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== 10.6#h(0.6em)Area


  #figure(figph[A diagram of a baseball field with its measurements marked.], alt: "A diagram of a baseball field with its measurements marked.", caption: [The area of a regulation baseball diamond must adhere to specific measurements to be legal.])


  
    
=== Learning Objectives


    After completing this section, you should be able to:


    + Calculate the area of triangles.
+ Calculate the area of quadrilaterals.
+ Calculate the area of other polygons.
+ Calculate the area of circles.


  
  Some areas carry more importance than other areas. Did you know that in a baseball game, when the player hits the ball and runs to first base that he must run within a 6-foot wide path? If he veers off slightly to the right, he is out. In other words, a few inches can be the difference in winning or losing a game. Another example is real estate. On Manhattan Island, one square foot of real estate is worth far more than real estate in practically any other area of the country. In other words, we place a value on area. As the context changes, so does the value.


  Area refers to a region measured in square units, like a square mile or a square foot. For example, to purchase tile for a kitchen floor, you would need to know how many square feet are needed because tile is sold by the square foot. Carpeting is sold by the square yard. As opposed to linear measurements like perimeter, which in in linear units. For example, fencing is sold in linear units, a linear foot or yard. Linear dimensions refer to an outline or a boundary. Square units refer to the area within that boundary. Different items may have different units, but either way, you must know the linear dimensions to calculate the area.


  Many geometric shapes have formulas for calculating areas, such as triangles, regular polygons, and circles. To calculate areas for many irregular curves or shapes, we need calculus. However, in this section, we will only look at geometric shapes that have known area formulas. The notation for area, as mentioned, is in square units and we write sq in or sq cm, or use an exponent, such as #math.equation(block: false, alt: "in squared")[$"in"^(2)$] or #math.equation(block: false, alt: "cm squared .")[$"cm"^(2) .$] Note that linear measurements have no exponent above the units or we can say that the exponent is 1.


  
    
=== Area of Triangles


    The formula for the area of a triangle is given as follows.


    #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
The area of a triangle is given as #math.equation(block: false, alt: "A equals the fraction 1 over 2 b h ,")[$A = frac(1, 2) b h ,$] where #math.equation(block: false, alt: "b")[$b$] represents the base and #math.equation(block: false, alt: "h")[$h$] represents the height.
]


    For example, consider the triangle in .


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  alt: "A right triangle with its legs marked 5 centimeters and 4 centimeters.", caption: [Triangle 1],
)


    The base measures 4 cm and the height measures 5 cm. Using the formula, we can calculate the area:


    #math.equation(block: true, alt: "A equals the fraction 1 over 2 open parenthesis 4 close parenthesis open parenthesis 5 close parenthesis equals the fraction 1 over 2 open parenthesis 20 close parenthesis equals 10 cm squared")[$A = frac(1, 2) ( 4 ) ( 5 ) = frac(1, 2) ( 20 ) = 10 #h(0.28em) "cm"^(2)$]
    
    In , the triangle has a base equal to 7 cm and a height equal to 3.5 cm. Notice that we can only find the height by dropping a perpendicular to the base. The area is then


    #math.equation(block: true, alt: "A equals the fraction 1 over 2 open parenthesis 7 close parenthesis open parenthesis 3.5 close parenthesis equals 12.25 cm squared .")[$A = frac(1, 2) ( 7 ) ( 3.5 ) = 12.25 #h(0.28em) "cm"^(2) .$]
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    #examplebox("Example 1")[Finding the Area of a Triangle][
Find the area of this triangle that has a base of 4 cm and the height is 6 cm .


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  alt: "A triangle with its base marked b equals 4 centimeters and height marked h equals 6 centimeters.", caption: none,
)


      
      #solutionbox[
Using the formula, we have


        #math.equation(block: true, alt: "A equals the fraction 1 over 2 open parenthesis 4 close parenthesis open parenthesis 6 close parenthesis equals the fraction 24 over 2 equals 12 cm squared .")[$A = frac(1, 2) ( 4 ) ( 6 ) = frac(24, 2) = 12 #h(0.28em) "cm"^(2) "."$]
]
]


    
  
  
    
=== Area of Quadrilaterals


    A #strong[quadrilateral] is a four-sided polygon with four vertices. Some quadrilaterals have either one or two sets of parallel sides. The set of quadrilaterals include the square, the rectangle, the parallelogram, the trapezoid, and the rhombus. The most common quadrilaterals are the square and the rectangle.


