Question

the length of a rectangle is 5 yd less than twice the width, and the area of the rectangle is 33 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 $l = 2w - 5$ yards.

Step2: Set up the area - formula equation

The area of a rectangle is $A=l\times w$. Given $A = 33$ square - yards, we substitute $l$ and $A$ into the formula: $(2w - 5)\times w=33$.

Step3: Expand the equation

Expand $(2w - 5)w$ to get $2w^{2}-5w = 33$. Rearrange it to the standard quadratic - form $2w^{2}-5w - 33 = 0$.

Step4: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 2$, $b=-5$, $c = - 33$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ or factor the equation. Factoring $2w^{2}-5w - 33=(2w - 11)(w + 3)=0$.
Setting each factor equal to zero gives $2w-11 = 0$ or $w + 3 = 0$. Solving $2w-11 = 0$ gives $w=\frac{11}{2}=5.5$ yards, and solving $w + 3 = 0$ gives $w=-3$ yards. Since the width cannot be negative, we take $w = 5.5$ yards.

Step5: Find the length

Substitute $w = 5.5$ into the length formula $l=2w - 5$. Then $l=2\times5.5-5=11 - 5 = 6$ yards.

Answer:

Length: 6 yd
Width: 5.5 yd