unit 5 - quadrilaterals
calculators may be used on this portion of the test.
1. what is the sum of the measures of the interior angles of a 27 - gon?
2. if the sum of the interior angles of a polygon is 2340°, how many sides does the polygon have?
3. what is the measure of each interior angle of a regular octagon?
for questions 4 - 6, each quadrilateral is a parallelogram.
4. find ( mangle a ).
(there is a parallelogram abcd with ( angle a=(13x - 25)° ) and ( angle c=(9x - 1)° ))
5. if ( xr = 4x - 9 ) and ( rz = 2x + 5 ), find ( xz ).
(there is a parallelogram wxyz with diagonals intersecting at r)
6. find the measure of each numbered angle.
(there is a rhombus pqrs with diagonals intersecting at t, ( mangle pqr = 106° ), some angles marked as 49°, 35° etc. and angles 1 - 6 numbered)

Question

Accepted Answer

### Question 1: Sum of interior angles of a 27 - gon
# Explanation:
## Step 1: Recall the formula for the sum of interior angles of a polygon
The formula for the sum of the interior angles of a polygon with $n$ sides is $S=(n - 2)\times180^{\circ}$, where $n$ is the number of sides of the polygon.
## Step 2: Substitute $n = 27$ into the formula
For a 27 - gon, $n = 27$. So we substitute $n=27$ into the formula $S=(n - 2)\times180^{\circ}$.
$S=(27- 2)\times180^{\circ}$
$S = 25\times180^{\circ}$
$S=4500^{\circ}$
# Answer:
The sum of the measures of the interior angles of a 27 - gon is $4500^{\circ}$

### Question 2: Number of sides of a polygon with sum of interior angles $2340^{\circ}$
# Explanation:
## Step 1: Use the formula for the sum of interior angles
We know that the sum of interior angles $S=(n - 2)\times180^{\circ}$, and we are given that $S = 2340^{\circ}$. So we set up the equation $(n - 2)\times180^{\circ}=2340^{\circ}$
## Step 2: Solve for $n$
First, divide both sides of the equation by $180^{\circ}$:
$\frac{(n - 2)\times180^{\circ}}{180^{\circ}}=\frac{2340^{\circ}}{180^{\circ}}$
$n-2 = 13$
Then, add 2 to both sides of the equation:
$n=13 + 2$
$n = 15$
# Answer:
The polygon has 15 sides.

### Question 3: Measure of each interior angle of a regular octagon
# Explanation:
## Step 1: Find the sum of interior angles of an octagon
An octagon has $n = 8$ sides. Using the formula $S=(n - 2)\times180^{\circ}$, we substitute $n = 8$:
$S=(8 - 2)\times180^{\circ}$
$S=6\times180^{\circ}$
$S = 1080^{\circ}$
## Step 2: Find the measure of each interior angle of a regular octagon
In a regular polygon, all interior angles are equal. So we divide the sum of interior angles by the number of sides ($n = 8$):
$\text{Each interior angle}=\frac{S}{n}=\frac{1080^{\circ}}{8}$
$\text{Each interior angle}=135^{\circ}$
# Answer:
The measure of each interior angle of a regular octagon is $135^{\circ}$

### Question 4: Find $m\angle A$ in parallelogram $ABCD$
# Explanation:
## Step 1: Recall the property of parallelogram angles
In a parallelogram, consecutive angles are supplementary, and opposite angles are equal. Also, in a parallelogram $ABCD$, $\angle A$ and $\angle C$ are opposite angles? No, wait, $\angle A$ and $\angle B$ are consecutive angles, $\angle A$ and $\angle D$ are consecutive angles, and $\angle A=\angle C$, $\angle B=\angle D$. Wait, looking at the diagram, $\angle A=(13x - 25)^{\circ}$ and $\angle C=(9x - 1)^{\circ}$. But in a parallelogram, opposite angles are equal, so $\angle A=\angle C$
So we set up the equation $13x-25=9x - 1$
## Step 2: Solve for $x$
Subtract $9x$ from both sides:
$13x-9x-25=9x - 9x-1$
$4x-25=- 1$
Add 25 to both sides:
$4x-25 + 25=-1 + 25$
$4x=24$
Divide both sides by 4:
$x = 6$
## Step 3: Find $m\angle A$
Substitute $x = 6$ into the expression for $\angle A$: $\angle A=(13x - 25)^{\circ}$
$\angle A=(13\times6-25)^{\circ}$
$\angle A=(78 - 25)^{\circ}$
$\angle A = 53^{\circ}$
# Answer:
$m\angle A = 53^{\circ}$

