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The length of a rectangle is four times its width. If the area of the rectangle is 100 in², find its perimeter.

Sagot :

GinnyAnsweridn
Certainly! Let's solve this step-by-step.

1. Define the variables:
- Let the width of the rectangle be [tex]\( w \)[/tex].
- According to the problem, the length of the rectangle is four times its width, so the length [tex]\( l \)[/tex] can be written as [tex]\( l = 4w \)[/tex].

2. Write the equation for the area:
- The area [tex]\( A \)[/tex] of the rectangle is given by the product of its length and width. Thus, we have:
[tex]\[ A = l \times w \][/tex]
- We know the area is 100 square inches, so:
[tex]\[ 100 = 4w \times w \][/tex]

3. Solve the equation for [tex]\( w \)[/tex]:
- Simplify the equation:
[tex]\[ 100 = 4w^2 \][/tex]
- Divide both sides by 4:
[tex]\[ 25 = w^2 \][/tex]
- Take the square root of both sides:
[tex]\[ w = \sqrt{25} \][/tex]
[tex]\[ w = 5 \][/tex]
- Hence, the width [tex]\( w \)[/tex] of the rectangle is 5 inches.

4. Determine the length:
- Since the length is four times the width:
[tex]\[ l = 4w = 4 \times 5 = 20 \][/tex]
- Thus, the length [tex]\( l \)[/tex] of the rectangle is 20 inches.

5. Calculate the perimeter:
- The perimeter [tex]\( P \)[/tex] of a rectangle is given by:
[tex]\[ P = 2 \times (l + w) \][/tex]
- Substitute the known values of length and width:
[tex]\[ P = 2 \times (20 + 5) \][/tex]
[tex]\[ P = 2 \times 25 \][/tex]
[tex]\[ P = 50 \][/tex]

Therefore, the perimeter of the rectangle is 50 inches.
Chibabykalu16idn

Answer:

50 in

Step-by-step explanation:

Given:

  • Length of a rectangle =  4 * width
  • Area of the rectangle = length * width = [tex]{100 \text{ in}}^2[/tex]

Where:

  • Length = L
  • Width = W

[tex]L \times W = \text{100 in$^2$}[/tex]

Substituting the length value into the formula for the area of a rectangle:

[tex](4W) \times W = \text{100 in$^2$}[/tex]

[tex]4{W}^{2} = \text{100 in$^2$}[/tex]

Divide both sides by 4 to isolate W²

[tex]\frac{4{W}^{2}}{4} = \frac{\text{100 in$^2$}}{4}[/tex]

[tex]{W}^{2} = \text{25 in$^2$}[/tex]

Square both sides to find the value of W

[tex]\sqrt{{W}^{2}} = \sqrt{\text{25 in$^2$}}[/tex]

[tex]W = \text{5 in}[/tex]

Substituting W = 5 in into the formula for the area of a rectangle

[tex]L \times \text{5 in} = \text{100 in$^2$}[/tex]

[tex]L = \frac{ \text{100 in$^2$} }{\text{5 in}}[/tex]

[tex]L = \text{20 in}[/tex]

Using the formula for the perimeter of a triangle:

[tex]P = 2 (L + W)[/tex]

L = 20 in and W = 5 in

[tex]P = 2 (\text{20 in + 5 in})[/tex]

[tex]P = 2 (\text{25 in})[/tex]

[tex]P = (\text{50 in})[/tex]

Therefore, the perimeter of the rectangle simplifies to 50 in

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