Question

1. geometry the length of a rectangle is 2 feet more than its width. find the dimensions of the rectangle if its area is 63 square feet. length = ____________ width = ____________

Explanation:

Step1: Let width be $x$ feet.

Since length is 2 feet more than width, length is $x + 2$ feet.

Step2: Set up area - formula equation.

The area of a rectangle $A=length\times width$. Given $A = 63$ square - feet, we have the equation $x(x + 2)=63$.

Step3: Expand and rearrange the equation.

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

Step4: Factor the quadratic equation.

We factor $x^{2}+2x−63=(x + 9)(x - 7)=0$.

Step5: Solve for $x$.

Setting each factor equal to zero gives $x+9 = 0$ or $x - 7=0$. So $x=-9$ or $x = 7$. Since width cannot be negative, we take $x = 7$.

Step6: Find the length.

If width $x = 7$ feet, then length $x + 2=7 + 2=9$ feet.

Answer:

Length = 9 feet, Width = 7 feet