Let's analyze the given image.

The problem states that a rectangle and a square have the same perimeter. The dimensions provided are:

1. Square: Each side is $2x + 8$.
2. Rectangle: Length is $4x + 8$ and width is $2x + 6$.

We need to find the condition under which these two shapes have equal perimeters.

Perimeter Calculation

1. Perimeter of the Square: $\text{Perimeter} = 4 \times \text{side} = 4(2x + 8) = 8x + 32$

2. Perimeter of the Rectangle: $\text{Perimeter} = 2 \times (\text{length} + \text{width}) = 2((4x + 8) + (2x + 6)) = 2(6x + 14) = 12x + 28$

Setting Perimeters Equal

Since the perimeters are equal: $8x + 32 = 12x + 28$

Simplifying this equation: $8x + 32 - 12x = 28 \implies -4x + 32 = 28$ $-4x = -4 \implies x = 1$

Conclusion

The value of $x$ that satisfies the condition of equal perimeters is $x = 1$.

Would you like a detailed explanation on any part of this solution, or have additional questions?

Related Questions:

1. How do you find the perimeter of a triangle given its side lengths?
2. What is the relationship between perimeter and area for similar shapes?
3. How would the problem change if the square and rectangle had the same area instead of perimeter?
4. Can you derive the area of the rectangle and square for the given value of $x$?
5. What are some real-world applications of perimeter and area calculations?

Tip: When comparing geometric properties like perimeter or area, always make sure to carefully evaluate the algebraic expressions given for each dimension.