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== 10.4#h(0.6em)Polygons, Perimeter, and Circumference


  #figure(figph[Two different patterns with triangles are shown.], alt: "Two different patterns with triangles are shown.", caption: [Geometric patterns are often used in fabrics due to the interest the shapes create.])


  
    
=== Learning Objectives


    After completing this section, you should be able to:


    + Identify polygons by their sides.
+ Identify polygons by their characteristics.
+ Calculate the perimeter of a polygon.
+ Calculate the sum of the measures of a polygon’s interior angles.
+ Calculate the sum of the measures of a polygon’s exterior angles.
+ Calculate the circumference of a circle.
+ Solve application problems involving perimeter and circumference.


  
  In our homes, on the road, everywhere we go, polygonal shapes are so common that we cannot count the many uses. Traffic signs, furniture, lighting, clocks, books, computers, phones, and so on, the list is endless. Many applications of polygonal shapes are for practical use, because the shapes chosen are the best for the purpose.


  Modern geometric patterns in fabric design have become more popular with time, and they are used for the beauty they lend to the material, the window coverings, the dresses, or the upholstery. This art is not done for any practical reason, but only for the interest these shapes can create, for the pure aesthetics of design.


  When designing fabrics, one has to consider the perimeter of the shapes, the triangles, the hexagons, and all polygons used in the pattern, including the circumference of any circular shapes. Additionally, it is the relationship of one object to another and experimenting with different shapes, changing perimeters, or changing angle measurements that we find the best overall design for the intended use of the fabric. In this section, we will explore these properties of polygons, the perimeter, the calculation of interior and exterior angles of polygons, and the circumference of a circle.


  
    
=== Identifying Polygons


    A #strong[polygon] is a closed, two-dimensional shape classified by the number of straight-line sides. See  for some examples. We show only up to eight-sided polygons, but there are many, many more.


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    If all the sides of a polygon have equal lengths and all the angles are equal, they are called #strong[regular polygons]. However, any shape with sides that are line segments can classify as a polygon. For example, the first two shapes, shown in  and , are both pentagons because they each have five sides and five vertices. The third shape  is a hexagon because it has six sides and six vertices. We should note here that the hexagon in  is a concave hexagon, as opposed to the first two shapes, which are convex pentagons. Technically, what makes a polygon concave is having an interior angle that measures greater than #math.equation(block: false, alt: "180 degrees")[$180^(∘)$]. They are hollowed out, or cave in, so to speak. Convex refers to the opposite effect where the shape is rounded out or pushed out.


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    While there are variations of all polygons, quadrilaterals contain an additional set of figures classified by angles and whether there are one or more pairs of parallel sides.


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    #examplebox("Example 1")[Identifying Polygons][
Identify each polygon.


        
      
      #solutionbox[
+ This shape has six sides. Therefore, it is a hexagon.
+ This shape has four sides, so it is a quadrilateral. It has two pairs of parallel sides making it a parallelogram.
+ This shape has eight sides making it an octagon.
+ This is an equilateral triangle, as all three sides are equal.
+ This is a rhombus; all four sides are equal.
+ This is a regular octagon, eight sides of equal length and equal angles.
]
]


    
    #examplebox("Example 2")[Determining Multiple Polygons][
What polygons make up ?


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      #solutionbox[
Shapes 1 and 5 are hexagons; shapes 2, 3, 4, 6, 7, 9, 10, 12, 13, 14, 15, and 16 are triangles; shapes 8 and 17 are parallelograms; and shape 11 is a trapezoid.
]
]


    
  
  
    
=== Perimeter


    Perimeter refers to the outside measurements of some area or region given in linear units. For example, to find out how much fencing you would need to enclose your backyard, you will need the perimeter. The general definition of #strong[perimeter] is the sum of the lengths of the sides of an enclosed region. For some geometric shapes, such as rectangles and circles, we have formulas. For other shapes, it is a matter of just adding up the side lengths.