    
      
==== Square


      In , a #math.equation(block: false, alt: "12 in times 12 in")[$12 #h(0.28em) "in" × 12 #h(0.28em) "in"$] grid is represented with twelve #math.equation(block: false, alt: "1 in times 1 in")[$1 #h(0.28em) "in" × 1 #h(0.28em) "in"$] squares across each row, and twelve #math.equation(block: false, alt: "1 in times 1 in")[$1 #h(0.28em) "in" × 1 #h(0.28em) "in"$] squares down each column. If you count the little squares, the sum equals 144 squares. Of course, you do not have to count little squares to find area—we have a formula. Thus, the formula for the area of a square, where #math.equation(block: false, alt: "s equals length of a side")[$s = "length of a side"$], is #math.equation(block: false, alt: "A equals s times s")[$A = s ⋅ s$]. The area of the square in  is #math.equation(block: false, alt: "A equals 12 in times 12 in equals 144 in squared .")[$A = 12 #h(0.28em) "in" × 12 #h(0.28em) "in" = 144 #h(0.28em) "in"^(2) .$]


      #figure(figph[A square grid with its sides marked s equals 12 inches.], alt: "A square grid with its sides marked s equals 12 inches.", caption: [Area of a Square])


      #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
The formula for the area of a square is #math.equation(block: false, alt: "A equals s times s")[$A = s ⋅ s$] or #math.equation(block: false, alt: "A equals s squared .")[$A = s^(2) .$]
]


    
  
  
    
=== Rectangle


    Similarly, the area for a rectangle is found by multiplying length times width. The rectangle in  has width equal to 5 in and length equal to 12 in. The area is #math.equation(block: false, alt: "A equals 5 open parenthesis 12 close parenthesis equals 60 in squared .")[$A = 5 ( 12 ) = 60 #h(0.28em) "in"^(2) .$]


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The area of a rectangle is given as #math.equation(block: false, alt: "A equals l w .")[$A = l w .$]
]


    Many everyday applications require the use of the perimeter and area formulas. Suppose you are remodeling your home and you want to replace all the flooring. You need to know how to calculate the area of the floor to purchase the correct amount of tile, or hardwood, or carpet. If you want to paint the rooms, you need to calculate the area of the walls and the ceiling to know how much paint to buy, and the list goes on. Can you think of other situations where you might need to calculate area?


    #examplebox("Example 2")[Finding the Area of a Rectangle][
You have a garden with an area of 196 square feet. How wide can the garden be if the length is 28 feet?


      
      #solutionbox[
The area of a rectangular region is #math.equation(block: false, alt: "A equals l w .")[$A = l w .$] Letting the width equal #math.equation(block: false, alt: "w")[$w$]:


        #math.equation(block: true, alt: "196, equals, 28 w; the fraction 196 over 28, equals, w equals 7 ft")[$196 & = & 28 w \ frac(196, 28) & = & w = 7 #h(0.28em) "ft"$]
]
]


    
    #examplebox("Example 3")[Determining the Cost of Floor Tile][
Jennifer is planning to install vinyl floor tile in her rectangular-shaped basement, which measures 29 ft by 16 ft. How many boxes of floor tile should she buy and how much will it cost if one box costs \$50 and covers #math.equation(block: false, alt: "20 ft squared ?")[$20 #h(0.28em) "ft"^(2) ?$]


      
      #solutionbox[
The area of the basement floor is #math.equation(block: false, alt: "A equals 29 open parenthesis 16 close parenthesis equals 464 ft squared .")[$A = 29 ( 16 ) = 464 #h(0.28em) "ft"^(2) .$] We will divide this area by #math.equation(block: false, alt: "20 ft .")[$20 #h(0.28em) "ft" .$] Thus, #math.equation(block: false, alt: "the fraction 464 over 20 equals 23.2 .")[$frac(464, 20) = 23.2 .$] Therefore, Jennifer will have to buy 24 boxes of tile at a cost \$1,200.
]
]


    
  
  
    
=== Parallelogram


    The area of a parallelogram can be found using the formula for the area of a triangle. Notice in , if we cut a diagonal across the parallelogram from one vertex to the opposite vertex, we have two triangles. If we multiply the area of a triangle by 2, we have the area of a parallelogram:


    #math.equation(block: true, alt: "A, equals, 2 open parenthesis the fraction 1 over 2 b h close parenthesis; A, equals, b h")[$A & = & 2 ( frac(1, 2) b h ) \ A & = & b h$]
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The area of a parallelogram is #math.equation(block: false, alt: "A equals b h .")[$A = b h .$]
]


