Question

the area of a rectangle is 77 ft², and the length of the rectangle is 10 ft less than three times the width. find the dimensions of the rectangle. length: ft width: ft

Explanation:

Step1: Define variables

Let the width of the rectangle be $w$ feet. Then the length $l = 3w - 10$ feet.

Step2: Set up the area - formula equation

The area of a rectangle is $A=l\times w$. Given $A = 77$ square - feet, we substitute $l$ and $A$ into the formula: $(3w - 10)w=77$.

Step3: Expand the equation

Expand the left - hand side: $3w^{2}-10w = 77$. Rearrange it to the standard quadratic form $3w^{2}-10w - 77 = 0$.

Step4: Factor the quadratic equation

We need to find two numbers that multiply to $3\times(-77)=-231$ and add up to $-10$. The numbers are $-21$ and $11$. So, we can rewrite the quadratic equation as $3w^{2}-21w + 11w-77 = 0$. Group the terms: $3w(w - 7)+11(w - 7)=0$, which factors to $(3w + 11)(w - 7)=0$.

Step5: Solve for $w$

Set each factor equal to zero:

• If $3w+11 = 0$, then $3w=-11$, and $w=-\frac{11}{3}$. But the width cannot be negative, so we discard this solution.
• If $w - 7 = 0$, then $w = 7$ feet.

Step6: Find the length

Substitute $w = 7$ into the length formula $l=3w - 10$. So, $l=3\times7 - 10=21 - 10 = 11$ feet.

Answer:

Length: 11 ft
Width: 7 ft