    A #strong[rectangle] is defined as part of the group known as quadrilaterals, or shapes with four sides. A rectangle has two sets of parallel sides with four angles. To find the perimeter of a rectangle, we use the following formula:


    #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
The formula for the perimeter #math.equation(block: false, alt: "P")[$P$] of a rectangle is #math.equation(block: false, alt: "P equals 2 L plus 2 W")[$P = 2 L + 2 W$], twice the length #math.equation(block: false, alt: "L")[$L$] plus twice the width #math.equation(block: false, alt: "W")[$W$].
]


    For example, to find the length of a rectangle that has a perimeter of 24 inches and a width of 4 inches, we use the formula. Thus,


    #math.equation(block: true, alt: "24, equals, 2 l plus 2 open parenthesis 4 close parenthesis; equals, 2 l plus 8; 24 minus 8, equals, 2 l; 16, equals, 2 l; 8, equals, l")[$24 & = & 2 l + 2 ( 4 ) \  & = & 2 l + 8 \ 24 − 8 & = & 2 l \ 16 & = & 2 l \ 8 & = & l$]
    The length is 8 units.


    The perimeter of a regular polygon with #math.equation(block: false, alt: "n")[$n$] sides is given as #math.equation(block: false, alt: "P equals n times s")[$P = n ⋅ s$]. For example, the perimeter of an equilateral triangle, a triangle with three equal sides, and a side length of 7 cm is #math.equation(block: false, alt: "P equals 3 open parenthesis 7 close parenthesis equals 21 cm")[$P = 3 ( 7 ) = 21 #h(0.28em) "cm"$].


    #examplebox("Example 3")[Finding the Perimeter of a Pentagon][
Find the perimeter of a regular pentagon with a side length of 7 cm .


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((40.00pt, 20.00pt), (160.00pt, 20.00pt), (180.00pt, 80.00pt), (100.00pt, 140.00pt), (20.00pt, 80.00pt), close: true, fill: none, stroke: 0.7pt + black)
  content((100.00pt, 11.00pt), anchor: "north", text(size: 9pt)[7 cm])
})]),
  alt: "A pentagon with one of its sides marked 7 centimeters.", caption: none,
)


      
      #solutionbox[
A regular pentagon has five equal sides. Therefore, the perimeter is equal to #math.equation(block: false, alt: "P equals 5 open parenthesis 7 close parenthesis equals 35 cm")[$P = 5 ( 7 ) = 35 #h(0.28em) "cm"$].
]
]


    
    #examplebox("Example 4")[Finding the Perimeter of an Octagon][
Find the perimeter of a regular octagon with a side length of 14 cm .


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((40.00pt, 20.00pt), (160.00pt, 20.00pt), (180.00pt, 60.00pt), (180.00pt, 100.00pt), (160.00pt, 140.00pt), (40.00pt, 140.00pt), (20.00pt, 100.00pt), (20.00pt, 60.00pt), close: true, fill: none, stroke: 0.7pt + black)
  content((100.00pt, 11.00pt), anchor: "north", text(size: 9pt)[14 centimeters])
  content((177.81pt, 35.53pt), anchor: "west", text(size: 9pt)[14 centimeters])
  content((189.00pt, 80.00pt), anchor: "west", text(size: 9pt)[14 centimeters])
  content((177.81pt, 124.47pt), anchor: "west", text(size: 9pt)[14 centimeters])
  content((100.00pt, 149.00pt), anchor: "south", text(size: 9pt)[14 centimeters])
  content((22.19pt, 124.47pt), anchor: "east", text(size: 9pt)[14 centimeters])
  content((11.00pt, 80.00pt), anchor: "east", text(size: 9pt)[14 centimeters])
  content((22.19pt, 35.53pt), anchor: "east", text(size: 9pt)[14 centimeters])
})]),
  alt: "An octagon with its sides marked 14 centimeters.", caption: none,
)


      
      #solutionbox[
A regular octagon has eight sides of equal length. Therefore, the perimeter of a regular octagon with a side length of 14 cm is #math.equation(block: false, alt: "P equals 8 open parenthesis 14 close parenthesis equals 112 cm")[$P = 8 ( 14 ) = 112 #h(0.28em) "cm"$].
]
]


    
  
  
    
=== Sum of Interior and Exterior Angles


    To find the sum of the measurements of interior angles of a regular polygon, we have the following formula.