    For example, if we have a parallelogram with the base be equal to 10 inches and the height equal to 5 inches, the area will be #math.equation(block: false, alt: "A equals open parenthesis 10 close parenthesis open parenthesis 5 close parenthesis equals 50 in squared .")[$A = ( 10 ) ( 5 ) = 50 #h(0.28em) "in"^(2) .$]


    #examplebox("Example 4")[Finding the Area of a Parallelogram][
In the parallelogram , if #math.equation(block: false, alt: "F B equals 10 , A D equals 15 ,")[$F B = 10 , A D = 15 ,$] find the exact area of the parallelogram.


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      #solutionbox[
Using the formula of #math.equation(block: false, alt: "A equals b h ,")[$A = b h ,$] we have


        #math.equation(block: true, alt: "A equals 10 open parenthesis 15 close parenthesis equals 150 cm squared .")[$A = 10 ( 15 ) = 150 #h(0.28em) "cm"^(2) .$]
]
]


    
    #examplebox("Example 5")[Finding the Area of a Parallelogram Park][
The boundaries of a city park form a parallelogram . The park takes up one city block, which is contained by two sets of parallel streets. Each street measures 55 yd long. The perpendicular distance between streets is 39 yd. How much sod, sold by the square foot, should the city purchase to cover the entire park and how much will it cost? The sod is sold for \$0.50 per square foot, installation is \$1.50 per square foot, and the cost of the equipment for the day is \$100.


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  alt: "A park is in the shape of a parallelogram. The sides measure 55 yards. The height measures 39 yards.", caption: none,
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      #solutionbox[
#strong[Step 1:] As sod is sold by the square foot, the first thing we have to do is translate the measurements of the park from yards to feet. There are 3 ft to a yard, so 55 yd is equal to 165 ft, and 39 yd is equal to 117 ft.


        #strong[Step 2:] The park has the shape of a parallelogram, and the formula for the area is #math.equation(block: false, alt: "A equals b h")[$A = b h$]:


        #math.equation(block: true, alt: "A equals 165 open parenthesis 117 close parenthesis equals 19,305 ft squared .")[$A = 165 ( 117 ) = 19,305 #h(0.28em) "ft"^(2) .$]
        
        #strong[Step 3:] The city needs to purchase #math.equation(block: false, alt: "19 , 305 ft squared")[$19 , 305 #h(0.28em) "ft"^(2)$] of sod. The cost will be \$0.50 per square foot for the sod and \$1.50 per square foot for installation, plus \$100 for equipment:


        #math.equation(block: true, alt: "19,305 open parenthesis $ 0.50 close parenthesis plus 19,305 open parenthesis $ 1.50 close parenthesis, equals, $ 9,652.5 plus $ 28,957.5 plus $ 100.00; equals, $ 38,710.00")[$19,305 ( "$" 0.50 ) + 19,305 ( "$" 1.50 ) & = & "$" 9,652.5 + "$" 28,957.5 + "$" 100.00 \  & = & "$" 38,710.00$]
]
]


    
  
  
    
=== Trapezoid


    Another quadrilateral is the trapezoid. A trapezoid has one set of parallel sides or bases. The formula for the area of a trapezoid with parallel bases #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] and height #math.equation(block: false, alt: "h")[$h$] is given here.


    #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
The formula for the area of a trapezoid is given as #math.equation(block: false, alt: "A equals the fraction 1 over 2 h open parenthesis a plus b close parenthesis .")[$A = frac(1, 2) h ( a + b ) .$]
]


    For example, find the area of the trapezoid in  that has base #math.equation(block: false, alt: "a")[$a$] equal to 8 cm, base #math.equation(block: false, alt: "b")[$b$] equal to 6 cm, and height equal to 6 cm.


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  alt: "A trapezoid with the bottom and top bases marked a and b. The height is marked h.", caption: [Area of a Trapezoid],
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    The area is #math.equation(block: false, alt: "A equals the fraction 1 over 2 open parenthesis 6 close parenthesis open parenthesis 6 plus 8 close parenthesis equals 42 cm squared")[$A = frac(1, 2) ( 6 ) ( 6 + 8 ) = 42 #h(0.28em) "cm"^(2)$].


