Question

A regular pentagon is inscribed in a circle as shown. 1. Find the measure of minor arc cut off by one of the diagonals.2. Find the length of the same minor arc in problem 16a, given the radius of the circle is 10 cm. Leave the answer is terms of pi.

Answer 1

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:

`\begin{gathered} (n-2)\text{ x 180} \\ (5-2)\text{ x 180} \\ 3\text{ x 180} \\ 540 \end{gathered}`

Each interior angle of a regular pentagon is

`(540)/(5)\text{ = 108}`

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.

`\text{The major arc = 2 x 108 = 216}`

Then recall,

`\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}`

Step 2:

We are to find the length of the same minor arc in problem;

`(\theta)/(360)\text{ x 2 }\pi\text{ r}`

Where our angle (titan) is 144 and radius is 10

`\begin{gathered} (144)/(360)\text{ x 2 x }\pi\text{ x 10} \\ \\ 8\pi \end{gathered}`

So the length of the minor arc, given that the radius is 10 cm is 8 pi