Question

the area of a rectangle is 63 m², and the length of the rectangle is 5 m more than twice the width. 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 $l = 2w + 5$ meters.

Step2: Set up area - formula equation

The area of a rectangle is $A=l\times w$. Given $A = 63$ square - meters, we substitute $l$ and $A$ into the formula: $(2w + 5)\times w=63$.

Step3: Expand the equation

Expand $(2w + 5)w$ to get $2w^{2}+5w = 63$. Rearrange it to the standard quadratic - form $2w^{2}+5w−63 = 0$.

Step4: 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. Factoring $2w^{2}+5w−63=(2w - 9)(w + 7)=0$.
Setting each factor equal to zero gives $2w−9 = 0$ or $w + 7 = 0$.
From $2w−9 = 0$, we get $w=\frac{9}{2}=4.5$. From $w + 7 = 0$, we get $w=-7$. Since the width cannot be negative, we take $w = 4.5$ meters.

Step5: Find the length

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

Answer:

Length: 14 m
Width: 4.5 m