    #examplebox("Example 6")[Finding the Area of a Trapezoid][
#math.equation(block: false, alt: "A B C D")[$A B C D$]  is a regular trapezoid with #math.equation(block: false, alt: "A B ¯ ‖ C D ¯ .")[$overline(A B) ‖ overline(C D) .$] Find the exact perimeter of #math.equation(block: false, alt: "A B C D")[$A B C D$], and then find the area.


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  import cetz.draw: *
  content((100.00pt, 86.73pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[Trapezoid ABCD])
  line((6.06pt, 6.06pt), (193.94pt, 6.06pt), (127.70pt, 66.67pt), (61.03pt, 66.67pt), close: true, fill: none, stroke: 0.7pt + black)
  content((100.83pt, -2.90pt), anchor: "north", text(size: 9pt)[31 in])
  content((93.53pt, 75.63pt), anchor: "south", text(size: 9pt)[11 in])
  content((24.55pt, 36.36pt), anchor: "east", text(size: 9pt)[13.5 in])
  line((61.03pt, 66.67pt), (61.03pt, 6.06pt), stroke: (paint: black, thickness: 0.6pt, dash: "dashed"))
  content((70.03pt, 36.36pt), anchor: "south", text(size: 9pt)[h = 10 in])
  line((127.70pt, 66.67pt), (127.70pt, 6.06pt), stroke: (paint: black, thickness: 0.6pt, dash: "dashed"))
  content((136.70pt, 36.36pt), anchor: "south", text(size: 9pt)[h = 10 in])
  content((10.06pt, 10.06pt), anchor: "south-west", text(size: 9pt)[D])
  content((189.94pt, 10.06pt), anchor: "south-east", text(size: 9pt)[C])
  content((123.70pt, 62.67pt), anchor: "north-east", text(size: 9pt)[B])
  content((65.03pt, 62.67pt), anchor: "north-west", text(size: 9pt)[A])
})]),
  alt: "A trapezoid, A B C D. A B measures 11 inches. A D measures 13.5 inches. D C measures 31 inches. Two dashed vertical lines are drawn from A and B to meet the side, D C. The dashes lines are labeled h equals 10 inches.", caption: none,
)


      
      #solutionbox[
The perimeter is the measure of the boundary of the shape, so we just add up the lengths of the sides. We have #math.equation(block: false, alt: "P equals 31 plus 13.5 plus 11 plus 13.5 equals 69 in .")[$P = 31 + 13.5 + 11 + 13.5 = 69 #h(0.28em) "in" .$] Then, the area of the trapezoid using the formula is #math.equation(block: false, alt: "A equals the fraction 1 over 2 open parenthesis 10 close parenthesis open parenthesis 11 plus 31 close parenthesis equals 210 in squared")[$A = frac(1, 2) ( 10 ) ( 11 + 31 ) = 210 #h(0.28em) "in"^(2)$].
]
]


    
    
      
==== Rhombus


      The rhombus has two sets of parallel sides. To find the area of a rhombus, there are two formulas we can use. One involves determining the measurement of the diagonals.


      #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
The area of a rhombus is found using one of these formulas:


        - #math.equation(block: false, alt: "A equals the fraction d sub 1 d sub 2 over 2 ,")[$A = frac(d_(1) d_(2), 2) ,$] where #math.equation(block: false, alt: "d sub 1")[$d_(1)$] and #math.equation(block: false, alt: "d sub 2")[$d_(2)$] are the diagonals.
- #math.equation(block: false, alt: "A equals the fraction 1 over 2 b h ,")[$A = frac(1, 2) b h ,$] where #math.equation(block: false, alt: "b")[$b$] is the base and #math.equation(block: false, alt: "h")[$h$] is the height.
]


      For our purposes here, we will use the formula that uses diagonals. For example, if the area of a rhombus is #math.equation(block: false, alt: "220 cm squared ,")[$220 #h(0.28em) "cm"^(2) ,$] and the measure of #math.equation(block: false, alt: "d sub 2 equals 11 ,")[$d_(2) = 11 ,$] find the measure of #math.equation(block: false, alt: "d sub 1 .")[$d_(1) .$] To solve this problem, we input the known values into the formula and solve for the unknown.


