Question

1. the perimeter of a rectangle is 40 inches, and its area is 96 in². find its length and width.

Explanation:

Step1: Set up equations

Let the length of the rectangle be $l$ and the width be $w$. The perimeter formula is $P = 2(l + w)$, so $2(l + w)=40$, which simplifies to $l + w=20$, and we can express $l = 20 - w$. The area formula is $A=lw$, so $lw = 96$.

Step2: Substitute and solve

Substitute $l = 20 - w$ into $lw = 96$, we get $(20 - w)w=96$. Expand to $20w-w^{2}=96$, or $w^{2}-20w + 96=0$.

Step3: Factor the quadratic equation

Factor $w^{2}-20w + 96=0$ as $(w - 8)(w - 12)=0$.

Step4: Find the values of width and length

Set each factor equal to zero: $w-8 = 0$ gives $w = 8$; $w - 12=0$ gives $w = 12$. When $w = 8$, $l=20 - 8=12$; when $w = 12$, $l=20 - 12 = 8$.

Answer:

The length is 12 inches and the width is 8 inches (or vice - versa).