The dimensions of a rectangle whose length is 4 times its width and whose perimeter is 94 cm Find the dimension round to the nearest tenth if necessary

1. The perimeter \( P \) of a rectangle is given by the formula:

P = 2 \times (\text{length} + \text{width})

Given \( P = 94 \ \text{cm} \).

2. Let the width be \( w \). Therefore, the length \( l \) is:

l = 4w

3. Substitute for \( l \) in the perimeter formula:

94 = 2 \times (4w + w)

94 = 2 \times 5w

94 = 10w

4. Solve for \( w \):

w = \frac{94}{10} = 9.4 \ \text{cm}

5. Find the length \( l \):

l = 4 \times 9.4 = 37.6 \ \text{cm}

6. Round the dimensions to the nearest tenth:

- Width: \( 9.4 \ \text{cm} \) rounded to the nearest tenth remains \( 9.4 \ \text{cm} \).

- Length: \( 37.6 \ \text{cm} \) rounded to the nearest tenth remains \( 37.6 \ \text{cm} \).

7. Correct the earlier confusion to ensure width \( w \) and length \( l \) match initial condition:

- Set equation correctly initially for perimiter

94 = 2(w +4w) = 10 w\ ,

- Hence final update using equation above

w = \frac{94}{20} = 4.7 \ \text{cm}

l = 4 \times 4.7 = 18.8 \ \text{cm}

- Final dimensions:

- Width: 4.7\ \text{cm}

- Length: 18.8\ \text{cm}