Question

the length of a rectangle is 4 yd less than three times the width, and the area of the rectangle is 55 yd². find the dimensions of the rectangle. length: yd width: yd

Explanation:

Step1: Define variables

Let the width of the rectangle be $w$ yards. Then the length $l = 3w - 4$ yards.

Step2: Use area formula

The area formula of a rectangle is $A=l\times w$. We know $A = 55$ square - yards. Substitute $l$ and $A$ into the formula: $55=(3w - 4)\times w$.

Step3: Expand the equation

Expand $(3w - 4)w$ to get $3w^{2}-4w=55$. Rearrange it to the standard quadratic - form $3w^{2}-4w - 55 = 0$.

Step4: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 3$, $b=-4$, $c = - 55$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ or factor. Factor $3w^{2}-4w - 55$: $3w^{2}-4w - 55=3w^{2}-15w + 11w-55=3w(w - 5)+11(w - 5)=(3w + 11)(w - 5)=0$.
Set each factor equal to zero:

• $3w+11 = 0$, then $3w=-11$, $w=-\frac{11}{3}$. But the width cannot be negative, so we discard this solution.
• $w - 5=0$, then $w = 5$ yards.

Step5: Find the length

Substitute $w = 5$ into the length formula $l=3w - 4$. So $l=3\times5-4=15 - 4 = 11$ yards.

Answer:

Length: 11 yd
Width: 5 yd