The problem asks us to find the perimeter of two squares formed by cutting a wire of 44 cm into two parts. The total area of the two squares is given as 65 cm².

Let’s break it down:

1. Let the lengths of the two parts of the wire be $x$ and $44 - x$.
2. Since each part of the wire forms the perimeter of a square, the perimeter of the first square is $x$, and the perimeter of the second square is $44 - x$.
3. The perimeter of a square is $4 \times \text{side length}$. Therefore, the side length of the first square is $\frac{x}{4}$, and for the second square, it is $\frac{44 - x}{4}$.

Area:

• The area of a square is given by $\text{side length}^2$.
• Therefore, the area of the first square is $\left( \frac{x}{4} \right)^2 = \frac{x^2}{16}$, and the area of the second square is $\left( \frac{44 - x}{4} \right)^2 = \frac{(44 - x)^2}{16}$.

Total area:

We know that the total area of the two squares is 65 cm², so: $\frac{x^2}{16} + \frac{(44 - x)^2}{16} = 65$ Multiplying the entire equation by 16 to eliminate the denominator: $x^2 + (44 - x)^2 = 1040$ Expanding $(44 - x)^2$: $x^2 + (1936 - 88x + x^2) = 1040$ Simplifying: $2x^2 - 88x + 1936 = 1040$ Subtract 1040 from both sides: $2x^2 - 88x + 896 = 0$ Dividing the entire equation by 2: $x^2 - 44x + 448 = 0$ This is a quadratic equation. Now, we can solve for $x$ using the quadratic formula: $x = \frac{-(-44) \pm \sqrt{(-44)^2 - 4(1)(448)}}{2(1)}$ $x = \frac{44 \pm \sqrt{1936 - 1792}}{2}$ $x = \frac{44 \pm \sqrt{144}}{2}$ $x = \frac{44 \pm 12}{2}$ Thus, $x = \frac{44 + 12}{2} = 28$ or $x = \frac{44 - 12}{2} = 16$.

Conclusion:

• The first piece of the wire is 28 cm, forming a square with a perimeter of 28 cm.
• The second piece of the wire is 16 cm, forming a square with a perimeter of 16 cm.

Hence, the perimeters of the squares are 28 cm and 16 cm.

Would you like any further details or clarification?

Related Questions:

1. How do you determine the side length of a square given its perimeter?
2. What is the relationship between the perimeter and area of a square?
3. How can we use the quadratic formula to solve general quadratic equations?
4. If the wire had a different total length, how would that affect the sizes of the squares?
5. What other shapes could we form using the two pieces of wire, and how would the areas compare?

Tip:

When solving word problems, always start by clearly defining your variables and relationships, making it easier to set up equations!