SOLUTION:
Step 1:
we are to find the measure of minor arc cut off by one of the diagonals;
The sum of interior angles in a pentagon is:
[tex]\begin{gathered} (n-2)\text{ x 180} \\ (5-2)\text{ x 180} \\ 3\text{ x 180} \\ 540 \end{gathered}[/tex]Each interior angle of a regular pentagon is
[tex]\frac{540}{5}\text{ = 108}[/tex]So the size of the major arc can be gotten by the circle theorem; the angle at the centre is twice the angle at the circumference.
[tex]\text{The major arc = 2 x 108 = 216}[/tex]Then recall,
[tex]\begin{gathered} \text{The major arc + the minor arc = 360 (sum of angles at a point)} \\ 216\text{ + the minor arc = 360} \\ \text{The minor arc = 360 - 216} \\ \text{The minor arc =144} \end{gathered}[/tex]Step 2:
We are to find the length of the same minor arc in problem;
[tex]\frac{\theta}{360}\text{ x 2 }\pi\text{ r}[/tex]Where our angle (titan) is 144 and radius is 10
[tex]\begin{gathered} \frac{144}{360}\text{ x 2 x }\pi\text{ x 10} \\ \\ 8\pi \end{gathered}[/tex]So the length of the minor arc, given that the radius is 10 cm is 8 pi