    #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
The sum of the interior angles of a polygon with #math.equation(block: false, alt: "n")[$n$] sides is given by


      #math.equation(block: true, alt: "S equals open parenthesis n minus 2 close parenthesis 180 degrees .")[$S = ( n − 2 ) 180^(∘) .$]
]


    For example, if we want to find the sum of the interior angles in a parallelogram, we have


    #math.equation(block: true, alt: "S, equals, open parenthesis 4 minus 2 close parenthesis 180 degrees; equals, 2 open parenthesis 180 close parenthesis equals 360 degrees .")[$S & = & ( 4 − 2 ) 180^(∘) \  & = & 2 ( 180 ) = 360^(∘) .$]
    Similarly, to find the sum of the interior angles inside a regular heptagon, we have


    #math.equation(block: true, alt: "S, equals, open parenthesis 7 minus 2 close parenthesis 180 degrees; equals, 5 open parenthesis 180 close parenthesis; equals, 900 degrees .")[$S & = & ( 7 − 2 ) 180^(∘) \  & = & 5 ( 180 ) \  & = & 900^(∘) .$]
    To find the measure of each interior angle of a regular polygon with #math.equation(block: false, alt: "n")[$n$] sides, we have the following formula.


    #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
The measure of each interior angle of a regular polygon with #math.equation(block: false, alt: "n")[$n$] sides is given by


      #math.equation(block: true, alt: "a equals the fraction open parenthesis n minus 2 close parenthesis 180 degrees over n .")[$a = frac(( n − 2 ) 180^(∘), n) "."$]
]


    For example, find the measure of an interior angle of a regular heptagon, as shown. We have


    #math.equation(block: true, alt: "a equals the fraction open parenthesis 7 minus 2 close parenthesis 180 degrees over 7 equals 128.57 degrees .")[$a = frac(( 7 − 2 ) 180^(∘), 7) = 128.57^(∘) "."$]
    #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((100.00pt, 20.00pt), (160.00pt, 40.00pt), (180.00pt, 100.00pt), (160.00pt, 160.00pt), (100.00pt, 180.00pt), (40.00pt, 160.00pt), (20.00pt, 100.00pt), close: true, fill: none, stroke: 0.7pt + black)
  arc((100.00pt, 20.00pt), start: 18.43deg, stop: 135.00deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((105.28pt, 42.38pt), text(size: 9pt)[128.57°])
})]),
  alt: "A heptagon with one of its angles marked 128.57 degrees.", caption: [Interior Angles],
)


    #examplebox("Example 5")[Calculating the Sum of Interior Angles][
Find the measure of an interior angle in a regular octagon using the formula, and then find the sum of all the interior angles using the sum formula.


      
      #solutionbox[
An octagon has eight sides, so #math.equation(block: false, alt: "n equals 8")[$n = 8$].


        #strong[Step 1:] Using the formula #math.equation(block: false, alt: "a equals the fraction open parenthesis n minus 2 close parenthesis 180 degrees over 8")[$a = frac(( n − 2 ) 180^(∘), 8)$]:


        #math.equation(block: true, alt: "a, equals, the fraction open parenthesis 8 minus 2 close parenthesis 180 degrees over 8; equals, the fraction open parenthesis 6 close parenthesis 180 degrees over 8; equals, 135 degrees .")[$a & = & frac(( 8 − 2 ) 180^(∘), 8) \  & = & frac(( 6 ) 180^(∘), 8) \  & = & 135^(∘) .$]
        So, the measure of each interior angle in a regular octagon is #math.equation(block: false, alt: "135 degrees")[$135^(∘)$].


        #strong[Step 2:] The sum of the angles inside an octagon, so using the formula:


        #math.equation(block: true, alt: "S, equals, open parenthesis n minus 2 close parenthesis 180 degrees; equals, open parenthesis 8 minus 2 close parenthesis 180 degrees; equals, 6 open parenthesis 180 close parenthesis; equals, 1,080 degrees .")[$S & = & ( n − 2 ) 180^(∘) \  & = & ( 8 − 2 ) 180^(∘) \  & = & 6 ( 180 ) \  & = & 1,080^(∘) .$]
        #strong[Step 3:] We can test this, as we already know the measure of each angle is #math.equation(block: false, alt: "135 degrees")[$135^(∘)$]. Thus, #math.equation(block: false, alt: "8 open parenthesis 135 degrees close parenthesis equals 1,080 degrees")[$8 ( 135^(∘) ) = 1,080^(∘)$].
]
]


    
    #examplebox("Example 6")[Calculating Interior Angles][
Use algebra to calculate the measure of each interior angle of the five-sided polygon .


