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.

Explanation:

Step1: Define variables

Let the width of the rectangle be \( w \) yards. Then the length \( l \) is \( 3w - 4 \) yards (since length is 4 yd less than three times the width).

Step2: Use area formula

The area of a rectangle is \( A = l \times w \). We know \( A = 55 \) yd², so substitute \( l = 3w - 4 \) into the area formula:
\[
(3w - 4)w = 55
\]
Expand the left side:
\[
3w^2 - 4w = 55
\]
Rearrange into standard quadratic form:
\[
3w^2 - 4w - 55 = 0
\]

Step3: Solve the quadratic equation

We can solve \( 3w^2 - 4w - 55 = 0 \) using the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -4 \), and \( c = -55 \).

First, calculate the discriminant \( D = b^2 - 4ac \):
\[
D = (-4)^2 - 4(3)(-55) = 16 + 660 = 676
\]
Then, find \( w \):
\[
w = \frac{-(-4) \pm \sqrt{676}}{2(3)} = \frac{4 \pm 26}{6}
\]
We have two solutions:

1. \( w = \frac{4 + 26}{6} = \frac{30}{6} = 5 \)
2. \( w = \frac{4 - 26}{6} = \frac{-22}{6} = -\frac{11}{3} \) (discard since width can't be negative)

Step4: Find the length

Now that \( w = 5 \) yards, substitute into the length formula \( l = 3w - 4 \):
\[
l = 3(5) - 4 = 15 - 4 = 11
\]

Answer:

Length: \( 11 \) yd, Width: \( 5 \) yd