for exercises 1–4, find the sum of the interior angles and the measure of each interior angle for the given regular polygons. round to the nearest tenth as needed.
1. 12 - gon
2. 102 - gon
3. 90 - gon
4. 36 - gon
for exercises 5–8, given the measure of an interior angle of a regular polygon, how many sides does each polygon have?
5. 174°
6. 156°
7. 165°
8. 177.5°
for exercises 9–12, given the measure of an exterior angle of a regular polygon, how many sides does the polygon have?
9. 12°
10. 45°
11. 18°
12. 14.4°
13. what is the measure of each exterior angle?
14. what is the value of x?
15. understand why is the sum of the interior angles of a polygon always a multiple of 180?

Question

Accepted Answer

### Exercise 1: 12 - gon
# Explanation:
## Step 1: Sum of interior angles
The formula for the sum of interior angles of a polygon is $S=(n - 2)\times180^{\circ}$, where $n$ is the number of sides. For a 12 - gon, $n = 12$.
$S=(12 - 2)\times180^{\circ}=10\times180^{\circ}=1800^{\circ}$
## Step 2: Measure of each interior angle
In a regular polygon, each interior angle $\theta=\frac{(n - 2)\times180^{\circ}}{n}$. For $n = 12$,
$\theta=\frac{1800^{\circ}}{12}=150^{\circ}$
# Answer:
Sum of interior angles: $1800^{\circ}$, Each interior angle: $150^{\circ}$

### Exercise 2: 102 - gon
# Explanation:
## Step 1: Sum of interior angles
Using $S=(n - 2)\times180^{\circ}$, with $n = 102$.
$S=(102 - 2)\times180^{\circ}=100\times180^{\circ}=18000^{\circ}$
## Step 2: Measure of each interior angle
$\theta=\frac{(n - 2)\times180^{\circ}}{n}=\frac{18000^{\circ}}{102}\approx176.5^{\circ}$ (rounded to the nearest tenth)
# Answer:
Sum of interior angles: $18000^{\circ}$, Each interior angle: $\approx176.5^{\circ}$

### Exercise 3: 90 - gon
# Explanation:
## Step 1: Sum of interior angles
Using $S=(n - 2)\times180^{\circ}$, $n = 90$.
$S=(90 - 2)\times180^{\circ}=88\times180^{\circ}=15840^{\circ}$
## Step 2: Measure of each interior angle
$\theta=\frac{(n - 2)\times180^{\circ}}{n}=\frac{15840^{\circ}}{90}=176^{\circ}$
# Answer:
Sum of interior angles: $15840^{\circ}$, Each interior angle: $176^{\circ}$

### Exercise 4: 36 - gon
# Explanation:
## Step 1: Sum of interior angles
Using $S=(n - 2)\times180^{\circ}$, $n = 36$.
$S=(36 - 2)\times180^{\circ}=34\times180^{\circ}=6120^{\circ}$
## Step 2: Measure of each interior angle
$\theta=\frac{(n - 2)\times180^{\circ}}{n}=\frac{6120^{\circ}}{36}=170^{\circ}$
# Answer:
Sum of interior angles: $6120^{\circ}$, Each interior angle: $170^{\circ}$

### Exercise 5: Interior angle $174^{\circ}$
# Explanation:
## Step 1: Recall the formula for interior angle
$\theta=\frac{(n - 2)\times180^{\circ}}{n}$, we know $\theta = 174^{\circ}$.
$174^{\circ}=\frac{(n - 2)\times180^{\circ}}{n}$
## Step 2: Solve for $n$
Multiply both sides by $n$: $174n=(n - 2)\times180$
Expand: $174n = 180n-360$
Subtract $174n$ from both sides: $0 = 6n - 360$
Add 360 to both sides: $6n=360$
Divide by 6: $n = 60$
# Answer:
The polygon has 60 sides.

### Exercise 6: Interior angle $156^{\circ}$
# Explanation:
## Step 1: Use the interior angle formula
$156^{\circ}=\frac{(n - 2)\times180^{\circ}}{n}$
## Step 2: Solve for $n$
Multiply both sides by $n$: $156n=(n - 2)\times180$
Expand: $156n=180n - 360$
Subtract $156n$: $0 = 24n-360$
Add 360: $24n = 360$
Divide by 24: $n = 15$
# Answer:
The polygon has 15 sides.

### Exercise 7: Interior angle $165^{\circ}$
# Explanation:
## Step 1: Use the interior angle formula
$165^{\circ}=\frac{(n - 2)\times180^{\circ}}{n}$
## Step 2: Solve for $n$
Multiply by $n$: $165n=(n - 2)\times180$
Expand: $165n=180n - 360$
Subtract $165n$: $0 = 15n-360$
Add 360: $15n = 360$
Divide by 15: $n = 24$
# Answer:
The polygon has 24 sides.

### Exercise 8: Interior angle $177.5^{\circ}$
# Explanation:
## Step 1: Use the interior angle formula
$177.5^{\circ}=\frac{(n - 2)\times180^{\circ}}{n}$
## Step 2: Solve for $n$
Multiply by $n$: $177.5n=(n - 2)\times180$
Expand: $177.5n=180n - 360$
Subtract $177.5n$: $0 = 2.5n-360$
Add 360: $2.5n = 360$
Divide by 2.5: $n=\frac{360}{2.5}=144$
# Answer:
The polygon has 144 sides.

### Exercise 9: Exterior angle $12^{\circ}$
# Explanation:
## Step 1: Recall exterior angle property
The sum of exterior angles of any polygon is $360^{\circ}$. For a regular polygon, each exterior angle $\phi=\frac{360^{\circ}}{n}$, where $n$ is the number of sides.
We know $\phi = 12^{\circ}$, so $n=\frac{360^{\circ}}{\phi}$
## Step 2: Calculate $n$
$n=\frac{360^{\circ}}{12^{\circ}} = 30$
# Answer:
The polygon has 30 sides.

