7. a) All triangles that fit the described configuration are isosceles triangles where the leg is 2 cm shorter than the base.

If we denote the length of the base by $a$, and the length of the legs by $b$, the triangle inequality must hold: $2b > a$.

1 POINT

Thus, we have the following triangles:

| $a$ | $b$ | $2b > a$ |
| :--- | :--- | :--- |
| $10 \text{ cm}$ | $8 \text{ cm}$ | YES |
| $8 \text{ cm}$ | $6 \text{ cm}$ | YES |
| $6 \text{ cm}$ | $4 \text{ cm}$ | YES |
| $4 \text{ cm}$ | $2 \text{ cm}$ | NO |

(table)

At most, three such triangles can be formed in the described manner.

b) The shape formed by folding the wire consists of three isosceles triangles with one common vertex.

Let's draw one such shape:

![](https://cdn.mathpix.com/cropped/2024_05_30_59f5bb0a569372aa766ag-6.jpg?height=531&width=766&top_left_y=1051&top_left_x=288)

The perimeter of this shape consists of one side of length $10 \text{ cm}$, three sides of length $8 \text{ cm}$, three sides of length $6 \text{ cm}$, and two sides of length $4 \text{ cm}$.

Thus, the sum of the perimeters of all the composed triangles is:

$o = 10 + 3 \cdot (8 + 6) + 2 \cdot 4$

1 POINT

$o = 60 \text{ cm}$

1 POINT

c) We need to calculate the area of a rectangle whose sides differ by $12 \text{ cm}$, and whose perimeter is equal to the length of the wire.

$o = 2(a + b)$

![](https://cdn.mathpix.com/cropped/2024_05_30_59f5bb0a569372aa766ag-6.jpg?height=239&width=465&top_left_y=2025&top_left_x=287)

$a$

Since the perimeter is $60 \text{ cm}$, the sum of the lengths of the adjacent sides of the rectangle is $a + b = 30 \text{ cm}$.

If $a - b = 12 \text{ cm}$, then $a = b + 12$, so we have:

$b + 12 + b = 30$.

Thus, $b = 9 \text{ cm}$, and $a = 21 \text{ cm}$.

The area of this rectangle is:

$\mathrm{P} = 21 \cdot 9 = 189 \text{ cm}^2$.
7. From a long piece of wire, isosceles triangles are formed in the following way:

A segment of length $10 \mathrm{~cm}$ and two segments of length $8 \mathrm{~cm}$ are measured and used to form an isosceles triangle, without cutting the wire. Then, the formation of isosceles triangles from the remaining wire continues such that the lengths of the sides of each subsequent triangle are $2 \mathrm{~cm}$ shorter than the corresponding sides of the previous triangle (all triangles share one common vertex). The process continues without cutting the wire, as long as possible.

a) How many such triangles can be formed at most if the process continues in the described manner?

b) When all possible triangles are formed in the described manner, the wire is cut. What is the sum of the perimeters of all the triangles thus formed?

c) From the same (cut) piece of wire, a rectangle is formed whose adjacent sides differ in length by $12 \mathrm{~cm}$. Calculate the area of this rectangle.