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Question
name:date:per: 3unit 8: polygons & quadrilateralshomework 1: angles of polygons this is a 2-page document! 1. what is the sum of the measures of the interior angles of an octagon?2. what is the sum of the measures of the interior angles of a 25-gon?3. what is the measure of each interior angle of a regular hexagon?4. what is the measure of each interior angle of a regular 16-gon?5. what is the sum of the measures of the exterior angles of a decagon?6. what is the measure of each exterior angle of a regular 30-gon?7. an exterior angle of a regular polygon measures $22.5^\circ$. how many sides does it have?8. an interior angle of a regular polygon measures $170^\circ$. how many sides does it have?9. if the sum of the measures of the interior angles of a polygon is $1980^\circ$, how many sides does the polygon have?10. if the sum of the measures of the interior angles of a polygon is $5400^\circ$, how many sides does the polygon have?11. the measure of the seven angles in a nonagon measure $138^\circ, 154^\circ, 145^\circ, 132^\circ, 128^\circ, 147^\circ$, and $130^\circ$. if the two remaining angles are equal in measure, what is the measure of each angle?12. find the value of $x$.13. find the value of $x$.
Step1: Sum of interior angles formula
The formula for the sum of interior angles of an $n$-sided polygon is $(n-2)\times180^\circ$.
Step2: Solve Q1 (octagon, $n=8$)
Substitute $n=8$ into the formula:
$(8-2)\times180^\circ = 6\times180^\circ = 1080^\circ$
Step3: Solve Q2 (25-gon, $n=25$)
Substitute $n=25$ into the formula:
$(25-2)\times180^\circ = 23\times180^\circ = 4140^\circ$
Step4: Solve Q3 (regular hexagon, $n=6$)
First find total interior angles: $(6-2)\times180^\circ=720^\circ$. Divide by $n$ for regular polygon:
$\frac{720^\circ}{6}=120^\circ$
Step5: Solve Q4 (regular 16-gon, $n=16$)
First find total interior angles: $(16-2)\times180^\circ=2520^\circ$. Divide by $n$:
$\frac{2520^\circ}{16}=157.5^\circ$
Step6: Solve Q5 (exterior angles sum)
Sum of exterior angles of any polygon is always $360^\circ$.
Step7: Solve Q6 (regular 30-gon, exterior angle)
Divide total exterior angles by $n$:
$\frac{360^\circ}{30}=12^\circ$
Step8: Solve Q7 (exterior angle $22.5^\circ$)
Number of sides $n=\frac{360^\circ}{\text{exterior angle}}$:
$\frac{360^\circ}{22.5^\circ}=16$
Step9: Solve Q8 (interior angle $170^\circ$)
First find exterior angle: $180^\circ-170^\circ=10^\circ$. Then find $n$:
$\frac{360^\circ}{10^\circ}=36$
Step10: Solve Q9 (sum $1980^\circ$)
Rearrange interior sum formula: $n=\frac{\text{sum}}{180^\circ}+2$
$n=\frac{1980^\circ}{180^\circ}+2=11+2=13$
Step11: Solve Q10 (sum $5400^\circ$)
Use rearranged formula:
$n=\frac{5400^\circ}{180^\circ}+2=30+2=32$
Step12: Solve Q11 (nonagon, $n=9$)
First find total interior angles: $(9-2)\times180^\circ=1260^\circ$. Sum given angles:
$138+154+145+132+128+147+130=974^\circ$
Let each missing angle be $x$. Set up equation:
$974+2x=1260$
$2x=1260-974=286$
$x=\frac{286}{2}=143^\circ$
Step13: Solve Q12 (pentagon, $n=5$)
Total interior angles: $(5-2)\times180^\circ=540^\circ$. Sum given angles:
$87+105+135+92=419^\circ$
Solve for $x$:
$x=540-419=121^\circ$
Step14: Solve Q13 (nonagon, $n=9$)
Total interior angles: $1260^\circ$. Sum given angles:
$136+124+141+158+116+129+132=936^\circ$
Solve for $x$:
$x=1260-936=324^\circ$
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- $1080^\circ$
- $4140^\circ$
- $120^\circ$
- $157.5^\circ$
- $360^\circ$
- $12^\circ$
- $16$
- $36$
- $13$
- $32$
- $143^\circ$
- $121^\circ$
- $324^\circ$