### Exercise 10: Exterior angle $45^{\circ}$
# Explanation:
## Step 1: Use exterior angle formula
$n=\frac{360^{\circ}}{\phi}$, $\phi = 45^{\circ}$
## Step 2: Calculate $n$
$n=\frac{360^{\circ}}{45^{\circ}}=8$
# Answer:
The polygon has 8 sides.

### Exercise 11: Exterior angle $18^{\circ}$
# Explanation:
## Step 1: Use exterior angle formula
$n=\frac{360^{\circ}}{\phi}$, $\phi = 18^{\circ}$
## Step 2: Calculate $n$
$n=\frac{360^{\circ}}{18^{\circ}} = 20$
# Answer:
The polygon has 20 sides.

### Exercise 12: Exterior angle $14.4^{\circ}$
# Explanation:
## Step 1: Use exterior angle formula
$n=\frac{360^{\circ}}{\phi}$, $\phi = 14.4^{\circ}$
## Step 2: Calculate $n$
$n=\frac{360^{\circ}}{14.4^{\circ}}=25$
# Answer:
The polygon has 25 sides.

### Exercise 13: Measure of each exterior angle
First, find the sum of interior angles of the pentagon (the figure has 5 sides? Wait, looking at the diagram: the interior angles are $124^{\circ}$, $118^{\circ}$, $100^{\circ}$, $100^{\circ}$, and $x^{\circ}$. The sum of interior angles of a pentagon ($n = 5$) is $(5 - 2)\times180^{\circ}=540^{\circ}$
# Explanation:
## Step 1: Find $x$
Sum of interior angles: $124 + 118+100 + 100+x=540$
$442+x = 540$
$x=540 - 442=98^{\circ}$
## Step 2: Find exterior angles
For a polygon, interior angle + exterior angle $=180^{\circ}$
- For the angle adjacent to $124^{\circ}$ (angle 1's adjacent interior angle): exterior angle $m\angle1=180 - 124 = 56^{\circ}$
- For $x = 98^{\circ}$ (angle 2's adjacent interior angle): exterior angle $m\angle2=180 - 98 = 82^{\circ}$
- For $118^{\circ}$ (angle 3's adjacent interior angle): exterior angle $m\angle3=180 - 118 = 62^{\circ}$
- For $100^{\circ}$ (angle 4's adjacent interior angle): exterior angle $m\angle4=180 - 100 = 80^{\circ}$
- For $100^{\circ}$ (angle 5's adjacent interior angle): exterior angle $m\angle5=180 - 100 = 80^{\circ}$
Wait, but maybe the question is about the sum of exterior angles? Wait, the problem says "What is the measure of each exterior angle?" But the figure has multiple exterior angles. Wait, maybe it's a pentagon, and we can also use the fact that sum of exterior angles of any polygon is $360^{\circ}$. Let's check the sum of the exterior angles we calculated: $56 + 82+62 + 80+80=360^{\circ}$, which matches. But if we consider a regular polygon, but this is not regular. Wait, maybe the question is mis - phrased, but if we assume it's about the exterior angles we calculated above, each exterior angle is as follows:
$m\angle1 = 56^{\circ}$, $m\angle2 = 82^{\circ}$, $m\angle3 = 62^{\circ}$, $m\angle4 = 80^{\circ}$, $m\angle5 = 80^{\circ}$
# Answer:
$m\angle1 = 56^{\circ}$, $m\angle2 = 82^{\circ}$, $m\angle3 = 62^{\circ}$, $m\angle4 = 80^{\circ}$, $m\angle5 = 80^{\circ}$

### Exercise 14: Value of $x$
The figure is a pentagon ($n = 5$), sum of interior angles $=(5 - 2)\times180^{\circ}=540^{\circ}$
# Explanation:
## Step 1: Set up the equation
The angles are $(4x + 15)^{\circ}$, $(4x + 15)^{\circ}$, $(4x-10)^{\circ}$, $(4x - 10)^{\circ}$, $(5x + 5)^{\circ}$
Sum: $(4x + 15)+(4x + 15)+(4x-10)+(4x - 10)+(5x + 5)=540$
## Step 2: Simplify the equation
Combine like terms:
$(4x+4x + 4x+4x+5x)+(15 + 15-10 - 10 + 5)=540$
$21x+(15)=540$
$21x=540 - 15=525$
## Step 3: Solve for $x$
$x=\frac{525}{21}=25$
# Answer:
$x = 25$

### Exercise 15: Why sum of interior angles is multiple of $180$
# Brief Explanations:
A polygon with $n$ sides can be divided into $(n - 2)$ non - overlapping triangles by drawing diagonals from one vertex. The sum of the interior angles of a triangle is $180^{\circ}$. So the sum of the interior angles of the polygon is $(n - 2)\times180^{\circ}$, which is a multiple of $180^{\circ}$ since $(n - 2)$ is an integer for $n\geq3$.
# Answer:
A polygon with $n$ sides can be divided into $(n - 2)$ triangles (by drawing diagonals from one vertex), and since the sum of angles in a triangle is $180^{\circ}$, the sum of interior angles of the polygon is $(n - 2)\times180^{\circ}$, hence a multiple of $180^{\circ}$.

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QUESTION IMAGE

Question

Explanation:

Exercise 1: 12 - gon

Step 1: Sum of interior angles

Step 2: Measure of each interior angle

Step 1: Sum of interior angles

Step 2: Measure of each interior angle

Step 1: Sum of interior angles

Step 2: Measure of each interior angle

Answer:

Exercise 2: 102 - gon