### Question 5: Find $XZ$ given $XR = 4x-9$ and $RZ=2x + 5$ in a parallelogram
# Explanation:
## Step 1: Recall the property of diagonals in a parallelogram
In a parallelogram, the diagonals bisect each other. So $XR=RZ$
## Step 2: Solve for $x$
Set $4x-9=2x + 5$
Subtract $2x$ from both sides:
$4x-2x-9=2x-2x + 5$
$2x-9=5$
Add 9 to both sides:
$2x-9 + 9=5 + 9$
$2x=14$
Divide both sides by 2:
$x = 7$
## Step 3: Find $XR$ and $RZ$
First, find $XR$: $XR=4x-9$, substitute $x = 7$
$XR=4\times7-9=28 - 9=19$
Find $RZ$: $RZ=2x + 5$, substitute $x = 7$
$RZ=2\times7+5=14 + 5=19$
## Step 4: Find $XZ$
Since $XZ=XR+RZ$ (because the diagonals bisect each other, $XZ$ is the sum of $XR$ and $RZ$)
$XZ=19 + 19=38$
# Answer:
The length of $XZ$ is 38

### Question 6: Find the measure of each numbered angle (assuming the diagram is a rhombus or a kite - like figure, we will use properties of parallelograms/rhombuses)
(Note: Since the diagram has some hand - written numbers, we assume the following: The quadrilateral $PQRS$ is a rhombus (since diagonals bisect angles and bisect each other). Given $m\angle PQR = 106^{\circ}$, and we have angles at $T$, $R$ etc.)

1. **Angle 1 and Angle 2**:
In a rhombus, the diagonal bisects the vertex angle. So if $\angle PQR=106^{\circ}$, the diagonal $QR$ (or the diagonal that splits $\angle PQR$) will split $\angle PQR$ into two equal angles? Wait, no, the diagonal $PR$ and $QS$ intersect at $T$. In a rhombus, diagonals bisect the angles. So $\angle 1=\angle 2$ and $\angle 1+\angle 2=\angle PQR = 106^{\circ}$, so $\angle 1=\angle 2=\frac{106^{\circ}}{2}=53^{\circ}$

2. **Angle 3**:
We know that in triangle $QTR$, we have some angles. Wait, we know that $\angle TRS = 35^{\circ}$, and in a rhombus, $PQ\parallel SR$, and $PR$ is a transversal. Also, in a rhombus, $PR$ bisects $\angle QRS$. If we assume that $\angle QRS$ and $\angle PQR$ are supplementary (since consecutive angles in a parallelogram are supplementary), $\angle QRS=180^{\circ}-\angle PQR=180 - 106 = 74^{\circ}$. Since the diagonal $PR$ bisects $\angle QRS$, $\angle 3=\frac{\angle QRS}{2}=\frac{74^{\circ}}{2}=37^{\circ}$? Wait, but we are given $\angle TRS = 35^{\circ}$, maybe my initial assumption is wrong. Alternatively, looking at the triangle with angle $49^{\circ}$ at $S$ and $35^{\circ}$ at $R$. In triangle $TSR$, the sum of angles is $180^{\circ}$, so $\angle STR=180-(49 + 35)=96^{\circ}$, and $\angle QTR$ is vertical to $\angle STR$, so $\angle QTR = 96^{\circ}$. Then in triangle $QTR$, with $\angle 2 = 53^{\circ}$ and $\angle QTR=96^{\circ}$, $\angle 3=180-(53 + 96)=31^{\circ}$? This part is a bit ambiguous due to the hand - written numbers, but assuming the standard properties:

- $\angle 1=\angle 2 = 53^{\circ}$ (since diagonal bisects the $106^{\circ}$ angle)
- $\angle 3 = 35^{\circ}$ (given or alternate interior angle)
- $\angle 4$: Since $\angle 4$ and $\angle STR$ are vertical angles? If $\angle STR = 180-(49+35)=96^{\circ}$, then $\angle 4 = 96^{\circ}$
- $\angle 5$: In a rhombus, $PS = PQ$, and the diagonal $QS$ bisects $\angle PSQ$ and $\angle PQR$. Also, $\angle 5$ and $\angle 3$ are equal (alternate interior angles) or $\angle 5 = 35^{\circ}$
- $\angle 6$: $\angle 6$ and $\angle 3$ are equal (since $PQ\parallel SR$ and $PR$ is a transversal), so $\angle 6 = 35^{\circ}$

(Note: Due to the ambiguity in the hand - written numbers, the above is a general approach. If we assume the following:

Given $m\angle PQR=106^{\circ}$, diagonal $QS$ bisects $\angle PQR$, so $\angle 1=\angle 2 = 53^{\circ}$

In triangle $TSR$, $\angle TSR = 49^{\circ}$, $\angle TRS=35^{\circ}$, so $\angle STR=180-(49 + 35)=96^{\circ}$, so $\angle 4=\angle STR = 96^{\circ}$ (vertical angles)

In triangle $PQT$, $\angle 1 = 53^{\circ}$, $\angle 4=96^{\circ}$, so $\angle 6=180-(53 + 96)=31^{\circ}$

Since $PQ = PS$ (rhombus), triangle $PSP$? No, $PQ = PS$, so triangle $PSQ$ is isosceles, but maybe $\angle 5=\angle 6 = 31^{\circ}$

And $\angle 3$: Since $\angle QRS=180 - 106 = 74^{\circ}$, and diagonal $PR$ bisects $\angle QRS$, $\angle 3=\frac{74^{\circ}}{2}=37^{\circ}$, but we are given $\angle TRS = 35^{\circ}$, there might be a typo. However, following the standard rhombus angle - bisecting property:

# Explanation (for $\angle 1$ and $\angle 2$):
## Step 1: Property of rhombus diagonals
In a rhombus, the diagonal bisects the vertex angle. So for $\angle PQR = 106^{\circ}$, the diagonal $QS$ bisects $\angle PQR$ into two equal angles.
## Step 2: Calculate $\angle 1$ and $\angle 2$
$\angle 1=\angle 2=\frac{\angle PQR}{2}=\frac{106^{\circ}}{2}=53^{\circ}$

# Answer (for $\angle 1$ and $\angle 2$):
$\angle 1 = 53^{\circ}$, $\angle 2=53^{\circ}$

(For other angles, due to the ambiguity in the diagram, we can only provide the calculation for the angles that can be determined with the given information. If we assume the above - mentioned steps, the angles are: $\angle 1 = 53^{\circ}$, $\angle 2 = 53^{\circ}$, $\angle 3=37^{\circ}$ (if $\angle QRS = 74^{\circ}$), $\angle 4 = 96^{\circ}$, $\angle 5 = 31^{\circ}$, $\angle 6=31^{\circ}$)

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QUESTION IMAGE

Question

Explanation:

Question 1: Sum of interior angles of a 27 - gon

Step 1: Recall the formula for the sum of interior angles of a polygon

Step 2: Substitute \(n = 27\) into the formula

Step 1: Use the formula for the sum of interior angles

Step 2: Solve for \(n\)

Step 1: Find the sum of interior angles of an octagon

Step 2: Find the measure of each interior angle of a regular octagon

Answer:

Question 2: Number of sides of a polygon with sum of interior angles \(2340^{\circ}\)