      #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((15.38pt, 76.92pt), (100.00pt, 15.38pt), (184.62pt, 76.92pt), (100.00pt, 138.46pt), close: true, fill: none, stroke: 0.7pt + black)
  line((100.00pt, 15.38pt), (100.00pt, 138.46pt), stroke: 0.7pt + black)
  content((91.00pt, 76.92pt), anchor: "south", text(size: 9pt)[d₁])
  line((15.38pt, 76.92pt), (184.62pt, 76.92pt), stroke: 0.7pt + black)
  content((100.00pt, 85.92pt), anchor: "south", text(size: 9pt)[d₂])
  content((100.00pt, 67.69pt), anchor: "center", text(size: 9pt)[11])
})]),
  alt: "A rhombus with two diagonal lines labeled d subscript 1 and d subscript 2. The line, d subscript 2 measures 11.", caption: [Area of a Rhombus],
)


      We have that


      #math.equation(block: true, alt: "220, equals, the fraction 11 d sub 1 over 2; 220 open parenthesis 2 close parenthesis, equals, 11 d sub 1; the fraction 440 over 11, equals, d sub 1 equals 40")[$220 & = & frac(11 d_(1), 2) \ 220 ( 2 ) & = & 11 d_(1) \ frac(440, 11) & = & d_(1) = 40$]
      
      #examplebox("Example 7")[Finding the Area of a Rhombus][
Find the measurement of the diagonal #math.equation(block: false, alt: "d sub 1")[$d_(1)$] if the area of the rhombus is #math.equation(block: false, alt: "240 cm squared ,")[$240 #h(0.28em) "cm"^(2) ,$] and the measure of #math.equation(block: false, alt: "d sub 2 equals 24 cm .")[$d_(2) = 24 #h(0.28em) "cm" .$]


        
        #solutionbox[
Use the formula with the known values:


          #math.equation(block: true, alt: "240, equals, the fraction d sub 1 open parenthesis 24 close parenthesis over 2; 240 open parenthesis 2 close parenthesis, equals, d sub 1 open parenthesis 24 close parenthesis; the fraction 480 over 24, equals, d sub 1 equals 20")[$240 & = & frac(d_(1) ( 24 ), 2) \ 240 ( 2 ) & = & d_(1) ( 24 ) \ frac(480, 24) & = & d_(1) = 20$]
]
]


      
      #examplebox("Example 8")[Finding the Area of a Rhombus][
You notice a child flying a rhombus-shaped kite on the beach. When it falls to the ground, it falls on a beach towel measuring #math.equation(block: false, alt: "36 in")[$36 #h(0.28em) "in"$] by #math.equation(block: false, alt: "72 in .")[$72 #h(0.28em) "in" .$] You notice that one of the diagonals of the kite is the same length as the #math.equation(block: false, alt: "36 in")[$36 #h(0.28em) "in"$] width of the towel. The second diagonal appears to be 2 in longer. What is the area of the kite ?


          #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  content((95.00pt, 214.00pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[Rhombus-shaped kite])
  line((5.00pt, 100.00pt), (95.00pt, 195.00pt), (185.00pt, 100.00pt), (95.00pt, 5.00pt), close: true, fill: none, stroke: 0.7pt + black)
  line((5.00pt, 100.00pt), (185.00pt, 100.00pt), stroke: 0.7pt + black)
  content((95.00pt, 109.00pt), anchor: "south", text(size: 9pt)[d₁ = 36 inches])
  line((95.00pt, 5.00pt), (95.00pt, 195.00pt), stroke: 0.7pt + black)
  content((86.00pt, 100.00pt), anchor: "south", text(size: 9pt)[d₂ = 38 inches])
})]),
  alt: "A rhombus-shaped kite with two diagonal lines. The horizontal diagonal line is labeled d subscript 1 equals 36 inches. The vertical diagonal line is labeled d subscript 2 equals 38 inches.", caption: none,
)


        
        #solutionbox[
Using the formula, we have:


          #math.equation(block: true, alt: "A, equals, the fraction d sub 1 d sub 2 over 2; equals, the fraction 36 open parenthesis 38 close parenthesis over 2 equals 684 in squared")[$A & = & frac(d_(1) d_(2), 2) \  & = & frac(36 ( 38 ), 2) = 684 #h(0.28em) "in"^(2)$]
]
]


      
    
  
  
    
=== Area of Polygons


    To find the area of a regular polygon, we need to learn about a few more elements. First, the apothem #math.equation(block: false, alt: "a")[$a$] of a regular polygon is a line segment that starts at the center and is perpendicular to a side. The radius #math.equation(block: false, alt: "r")[$r$] of a regular polygon is also a line segment that starts at the center but extends to a vertex.


