Question

the area of a rectangle is 21 yd², and the length of the rectangle is 1 yd less than twice the width. 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 \( l \) is \( 2w - 1 \) yards (since length is 1 yd less than twice the width).

Step2: Use area formula

The area of a rectangle is given by \( A = l \times w \). We know the area \( A = 21 \) \( \text{yd}^2 \), so we substitute \( l = 2w - 1 \) and \( A = 21 \) into the formula:
\[
21=(2w - 1)\times w
\]
Expand the right - hand side:
\[
21 = 2w^{2}-w
\]
Rearrange the equation to form a quadratic equation:
\[
2w^{2}-w - 21=0
\]

Step3: Solve the quadratic equation

We can solve the quadratic equation \( 2w^{2}-w - 21 = 0 \) using the quadratic formula \( w=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \), where \( a = 2 \), \( b=- 1 \), and \( c=-21 \).

First, calculate the discriminant \( \Delta=b^{2}-4ac=(-1)^{2}-4\times2\times(-21)=1 + 168 = 169 \)

Then,
\[
w=\frac{-(-1)\pm\sqrt{169}}{2\times2}=\frac{1\pm13}{4}
\]

We have two solutions for \( w \):

• \( w=\frac{1 + 13}{4}=\frac{14}{4}=\frac{7}{2}=3.5 \)
• \( w=\frac{1-13}{4}=\frac{-12}{4}=- 3 \)

Since the width cannot be negative, we take \( w = 3.5 \) yards.

Step4: Find the length

Now that we know the width \( w = 3.5 \) yards, we find the length \( l=2w - 1 \). Substitute \( w = 3.5 \) into the formula for length:
\[
l=2\times3.5-1=7 - 1=6
\]

Answer:

Length : \( 6 \) yd
Width : \( 3.5 \) yd (or \( \frac{7}{2} \) yd)