The area of a rectangle is52yd^2, and the length of the rectangle is 5yd less than twice the width. Find the dimensions of the rectangle.



Let the **width** of the rectangle be \( x \) yards.  
Then, the **length** of the rectangle is given as **5 yards less than twice the width**, which can be expressed as:  
\[
\text{Length} = 2x - 5
\]

The **area** of the rectangle is given as **52 square yards**, so we use the formula for the area of a rectangle:

\[
\text{Area} = \text{Length} \times \text{Width}
\]

Substituting the given values:

\[
x(2x - 5) = 52
\]

### Step 1: Expand the equation  
\[
2x^2 - 5x = 52
\]

### Step 2: Rearrange into standard quadratic form  
\[
2x^2 - 5x - 52 = 0
\]

### Step 3: Solve for \( x \) using the quadratic formula  
The quadratic formula is:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \( 2x^2 - 5x - 52 = 0 \), we have:
- \( a = 2 \)
- \( b = -5 \)
- \( c = -52 \)

First, compute the discriminant:

\[
b^2 - 4ac = (-5)^2 - 4(2)(-52)
\]
\[
= 25 + 416 = 441
\]

Now, take the square root:

\[
\sqrt{441} = 21
\]

Now, solve for \( x \):

\[
x = \frac{-(-5) \pm 21}{2(2)}
\]

\[
x = \frac{5 \pm 21}{4}
\]

### Step 4: Find possible values for \( x \)

\[
x = \frac{5 + 21}{4} = \frac{26}{4} = 6.5
\]

\[
x = \frac{5 - 21}{4} = \frac{-16}{4} = -4
\]

Since the width cannot be negative, we take:

\[
x = 6.5 \text{ yards (width)}
\]

### Step 5: Find the length

\[
\text{Length} = 2(6.5) - 5 = 13 - 5 = 8 \text{ yards}
\]

### Final Answer:  
- **Width** = 6.5 yards  
- **Length** = 8 yards  

Would you like a more detailed breakdown or verification of the steps?

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### Related Questions:
1. How do you solve quadratic equations by factoring instead of using the quadratic formula?
2. What are the properties of a rectangle that distinguish it from other quadrilaterals?
3. How can you determine the dimensions of a rectangle if given perimeter instead of area?
4. How does changing the width affect the area of a rectangle while keeping the length proportional?
5. What real-world applications involve finding dimensions of a rectangle given area constraints?

### Math Tip:  
Always check whether the quadratic equation can be factored easily before using the quadratic formula, as factoring can save time. 🚀

The area of a rectangle is 52yd^2, and the length of the rectangle is 5yd less than twice the width. Find the dimensions of the rectangle.

Learn how to solve for the dimensions of a rectangle when given its area and a relationship between length and width. This problem uses the quadratic equation to find the width and length of the rectangle. Ideal for students in Grades 7-9 studying algebra and geometry.

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