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  content((100.00pt, 194.00pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[pentagon ABCDE])
  line((40.00pt, 20.00pt), (160.00pt, 20.00pt), (180.00pt, 100.00pt), (100.00pt, 160.00pt), (20.00pt, 100.00pt), close: true, fill: none, stroke: 0.7pt + black)
  arc((40.00pt, 20.00pt), start: 0.00deg, stop: 104.04deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((54.15pt, 38.13pt), text(size: 9pt)[5(x + 7)°])
  arc((160.00pt, 20.00pt), start: 75.96deg, stop: 180.00deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((145.85pt, 38.13pt), text(size: 9pt)[120°])
  arc((180.00pt, 100.00pt), start: 143.13deg, stop: 255.96deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((158.33pt, 92.30pt), text(size: 9pt)[6x + 25°])
  arc((100.00pt, 160.00pt), start: 216.87deg, stop: 323.13deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((100.00pt, 137.00pt), text(size: 9pt)[5(2x + 5)°])
  arc((20.00pt, 100.00pt), start: 284.04deg, stop: 36.87deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((41.67pt, 92.30pt), text(size: 9pt)[5(3x - 5)°])
  content((44.00pt, 24.00pt), anchor: "south-west", text(size: 9pt)[A])
  content((156.00pt, 24.00pt), anchor: "south-east", text(size: 9pt)[B])
  content((176.00pt, 100.00pt), anchor: "east", text(size: 9pt)[C])
  content((100.00pt, 156.00pt), anchor: "north", text(size: 9pt)[D])
  content((24.00pt, 100.00pt), anchor: "west", text(size: 9pt)[E])
})]),
  alt: "A polygon, A B C D E. The angles A, B, C, D, and E measure 5 (x plus 7) degrees, 120 degrees, 6 x plus 25 degrees, 5 (2 x plus 5) degrees, and 5 (3 x minus 5) degrees.", caption: none,
)


      
      #solutionbox[
#strong[Step 1:] Let us find out what the total of the sum of the interior angles should be. Use the formula for the sum of the angles in a polygon with #math.equation(block: false, alt: "n")[$n$] sides: #math.equation(block: false, alt: "S equals open parenthesis n minus 2 close parenthesis 180 degrees")[$S = ( n − 2 ) 180^(∘)$]. So, #math.equation(block: false, alt: "S equals open parenthesis 5 minus 2 close parenthesis 180 degrees equals 540 degrees")[$S = ( 5 − 2 ) 180^(∘) = 540^(∘)$].


        #strong[Step 2:] We add up all the angles and solve for #math.equation(block: false, alt: "x")[$x$]:


        #math.equation(block: true, alt: "5 open parenthesis x plus 7 close parenthesis plus 120 plus open parenthesis 6 x plus 25 close parenthesis plus 5 open parenthesis 2 x plus 5 close parenthesis plus 5 open parenthesis 3 x minus 5 close parenthesis, equals, 540; 5 x plus 6 x plus 10 x plus 15 x plus 180, equals, 540; 36 x, equals, 360; x, equals, 10")[$5 ( x + 7 ) + 120 + ( 6 x + 25 ) + 5 ( 2 x + 5 ) + 5 ( 3 x − 5 ) & = & 540 \ 5 x + 6 x + 10 x + 15 x + 180 & = & 540 \ 36 x & = & 360 \ x & = & 10$]
        #strong[Step 3:] We can then find the measure of each interior angle:


        #math.equation(block: true, alt: "m ∡ A equals 5 open parenthesis 10 plus 7 close parenthesis equals 85 degrees; m ∡ B equals 120 degrees; m ∡ C equals 6 open parenthesis 10 close parenthesis plus 25 equals 85 degrees; m ∡ D equals 5 open parenthesis 2 * 10 plus 5 close parenthesis equals 125 degrees; m ∡ E equals 5 open parenthesis 3 * 10 minus 5 close parenthesis equals 125 degrees")[$m ∡ A = 5 ( 10 + 7 ) = 85^(∘) \ m ∡ B = 120^(∘) \ m ∡ C = 6 ( 10 ) + 25 = 85^(∘) \ m ∡ D = 5 ( 2 * 10 + 5 ) = 125^(∘) \ m ∡ E = 5 ( 3 * 10 − 5 ) = 125^(∘)$]
]
]


    
    An #strong[exterior angle] of a regular polygon is an angle formed by extending a side length beyond the closed figure. The measure of an exterior angle of a regular polygon with #math.equation(block: false, alt: "n")[$n$] sides is found using the following formula:


