Question

the length of a rectangle is 5 yd more than twice the width, and the area of the rectangle is 63 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 \( 2w + 5 \) yards (since length is 5 yd more than twice the width).

Step2: Use area formula

The area of a rectangle is \( A = l \times w \). We know \( A = 63 \) yd², so substitute \( l = 2w + 5 \) into the area formula:
\[
(2w + 5) \times w = 63
\]
Expand the left side:
\[
2w^2 + 5w = 63
\]
Rearrange into standard quadratic form:
\[
2w^2 + 5w - 63 = 0
\]

Step3: Solve the quadratic equation

We can solve \( 2w^2 + 5w - 63 = 0 \) using the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 5 \), and \( c = -63 \).

First, calculate the discriminant \( D = b^2 - 4ac \):
\[
D = 5^2 - 4 \times 2 \times (-63) = 25 + 504 = 529
\]
Then, find the square root of the discriminant: \( \sqrt{529} = 23 \)

Now, substitute into the quadratic formula:
\[
w = \frac{-5 \pm 23}{2 \times 2}
\]
We have two solutions:

1. \( w = \frac{-5 + 23}{4} = \frac{18}{4} = 4.5 \)
2. \( w = \frac{-5 - 23}{4} = \frac{-28}{4} = -7 \)

Since width cannot be negative, we take \( w = 4.5 \) yards.

Step4: Find the length

Substitute \( w = 4.5 \) into the length formula \( l = 2w + 5 \):
\[
l = 2 \times 4.5 + 5 = 9 + 5 = 14
\]

Answer:

Length: \( 14 \) yd
Width: \( 4.5 \) yd