learning target: i can apply the properties of triangles to solve angles in polygons.
10) chebse drew a 16 - sided polygon.
score:
4) calculate the sum of the interior angles of the figure.

b) suppose the figure is a regular polygon. use the formula to calculate each interior angle measure.

11) what is the sum of the exterior angles of a 58 - gon?

12) what is the measure of each exterior angle of a regular icosaqen (20 - gon)? round answers to the nearest whole degree.

learning target: i can apply properties, postulates, and theorems to write proofs about parallel lines and congruent triangles.
score:
13) complete the proof.
given: ( overline{ya}congoverline{ba})
( angle bcongangle y)
prove: ( overline{az}congoverline{ac})

Question

learning target: i can apply the properties of triangles to solve angles in polygons.
10) chebse drew a 16 - sided polygon.
score:
4) calculate the sum of the interior angles of the figure.

b) suppose the figure is a regular polygon. use the formula to calculate each interior angle measure.

11) what is the sum of the exterior angles of a 58 - gon?

12) what is the measure of each exterior angle of a regular icosaqen (20 - gon)? round answers to the nearest whole degree.

learning target: i can apply properties, postulates, and theorems to write proofs about parallel lines and congruent triangles.
score:
13) complete the proof.
given: ( overline{ya}congoverline{ba})
( angle bcongangle y)
prove: ( overline{az}congoverline{ac})

Accepted Answer

### 10a)
# Explanation:
## Step1: Recall the formula for the sum of interior angles of a polygon.
The formula for the sum of the interior angles of an $ n $-sided polygon is $ S=(n - 2)\times180^{\circ} $, where $ n $ is the number of sides.
## Step2: Substitute $ n = 16 $ into the formula.
For a 16 - sided polygon, $ n=16 $. So we have $ S=(16 - 2)\times180^{\circ} $.
First, calculate $ 16-2 = 14 $. Then, $ 14\times180^{\circ}=2520^{\circ} $.
# Answer:
The sum of the interior angles is $ 2520^{\circ} $.

### 10b)
# Explanation:
## Step1: Recall the formula for each interior angle of a regular polygon.
For a regular $ n $-sided polygon, each interior angle $ I=\frac{(n - 2)\times180^{\circ}}{n} $.
## Step2: Substitute $ n = 16 $ into the formula.
We know from part (a) that $ (n - 2)\times180^{\circ}=2520^{\circ} $. Now, divide this sum by the number of sides $ n = 16 $. So $ I=\frac{2520^{\circ}}{16}=157.5^{\circ} $.
# Answer:
Each interior angle measures $ 157.5^{\circ} $.

### 11)
# Explanation:
## Step1: Recall the property of the sum of exterior angles of a polygon.
The sum of the exterior angles of any convex polygon (regardless of the number of sides $ n $) is always $ 360^{\circ} $.
## Step2: Apply the property to a 58 - gon.
Since the sum of exterior angles of any polygon is $ 360^{\circ} $, for a 58 - gon, the sum of exterior angles is also $ 360^{\circ} $.
# Answer:
The sum of the exterior angles of a 58 - gon is $ 360^{\circ} $.

### 12)
# Explanation:
## Step1: Recall the formula for each exterior angle of a regular polygon.
For a regular $ n $-sided polygon, each exterior angle $ E=\frac{360^{\circ}}{n} $.
## Step2: Substitute $ n = 20 $ into the formula.
For a regular 20 - gon ($ n = 20 $), we have $ E=\frac{360^{\circ}}{20}=18^{\circ} $.
# Answer:
Each exterior angle of a regular 20 - gon measures $ 18^{\circ} $.

### 13)
# Brief Explanations:
1. **Step 1 (Statements and Reasons)**:
   - Given: $ \overline{YA}\cong\overline{BA} $ and $ \angle B\cong\angle Y $ (from the problem's "Given" section).
2. **Step 2 (Reason for $ \angle YAZ\cong\angle BAC $)**:
   - $ \angle YAZ $ and $ \angle BAC $ are vertical angles. By the Vertical Angles Theorem, vertical angles are congruent. So the reason for $ \angle YAZ\cong\angle BAC $ is "Vertical Angles are Congruent".
3. **Step 3 (Reason for $ \triangle AZY\cong\triangle ACB $)**:
   - We have $ \overline{YA}\cong\overline{BA} $ (given), $ \angle Y\cong\angle B $ (given), and $ \angle YAZ\cong\angle BAC $ (vertical angles). So by the Angle - Side - Angle (ASA) Congruence Postulate, $ \triangle AZY\cong\triangle ACB $. The reason is "ASA Congruence Postulate".
4. **Step 4 (Reason for $ \overline{AZ}\cong\overline{AC} $)**:
   - Corresponding parts of congruent triangles are congruent (CPCTC). Since $ \triangle AZY\cong\triangle ACB $, their corresponding sides $ \overline{AZ} $ and $ \overline{AC} $ are congruent. The reason is "CPCTC".

# Answer:
- 2) Reason: Vertical Angles are Congruent
- 3) Reason: ASA Congruence Postulate
- 4) Reason: CPCTC (Corresponding Parts of Congruent Triangles are Congruent)

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QUESTION IMAGE

Question

Explanation:

10a)

Step1: Recall the formula for the sum of interior angles of a polygon.

Step2: Substitute \( n = 16 \) into the formula.

Step1: Recall the formula for each interior angle of a regular polygon.

Step2: Substitute \( n = 16 \) into the formula.

Step1: Recall the property of the sum of exterior angles of a polygon.

Step2: Apply the property to a 58 - gon.

Answer:

10b)