    #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
To find the measure of an exterior angle of a regular polygon with #math.equation(block: false, alt: "n")[$n$] sides we use the formula


      #math.equation(block: true, alt: "b equals the fraction 360 degrees over n .")[$b = frac(360^(∘), n) "."$]
]


    In , we have a regular hexagon #math.equation(block: false, alt: "ABCDEF")[$italic("ABCDEF")$]. By extending the lines of each side, an angle is formed on the exterior of the hexagon at each vertex. The measure of each exterior angle is found using the formula, #math.equation(block: false, alt: "b equals the fraction 360 degrees over 6 equals 60 degrees")[$b = frac(360^(∘), 6) = 60^(∘)$].


    #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((140.00pt, 80.00pt), (120.00pt, 140.00pt), (60.00pt, 140.00pt), (20.00pt, 80.00pt), (60.00pt, 20.00pt), (120.00pt, 20.00pt), close: true, fill: none, stroke: 0.7pt + black)
  arc((140.00pt, 80.00pt), start: 71.57deg, stop: 108.43deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((140.00pt, 103.00pt), text(size: 9pt)[1])
  arc((120.00pt, 140.00pt), start: 135.00deg, stop: 180.00deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  arc((60.00pt, 140.00pt), start: 180.00deg, stop: 236.31deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  arc((20.00pt, 80.00pt), start: 251.57deg, stop: 303.69deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  arc((60.00pt, 20.00pt), start: 315.00deg, stop: 0.00deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  arc((120.00pt, 20.00pt), start: 0.00deg, stop: 71.57deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((136.00pt, 80.00pt), anchor: "east", text(size: 9pt)[A])
  content((120.00pt, 136.00pt), anchor: "north", text(size: 9pt)[B])
  content((60.00pt, 136.00pt), anchor: "north", text(size: 9pt)[C])
  content((24.00pt, 80.00pt), anchor: "west", text(size: 9pt)[D])
  content((60.00pt, 24.00pt), anchor: "south", text(size: 9pt)[E])
  content((120.00pt, 24.00pt), anchor: "south", text(size: 9pt)[F])
})]),
  alt: "A hexagon, A B C D E F. The exterior angles are marked at each vertex. The exterior angle at A is marked 1.", caption: [Exterior Angles],
)


    Now, an important point is that the #strong[sum] of the exterior angles of a regular polygon with #math.equation(block: false, alt: "n")[$n$] sides equals #math.equation(block: false, alt: "360 degrees .")[$360^(∘) .$] This implies that when we multiply the measure of one exterior angle by the number of sides of the regular polygon, we should get #math.equation(block: false, alt: "360 degrees .")[$360^(∘) .$] For the example in , we multiply the measure of each exterior angle, #math.equation(block: false, alt: "60 degrees")[$60^(∘)$], by the number of sides, six. Thus, the sum of the exterior angles is #math.equation(block: false, alt: "6 open parenthesis 60 degrees close parenthesis equals 360 degrees .")[$6 ( 60^(∘) ) = 360^(∘) .$]


    #examplebox("Example 7")[Calculating the Sum of Exterior Angles][
Find the sum of the measure of the exterior angles of the pentagon .


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((72.73pt, 54.55pt), (145.45pt, 54.55pt), (163.64pt, 109.09pt), (109.09pt, 145.45pt), (54.55pt, 109.09pt), close: true, fill: none, stroke: 0.7pt + black)
  arc((72.73pt, 54.55pt), start: 288.43deg, stop: 0.00deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  arc((145.45pt, 54.55pt), start: 0.00deg, stop: 71.57deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  arc((163.64pt, 109.09pt), start: 71.57deg, stop: 146.31deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  arc((109.09pt, 145.45pt), start: 146.31deg, stop: 213.69deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  arc((54.55pt, 109.09pt), start: 213.69deg, stop: 288.43deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((68.73pt, 50.55pt), anchor: "north-east", text(size: 9pt)[A])
  content((149.45pt, 50.55pt), anchor: "north-west", text(size: 9pt)[B])
  content((167.64pt, 109.09pt), anchor: "west", text(size: 9pt)[C])
  content((109.09pt, 149.45pt), anchor: "south", text(size: 9pt)[D])
  content((50.55pt, 109.09pt), anchor: "east", text(size: 9pt)[E])
})]),
  alt: "A pentagon, A B C D E. The exterior angles are marked at each vertex.", caption: none,
)


