7. Dijana used sticks of length $5 \mathrm{~cm}$ to form a row of equilateral triangles (as shown in the figure). If Dijana used 99 sticks, what is the distance between the two farthest points in the thus formed row of triangles?

![](https://cdn.mathpix.com/cropped/2024_06_03_acca65fe0c81619e2622g-04.jpg?height=197&width=916&top_left_y=1244&top_left_x=570)

The use of a pocket calculator or any reference materials is not allowed.

## SCHOOL/CITY COMPETITION IN MATHEMATICS
Another way:

Based on the image

![](https://cdn.mathpix.com/cropped/2024_05_30_494875a1129ddb904645g-07.jpg?height=140&width=660&top_left_y=384&top_left_x=732)

we conclude:

| Number of triangles | 1 | 2 | 3 | 4 | 5 | $\cdots$ | $x$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Number of matches | 3 | 5 | 7 | 9 | 11 | $\cdots$ | $2 x+1$ |

i.e., to create a sequence containing $x$ triangles, $2 x+1$ matches are needed.

If Diana used 99 matches, she formed $(99-1): 2=49$ triangles in the sequence. 2 POINTS

Consider sequences composed of an odd number of triangles.
![](https://cdn.mathpix.com/cropped/2024_05_30_494875a1129ddb904645g-07.jpg?height=156&width=1516&top_left_y=1095&top_left_x=290)

If we denote the length of the side of a triangle by $a$, we can write:

| Number of triangles | 1 | 3 | 5 | 7 | $\ldots$ | $n$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Distance between the farthest points | $1 \cdot a$ | $2 \cdot a$ | $3 \cdot a$ | $4 \cdot a$ | $\cdots$ | $((n+1): 2) \cdot a$ |

Thus, if the sequence contains an odd number of triangles, the distance between the farthest points is $((\text{number of triangles} + 1): 2) \cdot$ length of the side of the triangle.

For a sequence of 49 triangles, the required distance is $((49+1): 2) \cdot$ length of the side of the triangle.

If each match is $a=5 \mathrm{~cm}$ long, then the distance between the two farthest points is $(50: 2) \cdot 5=25 \cdot 5=125 \mathrm{~cm}$.