Question

the length of a rectangle is 5 ft less than double the width, and the area of the rectangle is 52 ft². 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 $l = 2w - 5$ feet.

Step2: Set up the area - formula equation

The area of a rectangle is $A=l\times w$. Given $A = 52$, we substitute $l$ and $A$ into the formula: $(2w - 5)\times w=52$.

Step3: Expand the equation

Expand $(2w - 5)w$ to get $2w^{2}-5w = 52$. Rearrange it to the standard quadratic - form $2w^{2}-5w - 52 = 0$.

Step4: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 2$, $b=-5$, $c = - 52$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-5)^{2}-4\times2\times(-52)=25 + 416 = 441$. Then $w=\frac{5\pm\sqrt{441}}{4}=\frac{5\pm21}{4}$. We have two solutions for $w$: $w_1=\frac{5 + 21}{4}=\frac{26}{4}=6.5$ and $w_2=\frac{5-21}{4}=\frac{-16}{4}=-4$. Since the width cannot be negative, we take $w = 6.5$ feet.

Step5: Find the length

Substitute $w = 6.5$ into the length formula $l = 2w-5$. Then $l=2\times6.5 - 5=13 - 5 = 8$ feet.

Answer:

Length: 8 ft
Width: 6.5 ft