      
      #solutionbox[
Each individual angle measures #math.equation(block: false, alt: "the fraction 360 over 5 equals 72 degrees .")[$frac(360, 5) = 72^(∘) .$] Then, the sum of the exterior angles is #math.equation(block: false, alt: "5 open parenthesis 72 degrees close parenthesis equals 360 degrees .")[$5 ( 72^(∘) ) = 360^(∘) .$]
]
]


    
  
  
    
=== Circles and Circumference


    The perimeter of a circle is called the #strong[circumference]. To find the circumference, we use the formula #math.equation(block: false, alt: "C equals π d ,")[$C = π d ,$] where #math.equation(block: false, alt: "d")[$d$] is the diameter, the distance across the center, or #math.equation(block: false, alt: "C equals 2 π r ,")[$C = 2 π r ,$] where #math.equation(block: false, alt: "r")[$r$] is the radius.


    #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
The circumference of a circle is found using the formula #math.equation(block: false, alt: "C equals π d ,")[$C = π d ,$] where #math.equation(block: false, alt: "d")[$d$] is the diameter of the circle, or #math.equation(block: false, alt: "C equals 2 π r ,")[$C = 2 π r ,$] where #math.equation(block: false, alt: "r")[$r$] is the radius.
]


    The radius is ½ of the diameter of a circle. The symbol #math.equation(block: false, alt: "π equals 3.141592654 and so on")[$π = 3.141592654 …$] is the ratio of the circumference to the diameter. Because this ratio is constant, our formula is accurate for any size circle.


    #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  circle((100.00pt, 100.00pt), radius: 60.00pt, fill: none, stroke: 0.7pt + black)
  line((100.00pt, 100.00pt), (160.00pt, 100.00pt), stroke: 0.7pt + black)
  content((130.00pt, 109.00pt), anchor: "south", text(size: 9pt)[r])
  line((40.00pt, 100.00pt), (160.00pt, 100.00pt), stroke: 0.7pt + black)
  content((100.00pt, 109.00pt), anchor: "south", text(size: 9pt)[d])
  circle((100.00pt, 100.00pt), radius: 1.8pt, fill: black, stroke: none)
  content((104.00pt, 104.00pt), anchor: "south-west", text(size: 9pt)[O])
  circle((160.00pt, 100.00pt), radius: 1.8pt, fill: black, stroke: none)
  content((164.00pt, 100.00pt), anchor: "west", text(size: 9pt)[A])
  circle((40.00pt, 100.00pt), radius: 1.8pt, fill: black, stroke: none)
  content((36.00pt, 100.00pt), anchor: "east", text(size: 9pt)[B])
})]),
  alt: "A circle with its diameter and radius marked.", caption: [Circle Diameter and Radius],
)


    Let the radius be equal to 3.5 inches. Then, the circumference is


    #math.equation(block: true, alt: "C, equals, 2 π open parenthesis 3.5 close parenthesis; equals, 21.99 in .")[$C & = & 2 π ( 3.5 ) \  & = & 21.99 #h(0.28em) "in" .$]
    #examplebox("Example 8")[Finding Circumference with Diameter][
Find the circumference of a circle with diameter 10 cm.


      
      #solutionbox[
If the diameter is 10 cm, the circumference is #math.equation(block: false, alt: "C equals 10 π equals 31.42 cm .")[$C = 10 π = 31.42 #h(0.28em) "cm" .$]
]
]


    
    #examplebox("Example 9")[Finding Circumference with Radius][
Find the radius of a circle with a circumference of 12 in.


      
      #solutionbox[
If the circumference is 12 in, then the radius is


        #math.equation(block: true, alt: "12, equals, 2 π r; the fraction 12 over 2 π, equals, r equals 1.91 in .")[$12 & = & 2 π r \ frac(12, 2 π) & = & r = 1.91 #h(0.28em) "in" .$]
]
]


    
    #examplebox("Example 10")[Calculating Circumference for the Real World][
You decide to make a trim for the window in . How many feet of trim do you need to buy?


