Question

sum of exterior angles of a polygon 360
exterior angle of a regular polygon ( e_n=\frac{360}{n})
number of sides of a regular polygon ( n = \frac{360}{e_n})
practice questions

1. what is the sum of the measures of the interior angles of a pentagon?
2. what is the sum of the measures of the interior angles of a 27 - gon?
3. what is the measure of each interior angle of a regular octagon? ( s = 180(8 - 2))
4. what is the measure of each interior angle of a regular 20 - gon?
5. five angles of a hexagon measure 119°, 129°, 104°, 139°, and 95°. what is the measure of the sixth angle?
6. the sum of the interior angles of a polygon is 1620°. how many sides does the polygon have?
7. the sum of the interior angles of a polygon is 3960°. how many sides does the polygon have?
8. what is the sum of the measures of the exterior angles of a nonagon?
9. what is the measure of each exterior angle of a 20 - gon?

Explanation:

Step1: Recall interior - angle sum formula

The sum of the interior angles of a polygon is given by $S=(n - 2)\times180^{\circ}$, where $n$ is the number of sides.

Step2: Solve question 1

For a pentagon, $n = 5$. Then $S=(5 - 2)\times180^{\circ}=3\times180^{\circ}=540^{\circ}$.

Step3: Solve question 2

For a 27 - gon, $n = 27$. Then $S=(27 - 2)\times180^{\circ}=25\times180^{\circ}=4500^{\circ}$.

Step4: Solve question 3

For a regular octagon, $n = 8$. The sum $S=(8 - 2)\times180^{\circ}=1080^{\circ}$. Each interior angle $A=\frac{(8 - 2)\times180^{\circ}}{8}=\frac{1080^{\circ}}{8}=135^{\circ}$.

Step5: Solve question 4

For a regular 20 - gon, $n = 20$. The sum $S=(20 - 2)\times180^{\circ}=3240^{\circ}$. Each interior angle $A=\frac{(20 - 2)\times180^{\circ}}{20}=\frac{3240^{\circ}}{20}=162^{\circ}$.

Step6: Solve question 5

The sum of the interior angles of a hexagon ($n = 6$) is $S=(6 - 2)\times180^{\circ}=720^{\circ}$. The sum of the five given angles is $119^{\circ}+129^{\circ}+104^{\circ}+139^{\circ}+95^{\circ}=586^{\circ}$. The sixth angle is $720^{\circ}-586^{\circ}=134^{\circ}$.

Step7: Solve question 6

Given $S = 1620^{\circ}$, we use the formula $S=(n - 2)\times180^{\circ}$. So, $1620^{\circ}=(n - 2)\times180^{\circ}$. Then $n-2=\frac{1620^{\circ}}{180^{\circ}} = 9$, and $n=9 + 2=11$.

Step8: Solve question 7

Given $S = 3960^{\circ}$, we use the formula $S=(n - 2)\times180^{\circ}$. So, $3960^{\circ}=(n - 2)\times180^{\circ}$. Then $n - 2=\frac{3960^{\circ}}{180^{\circ}}=22$, and $n=22 + 2=24$.

Step9: Recall exterior - angle sum property

The sum of the exterior angles of any polygon is always $360^{\circ}$. So for a non - agon, the sum of exterior angles is $360^{\circ}$.

Step10: Solve question 9

For a regular 20 - gon, since the sum of exterior angles is $360^{\circ}$, each exterior angle $A=\frac{360^{\circ}}{20}=18^{\circ}$.

Answer:

1. $540^{\circ}$
2. $4500^{\circ}$
3. $135^{\circ}$
4. $162^{\circ}$
5. $134^{\circ}$
6. $11$
7. $24$
8. $360^{\circ}$
9. $18^{\circ}$