width cannot be negative, we discard $w=-7$.

Step4: Find the length

Substitute $w = 4.5$ into the length formula $l = 2w+5$.
$l=2\times4.5 + 5=9 + 5=14$ yards.

Answer:

Length: 14 yd
Width: 4.5 yd