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((14.29pt, 14.29pt), (185.71pt, 14.29pt), (185.71pt, 100.00pt), (14.29pt, 100.00pt), close: true, fill: none, stroke: 0.7pt + black)
  content((100.00pt, 5.29pt), anchor: "north", text(size: 9pt)[12 ft])
  content((194.71pt, 57.14pt), anchor: "west", text(size: 9pt)[6 ft])
  content((100.00pt, 157.14pt), anchor: "center", text(size: 9pt)[semicircle (r = 6 ft)])
})]),
  alt: "A polygon shows a semicircle resting on top of a rectangle. The length and width of the rectangle measure 12 feet and 6 feet. The radius of the semicircle measures 6 feet.", caption: none,
)


      
      #solutionbox[
The trim will cover the 6 feet along the bottom and the two 12-ft sides plus the half circle on top. The circumference of a semicircle is ½ the circumference of a circle. The diameter of the semicircle is 6 ft. Then, the circumference of the semicircle would be #math.equation(block: false, alt: "the fraction 1 over 2 π d equals the fraction 1 over 2 π open parenthesis 6 close parenthesis equals 3 π ft equals 9.4 ft .")[$frac(1, 2) π d = frac(1, 2) π ( 6 ) = 3 π #h(0.28em) "ft" = 9.4 #h(0.28em) "ft" .$]


        Therefore, the total perimeter of the window is #math.equation(block: false, alt: "6 plus 12 plus 12 plus 9.4 equals 39.4 ft .")[$6 + 12 + 12 + 9.4 = 39.4 #h(0.28em) "ft" .$] You need to buy 39.4 ft of trim.
]
]


    
    #notebox("People in Mathematics", rgb("#183B6F"), rgb("#183B6F"), rgb("#EFF1F5"))[
#emph[Archimedes]


      #figure(figph[A portrait of Archimedes.], alt: "A portrait of Archimedes.", caption: [Archimedes])


      The overwhelming consensus is that Archimedes (287–212 BCE) was the greatest mathematician of classical antiquity, if not of all time. A Greek scientist, inventor, philosopher, astronomer, physicist, and mathematician, Archimedes flourished in Syracuse, Sicily. He is credited with the invention of various types of pulley systems and screw pumps based on the center of gravity. He advanced numerous mathematical concepts, including theorems for finding surface area and volume. Archimedes anticipated modern calculus and developed the idea of the “infinitely small” and the method of exhaustion. The method of exhaustion is a technique for finding the area of a shape inscribed within a sequence of polygons. The areas of the polygons converge to the area of the inscribed shape. This technique evolved to the concept of limits, which we use today.


      One of the more interesting achievements of Archimedes is the way he estimated the number pi, the ratio of the circumference of a circle to its diameter. He was the first to find a valid approximation. He started with a circle having a diameter of 1 inch. His method involved drawing a polygon inscribed inside this circle and a polygon circumscribed around this circle. He knew that the perimeter of the inscribed polygon was smaller than the circumference of the circle, and the perimeter of the circumscribed polygon was larger than the circumference of the circle. This is shown in the drawing of an eight-sided polygon. He increased the number of sides of the polygon each time as he got closer to the real value of pi. The following table is an example of how he did this.


      #figure(table(
  columns: 3, align: left, inset: 6pt,
  table.header([Sides], [Inscribed Perimeter], [Circumscribed Perimeter]),
  [4], [2.8284], [4.00],
  [8], [3.0615], [3.3137],
  [16], [3.1214], [3.1826],
  [32], [3.1365], [3.1517],
  [64], [3.1403], [3.1441],
))


      #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  content((100.00pt, 214.00pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[Concentric octagons and circle])
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  alt: "Two concentric octagons. A circle is inscribed about the outer octagon. The circle touches the vertices of the inner octagon and it touches the center points of the edges of the outer octagon.", caption: none,
)


      Archimedes settled on an approximation of #math.equation(block: false, alt: "π approximately equals 3.1416")[$π ≈ 3.1416$] after an iteration of 96 sides. Because pi is an irrational number, it cannot be written exactly. However, the capability of the supercomputer can compute pi to billions of decimal digits. As of 2002, the most precise approximation of pi includes 1.2 trillion decimal digits.
]


    #notebox("Who Knew?", rgb("#183B6F"), rgb("#183B6F"), rgb("#EFF1F5"))[
#emph[The Platonic Solids]


      The Platonic solids  have been known since antiquity. A polyhedron is a three-dimensional object constructed with congruent regular polygonal faces. Named for the philosopher, Plato believed that each one of the solids is associated with one of the four elements: Fire is associated with the tetrahedron or pyramid, earth with the cube, air with the octahedron, and water with the icosahedron. Of the fifth Platonic solid, the dodecahedron, Plato said, “… God used it for arranging the constellations on the whole heaven.”


