Péta constructed several plane figures from identical triangles, see the image. The perimeters of the first three are $8 \mathrm{~cm}, 11.4 \mathrm{~cm}$, and $14.7 \mathrm{~cm}$, respectively.

Determine the perimeter of the fourth figure.

(E. Semerádová)
![](https://cdn.mathpix.com/cropped/2024_04_17_d675c96fae1a3d3b24cag-6.jpg?height=352&width=1510&top_left_y=236&top_left_x=272)

Note: the image is for illustration only.
In the perimeters of the first, second, and fourth figures, two out of the three sides of the basic triangle are always included, such that each of the two sides is counted twice, and in the perimeters of different figures, different pairs of sides are included.

In the perimeter of the third figure, all sides of the basic triangle are included, and again each is counted twice.

Thus, the sum of the perimeters of the first, second, and fourth figures is equal to twice the perimeter of the third figure. This means that the unknown perimeter of the fourth figure is equal to the difference between the perimeters of the first and second figures and twice the perimeter of the third:

$$
2 \cdot 14.7 - 8 - 11.4 = 29.4 - 19.4 = 10(\mathrm{~cm}).
$$

Notes. From the problem statement, the lengths of the sides of the basic triangle, which we will denote as \(a, b, c\), can be derived:
![](https://cdn.mathpix.com/cropped/2024_04_17_d675c96fae1a3d3b24cag-6.jpg?height=404&width=1514&top_left_y=1382&top_left_x=270)

The difference in the perimeters of the third and first figures is equal to twice \(c\), thus

$$
c = \frac{1}{2}(14.7 - 8) = 3.35(\mathrm{~cm})
$$

The difference in the perimeters of the third and second figures is equal to twice \(b\), thus

$$
b = \frac{1}{2}(14.7 - 11.4) = 1.65(\mathrm{~cm})
$$

From this and the known perimeters of the first three figures, the length of the last side of the basic triangle can be expressed:

$$
a = \frac{8}{2} - 1.65 = \frac{11.4}{2} - 3.35 = \frac{14.7}{2} - 1.65 - 3.35 = 2.35(\mathrm{~cm}).
$$

The values of \(a, b, c\) satisfy the triangle inequalities \((1.65 + 2.35 > 3.35\) etc.), so a triangle with sides of these lengths actually exists and (up to scale) looks like this:

![](https://cdn.mathpix.com/cropped/2024_04_17_d675c96fae1a3d3b24cag-7.jpg?height=232&width=377&top_left_y=341&top_left_x=842)

From the above, we can derive the perimeter of the fourth figure for verification:

$$
2(b + c) = 2(1.65 + 3.35) = 10(\mathrm{~cm}).
$$