Question

the area of a rectangle is 28 ft², and the length of the rectangle is 5 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 $l = 3w - 5$ feet.

Step2: Set up area - formula equation

The area of a rectangle is $A=l\times w$. Given $A = 28$ square - feet, we substitute $l$ and $A$ into the formula: $(3w - 5)w=28$.

Step3: Expand and rearrange the equation

Expand $(3w - 5)w$ to get $3w^{2}-5w = 28$. Rearrange it to the standard quadratic - form $3w^{2}-5w - 28 = 0$.

Step4: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 3$, $b=-5$, $c = - 28$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ or factor. Factoring $3w^{2}-5w - 28$ gives $(3w + 7)(w - 4)=0$.
Setting each factor equal to zero:

• $3w+7 = 0$ gives $w=-\frac{7}{3}$. But since the width cannot be negative, we discard this solution.
• $w - 4 = 0$ gives $w = 4$ feet.

Step5: Find the length

Substitute $w = 4$ into the length formula $l=3w - 5$. Then $l=3\times4-5=12 - 5 = 7$ feet.

Answer:

Length: 7 ft
Width: 4 ft