Math / Algebra

QuestionThe length of a rectangle is 1 yd more than twice the width, and the area of the rectangle is $66 \mathrm{yd}^{2}$ . Find the dimensions of the rectangle.
Length $\square$ yd
Width : $\square$ yd

Studdy Solution

STEP 1

1. The rectangle's length $L$ is expressed in terms of its width $W$ .
2. The area of the rectangle is given as $66 \, \text{yd}^2$ .
3. The relationship between the length and width can be used to form an equation.

STEP 2

1. Express the length in terms of the width.
2. Set up the equation for the area of the rectangle.
3. Solve the quadratic equation for the width.
4. Calculate the length using the width.
5. Verify the solution by checking the area.

STEP 3

Express the length in terms of the width. According to the problem, the length is 1 yard more than twice the width:
$L = 2W + 1$

STEP 4

Set up the equation for the area of the rectangle. The area $A$ is given by:
$A = L \times W = 66$
Substitute the expression for $L$ from STEP_1:
$(2W + 1) \times W = 66$

STEP 5

Solve the quadratic equation for the width. First, expand the equation:
$2W^2 + W = 66$
Rearrange to form a standard quadratic equation:
$2W^2 + W - 66 = 0$

STEP 6

Factor the quadratic equation if possible, or use the quadratic formula. Let's use the quadratic formula:
The quadratic formula is given by:
$W = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
For our equation $2W^2 + W - 66 = 0$ , $a = 2$ , $b = 1$ , and $c = -66$ .
Calculate the discriminant:
$b^2 - 4ac = 1^2 - 4 \times 2 \times (-66) = 1 + 528 = 529$
Calculate the width:
$W = \frac{-1 \pm \sqrt{529}}{4}$ $W = \frac{-1 \pm 23}{4}$
This gives two potential solutions:
$W = \frac{22}{4} = 5.5$ $W = \frac{-24}{4} = -6$
Since width cannot be negative, we have:
$W = 5.5$

STEP 7

Calculate the length using the width found in STEP_4:
$L = 2W + 1 = 2(5.5) + 1 = 11 + 1 = 12$

STEP 8

Verify the solution by checking the area:
$A = L \times W = 12 \times 5.5 = 66$
The calculated area matches the given area, confirming the solution is correct.
The dimensions of the rectangle are:
Length: $12$ yd
Width: $5.5$ yd

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QuestionThe length of a rectangle is 1 yd more than twice the width, and the area of the rectangle is $66 \mathrm{yd}^{2}$ . Find the dimensions of the rectangle.
Length $\square$ yd
Width : $\square$ yd

Studdy Solution

STEP 1

1. The rectangle's length $L$ is expressed in terms of its width $W$ .
2. The area of the rectangle is given as $66 \, \text{yd}^2$ .
3. The relationship between the length and width can be used to form an equation.

STEP 2

1. Express the length in terms of the width.
2. Set up the equation for the area of the rectangle.
3. Solve the quadratic equation for the width.
4. Calculate the length using the width.
5. Verify the solution by checking the area.

STEP 3

Express the length in terms of the width. According to the problem, the length is 1 yard more than twice the width:
$L = 2W + 1$

STEP 4

Set up the equation for the area of the rectangle. The area $A$ is given by:
$A = L \times W = 66$
Substitute the expression for $L$ from STEP_1:
$(2W + 1) \times W = 66$

STEP 5

Solve the quadratic equation for the width. First, expand the equation:
$2W^2 + W = 66$
Rearrange to form a standard quadratic equation:
$2W^2 + W - 66 = 0$

STEP 6

STEP 7

Calculate the length using the width found in STEP_4:
$L = 2W + 1 = 2(5.5) + 1 = 11 + 1 = 12$

STEP 8

Verify the solution by checking the area:
$A = L \times W = 12 \times 5.5 = 66$
The calculated area matches the given area, confirming the solution is correct.
The dimensions of the rectangle are:
Length: $12$ yd
Width: $5.5$ yd

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QuestionThe length of a rectangle is 1 yd more than twice the width, and the area of the rectangle is 66yd266 \mathrm{yd}^{2}66yd2. Find the dimensions of the rectangle.Length □\square□ ydWidth : □\square□ yd

Studdy Solution

STEP 1

1. The rectangle's length L L L is expressed in terms of its width W W W.2. The area of the rectangle is given as 66 yd2 66 \, \text{yd}^2 66yd2.3. The relationship between the length and width can be used to form an equation.

STEP 2

1. Express the length in terms of the width.2. Set up the equation for the area of the rectangle.3. Solve the quadratic equation for the width.4. Calculate the length using the width.5. Verify the solution by checking the area.

STEP 3

Express the length in terms of the width. According to the problem, the length is 1 yard more than twice the width:L=2W+1 L = 2W + 1 L=2W+1

STEP 4

Set up the equation for the area of the rectangle. The area A A A is given by:A=L×W=66 A = L \times W = 66 A=L×W=66Substitute the expression for L L L from STEP_1:(2W+1)×W=66 (2W + 1) \times W = 66 (2W+1)×W=66

STEP 5

Solve the quadratic equation for the width. First, expand the equation:2W2+W=66 2W^2 + W = 66 2W2+W=66Rearrange to form a standard quadratic equation:2W2+W−66=0 2W^2 + W - 66 = 0 2W2+W−66=0

STEP 6

STEP 7

Calculate the length using the width found in STEP_4:L=2W+1=2(5.5)+1=11+1=12 L = 2W + 1 = 2(5.5) + 1 = 11 + 1 = 12 L=2W+1=2(5.5)+1=11+1=12

STEP 8

Verify the solution by checking the area:A=L×W=12×5.5=66 A = L \times W = 12 \times 5.5 = 66 A=L×W=12×5.5=66The calculated area matches the given area, confirming the solution is correct. The dimensions of the rectangle are:Length: 12 12 12 ydWidth: 5.5 5.5 5.5 yd

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QuestionThe length of a rectangle is 1 yd more than twice the width, and the area of the rectangle is $66 \mathrm{yd}^{2}$ . Find the dimensions of the rectangle.
Length $\square$ yd
Width : $\square$ yd

1. The rectangle's length $L$ is expressed in terms of its width $W$ .
2. The area of the rectangle is given as $66 \, \text{yd}^2$ .
3. The relationship between the length and width can be used to form an equation.

1. Express the length in terms of the width.
2. Set up the equation for the area of the rectangle.
3. Solve the quadratic equation for the width.
4. Calculate the length using the width.
5. Verify the solution by checking the area.

Express the length in terms of the width. According to the problem, the length is 1 yard more than twice the width:
$L = 2W + 1$

Set up the equation for the area of the rectangle. The area $A$ is given by:
$A = L \times W = 66$
Substitute the expression for $L$ from STEP_1:
$(2W + 1) \times W = 66$

Solve the quadratic equation for the width. First, expand the equation:
$2W^2 + W = 66$
Rearrange to form a standard quadratic equation:
$2W^2 + W - 66 = 0$

Calculate the length using the width found in STEP_4:
$L = 2W + 1 = 2(5.5) + 1 = 11 + 1 = 12$

Verify the solution by checking the area:
$A = L \times W = 12 \times 5.5 = 66$
The calculated area matches the given area, confirming the solution is correct.
The dimensions of the rectangle are:
Length: $12$ yd
Width: $5.5$ yd