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.

Explanation:

Step1: Let the width be \(w\) yards.

The length \(l = 2w - 5\) yards.

Step2: Use the area formula \(A=l\times w\).

We know \(A = 33\) and \(l=2w - 5\), so \(33=(2w - 5)\times w\).
Expanding gives \(2w^{2}-5w - 33 = 0\).

Step3: Solve the quadratic equation \(2w^{2}-5w - 33 = 0\).

We can factor it as \((2w - 11)(w + 3)=0\).
Setting each factor equal to zero:
If \(2w-11 = 0\), then \(2w=11\), \(w=\frac{11}{2}=5.5\).
If \(w + 3=0\), then \(w=- 3\). Since width cannot be negative, we discard \(w=-3\).

Step4: Find the length.

Substitute \(w = 5.5\) into \(l=2w - 5\), so \(l=2\times5.5-5=11 - 5 = 6\).

Answer:

Width: \(5.5\) yd, Length: \(6\) yd