Let's analyze the mathematical problems provided in the uploaded image.

Observations:

The task involves solving for the dimensions (length and width) of rectangles given specific conditions:

1. Task c: A rectangle with a perimeter of $42$, where one side is twice as long as the other.
2. Task d: A rectangle with a perimeter of $24$, where one side is $\frac{1}{2}$ of the other.
3. Task 3a: A rectangle with a perimeter of $94$, where one side is $5$ cm longer than the other.
4. Task 3b: A rectangle with a perimeter of $336$, where the length is $3$ times the width.

I'll calculate each problem step-by-step.

Task c:

• Perimeter formula: $u = 2 \cdot (\text{Length} + \text{Width})$
• Here, $\text{Length} = 2x$, $\text{Width} = x$, and $u = 42$.

$2 \cdot (2x + x) = 42$ $2 \cdot 3x = 42 \implies 6x = 42 \implies x = 7$

Thus:

• Length: $2x = 14$
• Width: $x = 7$

Task d:

• Perimeter formula: $u = 2 \cdot (\text{Length} + \text{Width})$
• Here, $\text{Length} = x$, $\text{Width} = \frac{1}{2}x$, and $u = 24$.

$2 \cdot (x + \frac{1}{2}x) = 24$ $2 \cdot \frac{3}{2}x = 24 \implies 3x = 24 \implies x = 8$

Thus:

• Length: $x = 8$
• Width: $\frac{1}{2}x = 4$

Task 3a:

• Perimeter formula: $u = 2 \cdot (\text{Length} + \text{Width})$
• Here, $\text{Length} = x + 5$, $\text{Width} = x$, and $u = 94$.

$2 \cdot (x + (x + 5)) = 94$ $2 \cdot (2x + 5) = 94 \implies 4x + 10 = 94 \implies 4x = 84 \implies x = 21$

Thus:

• Width: $x = 21$
• Length: $x + 5 = 26$

Task 3b:

• Perimeter formula: $u = 2 \cdot (\text{Length} + \text{Width})$
• Here, $\text{Length} = 3x$, $\text{Width} = x$, and $u = 336$.

$2 \cdot (3x + x) = 336$ $2 \cdot 4x = 336 \implies 8x = 336 \implies x = 42$

Thus:

• Width: $x = 42$
• Length: $3x = 126$

Summary of Solutions:

1. Task c: Length = $14$, Width = $7$
2. Task d: Length = $8$, Width = $4$
3. Task 3a: Length = $26$, Width = $21$
4. Task 3b: Length = $126$, Width = $42$

Would you like further clarifications or detailed explanations for any of the steps?

5 Related Questions:

1. How do you generalize the relationship between length, width, and perimeter?
2. Can you explain why dividing by 2 is necessary in perimeter equations?
3. How would the equation change if one side were $4$ times the other?
4. Can we derive area formulas from these length and width values?
5. What happens if the length and width are expressed in terms of fractions?

Tip:

When solving geometry problems, always start by writing the formula clearly and substituting values step-by-step to avoid mistakes.