## Task 5 - 030835

Given are the segments $s-a=3 \mathrm{~cm}, s-b=2 \mathrm{~cm}, s-c=1 \mathrm{~cm}$, where $2 s=a+b+c$ is the perimeter of the triangle.

a) Construct the triangle!

b) Justify the construction!
It holds that:

$$
\begin{aligned}
a & =a+b+c-(b+c)=2 s-b-c=(s-b)+(s-c) \\
b & =a+b+c-(a+c)=2 s-a-c=(s-a)+(s-c) \\
c & =a+b+c-(a+b)=2 s-a-b=(s-a)+(s-b)
\end{aligned}
$$

From the three sides described in this way, one constructs the triangle in the usual manner.

Note: For $a, b, c$, the side lengths result in 3, 4, and $5 \text{ cm}$, which corresponds to a right-angled triangle.

## Solution adopted from $[5]$