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  alt: "Five polygons. A tetrahedron has four faces. A cube has six faces. An octahedron has eight faces. A dodecahedron has twelve faces. An icosahedron has twenty faces.", caption: [Platonic Solids],
)


      Plato believed that the combination of these five polyhedra formed all matter in the universe. Later, Euclid proved that exactly five regular polyhedra exist and devoted the last book of the #emph[Elements] to this theory. These ideas were resuscitated by Johannes Kepler about 2,000 years later. Kepler used the solids to explain the geometry of the universe. The beauty and symmetry of the Platonic solids have inspired architects and artists from antiquity to the present.
]


  
  
  
  
    
=== Key Terms


    - perimeter
- polygon
- pentagon
- hexagon
- heptagon
- octagon
- quadrilateral
- trapezoid
- parallelogram
- circumference


  
  
    
=== Key Concepts


    - Regular polygons are closed, two-dimensional figures that have equal side lengths. They are named for the number of their sides.
- The perimeter of a polygon is the measure of the outline of the shape. We determine a shape’s perimeter by calculating the sum of the lengths of its sides.
- The sum of the interior angles of a regular polygon with #math.equation(block: false, alt: "n")[$n$] sides is found using the formula #math.equation(block: false, alt: "S equals open parenthesis n minus 2 close parenthesis 180 degrees .")[$S = ( n − 2 ) 180^(∘) .$] The measure of a single interior angle of a regular polygon with #math.equation(block: false, alt: "n")[$n$] sides is determined using the formula #math.equation(block: false, alt: "a equals the fraction open parenthesis n minus 2 close parenthesis 180 degrees over n .")[$a = frac(( n − 2 ) 180^(∘), n) .$]
- The sum of the exterior angles of a regular polygon is #math.equation(block: false, alt: "360 degrees .")[$360^(∘) .$] The measure of a single exterior angle of a regular polygon with #math.equation(block: false, alt: "n")[$n$] sides is found using the formula #math.equation(block: false, alt: "b equals the fraction 360 degrees over n .")[$b = frac(360^(∘), n) .$]
- The circumference of a circle is #math.equation(block: false, alt: "C equals 2 π r ,")[$C = 2 π r ,$] where #math.equation(block: false, alt: "r")[$r$] is the radius, or #math.equation(block: false, alt: "C equals π d ,")[$C = π d ,$] and #math.equation(block: false, alt: "d")[$d$] is the diameter.


  
  
    
=== Formulas


    The formula for the perimeter #math.equation(block: false, alt: "P")[$P$] of a rectangle is #math.equation(block: false, alt: "P equals 2 L plus 2 W")[$P = 2 L + 2 W$], twice the length #math.equation(block: false, alt: "L")[$L$] plus twice the width #math.equation(block: false, alt: "W")[$W$].


    The sum of the interior angles of a polygon with #math.equation(block: false, alt: "n")[$n$] sides is given by


    #math.equation(block: true, alt: "S equals open parenthesis n minus 2 close parenthesis 180 degrees .")[$S = ( n − 2 ) 180^(∘) .$]
    The measure of each interior angle of a regular polygon with #math.equation(block: false, alt: "n")[$n$] sides is given by


    #math.equation(block: true, alt: "a equals the fraction open parenthesis n minus 2 close parenthesis 180 degrees over n .")[$a = frac(( n − 2 ) 180^(∘), n) "."$]
    To find the measure of an exterior angle of a regular polygon with #math.equation(block: false, alt: "n")[$n$] sides we use the formula


    #math.equation(block: true, alt: "b equals the fraction 360 degrees over n .")[$b = frac(360^(∘), n) "."$]
    The circumference of a circle is found using the formula #math.equation(block: false, alt: "C equals π d ,")[$C = π d ,$] where #math.equation(block: false, alt: "d")[$d$] is the diameter of the circle, or #math.equation(block: false, alt: "C equals 2 π r ,")[$C = 2 π r ,$] where #math.equation(block: false, alt: "r")[$r$] is the radius.