Math / Geometry

QuestionThe length of a rectangle is 1 m more than double the width, and the area of the rectangle is $28 \mathrm{~m}^{2}$ . Find the dimensions of the rectangle.
Length : $\square$ m
Width : $\square$ m
$\square$

Studdy Solution

STEP 1

1. The length of the rectangle is expressed in terms of the width.
2. The area of the rectangle is $28 \, \text{m}^2$ .

STEP 2

1. Express the length in terms of the width.
2. Write the equation for the area of the rectangle.
3. Solve the equation for the width.
4. Calculate the length using the width.
5. Verify the solution.

STEP 3

Express the length in terms of the width. Let $w$ be the width of the rectangle. Then the length $l$ is given by:
$l = 2w + 1$

STEP 4

Write the equation for the area of the rectangle. The area $A$ is given by:
$A = l \times w$
Substitute the expression for $l$ :
$28 = (2w + 1) \times w$

STEP 5

Solve the equation for the width $w$ :
$28 = 2w^2 + w$
Rearrange the equation:
$2w^2 + w - 28 = 0$
Use the quadratic formula $w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 2$ , $b = 1$ , and $c = -28$ :
$w = \frac{-1 \pm \sqrt{1^2 - 4 \times 2 \times (-28)}}{2 \times 2}$ $w = \frac{-1 \pm \sqrt{1 + 224}}{4}$ $w = \frac{-1 \pm \sqrt{225}}{4}$ $w = \frac{-1 \pm 15}{4}$
Calculate the two potential solutions:
$w = \frac{14}{4} = 3.5$ $w = \frac{-16}{4} = -4$
Since width cannot be negative, we have:
$w = 3.5$

STEP 6

Calculate the length using the width:
$l = 2w + 1$ $l = 2(3.5) + 1$ $l = 7 + 1$ $l = 8$

STEP 7

Verify the solution by checking the area:
$A = l \times w$ $A = 8 \times 3.5$ $A = 28$
The calculated area matches the given area, confirming the solution is correct.
The dimensions of the rectangle are:
Length : $\boxed{8} \, \text{m}$
Width : $\boxed{3.5} \, \text{m}$

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QuestionThe length of a rectangle is 1 m more than double the width, and the area of the rectangle is $28 \mathrm{~m}^{2}$ . Find the dimensions of the rectangle.
Length : $\square$ m
Width : $\square$ m
$\square$

Studdy Solution

STEP 1

1. The length of the rectangle is expressed in terms of the width.
2. The area of the rectangle is $28 \, \text{m}^2$ .

STEP 2

1. Express the length in terms of the width.
2. Write the equation for the area of the rectangle.
3. Solve the equation for the width.
4. Calculate the length using the width.
5. Verify the solution.

STEP 3

Express the length in terms of the width. Let $w$ be the width of the rectangle. Then the length $l$ is given by:
$l = 2w + 1$

STEP 4

Write the equation for the area of the rectangle. The area $A$ is given by:
$A = l \times w$
Substitute the expression for $l$ :
$28 = (2w + 1) \times w$

STEP 5

STEP 6

Calculate the length using the width:
$l = 2w + 1$ $l = 2(3.5) + 1$ $l = 7 + 1$ $l = 8$

STEP 7

Verify the solution by checking the area:
$A = l \times w$ $A = 8 \times 3.5$ $A = 28$
The calculated area matches the given area, confirming the solution is correct.
The dimensions of the rectangle are:
Length : $\boxed{8} \, \text{m}$
Width : $\boxed{3.5} \, \text{m}$

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QuestionThe length of a rectangle is 1 m more than double the width, and the area of the rectangle is 28 m228 \mathrm{~m}^{2}28 m2. Find the dimensions of the rectangle.Length : □\square□ mWidth : □\square□ m□\square□

Studdy Solution

STEP 1

1. The length of the rectangle is expressed in terms of the width.2. The area of the rectangle is 28 m2 28 \, \text{m}^2 28m2.

STEP 2

1. Express the length in terms of the width.2. Write the equation for the area of the rectangle.3. Solve the equation for the width.4. Calculate the length using the width.5. Verify the solution.

STEP 3

Express the length in terms of the width. Let w w w be the width of the rectangle. Then the length l l l is given by:l=2w+1 l = 2w + 1 l=2w+1

STEP 4

Write the equation for the area of the rectangle. The area A A A is given by:A=l×w A = l \times w A=l×wSubstitute the expression for l l l:28=(2w+1)×w 28 = (2w + 1) \times w 28=(2w+1)×w

STEP 5

STEP 6

Calculate the length using the width:l=2w+1 l = 2w + 1 l=2w+1 l=2(3.5)+1 l = 2(3.5) + 1 l=2(3.5)+1 l=7+1 l = 7 + 1 l=7+1 l=8 l = 8 l=8

STEP 7

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QuestionThe length of a rectangle is 1 m more than double the width, and the area of the rectangle is $28 \mathrm{~m}^{2}$ . Find the dimensions of the rectangle.
Length : $\square$ m
Width : $\square$ m
$\square$

1. The length of the rectangle is expressed in terms of the width.
2. The area of the rectangle is $28 \, \text{m}^2$ .

1. Express the length in terms of the width.
2. Write the equation for the area of the rectangle.
3. Solve the equation for the width.
4. Calculate the length using the width.
5. Verify the solution.

Express the length in terms of the width. Let $w$ be the width of the rectangle. Then the length $l$ is given by:
$l = 2w + 1$

Write the equation for the area of the rectangle. The area $A$ is given by:
$A = l \times w$
Substitute the expression for $l$ :
$28 = (2w + 1) \times w$

Calculate the length using the width:
$l = 2w + 1$ $l = 2(3.5) + 1$ $l = 7 + 1$ $l = 8$