    #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((175.00pt, 100.00pt), (137.50pt, 165.00pt), (62.50pt, 165.00pt), (25.00pt, 100.00pt), (62.50pt, 35.00pt), (137.50pt, 35.00pt), close: true, fill: none, stroke: 0.7pt + black)
  line((100.00pt, 100.00pt), (175.00pt, 100.00pt), stroke: 0.7pt + black)
  content((137.50pt, 109.00pt), anchor: "south", text(size: 9pt)[r])
  line((100.00pt, 100.00pt), (156.25pt, 132.50pt), stroke: 0.7pt + black)
  content((123.62pt, 124.04pt), anchor: "south", text(size: 9pt)[a])
  circle((100.00pt, 100.00pt), radius: 1.8pt, fill: black, stroke: none)
})]),
  alt: "A hexagon with its radius marked r and apothem marked a.", caption: [Apothem and Radius of a Polygon],
)


    #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
The area of a regular polygon is found with the formula #math.equation(block: false, alt: "A equals the fraction 1 over 2 a p ,")[$A = frac(1, 2) a p ,$] where #math.equation(block: false, alt: "a")[$a$] is the apothem and #math.equation(block: false, alt: "p")[$p$] is the perimeter.
]


    For example, consider the regular hexagon shown in  with a side length of 4 cm, and the apothem measures #math.equation(block: false, alt: "a equals 2 the square root of 3 .")[$a = 2 sqrt(3) .$]


    #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((60.00pt, 30.72pt), (140.00pt, 30.72pt), (180.00pt, 100.00pt), (140.00pt, 169.28pt), (60.00pt, 169.28pt), (20.00pt, 100.00pt), close: true, fill: none, stroke: 0.7pt + black)
  content((100.00pt, 21.72pt), anchor: "north", text(size: 9pt)[4 cm])
  content((167.79pt, 60.86pt), anchor: "west", text(size: 9pt)[4 cm])
  content((167.79pt, 139.14pt), anchor: "west", text(size: 9pt)[4 cm])
  content((100.00pt, 178.28pt), anchor: "south", text(size: 9pt)[4 cm])
  content((32.21pt, 139.14pt), anchor: "east", text(size: 9pt)[4 cm])
  content((32.21pt, 60.86pt), anchor: "east", text(size: 9pt)[4 cm])
  line((100.00pt, 100.00pt), (100.00pt, 30.72pt), stroke: 0.7pt + black)
  content((109.00pt, 65.36pt), anchor: "south", text(size: 9pt)[a = 2√3])
  circle((100.00pt, 100.00pt), radius: 1.8pt, fill: black, stroke: none)
  content((96.00pt, 96.00pt), anchor: "north-east", text(size: 9pt)[C])
})]),
  alt: "A hexagon with its center marked C. Each side measures 4 centimeters. The apothem is marked a equals 2 times square root of 3.", caption: [Area of a Hexagon],
)


    We have the perimeter, #math.equation(block: false, alt: "p equals 6 open parenthesis 4 close parenthesis equals 24 cm .")[$p = 6 ( 4 ) = 24 #h(0.28em) "cm" .$] We have the apothem as #math.equation(block: false, alt: "a equals 2 the square root of 3 .")[$a = 2 sqrt(3) .$] Then, the area is:


    #math.equation(block: true, alt: "A, equals, the fraction 1 over 2 open parenthesis 2 the square root of 3 close parenthesis open parenthesis 24 close parenthesis; equals, 24 the square root of 3 equals 41.57 cm squared")[$A & = & frac(1, 2) ( 2 sqrt(3) ) ( 24 ) \  & = & 24 sqrt(3) = 41.57 #h(0.28em) "cm"^(2)$]
    #examplebox("Example 9")[Finding the Area of a Regular Octagon][
Find the area of a regular octagon with the apothem equal to 18 cm and a side length equal to 13 cm .


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((175.00pt, 130.00pt), (130.00pt, 175.00pt), (70.00pt, 175.00pt), (25.00pt, 130.00pt), (25.00pt, 70.00pt), (70.00pt, 25.00pt), (130.00pt, 25.00pt), (175.00pt, 70.00pt), close: true, fill: none, stroke: 0.7pt + black)
  content((158.86pt, 158.86pt), anchor: "west", text(size: 9pt)[13 cm])
  content((100.00pt, 184.00pt), anchor: "south", text(size: 9pt)[13 cm])
  content((41.14pt, 158.86pt), anchor: "east", text(size: 9pt)[13 cm])
  content((16.00pt, 100.00pt), anchor: "east", text(size: 9pt)[13 cm])
  content((41.14pt, 41.14pt), anchor: "east", text(size: 9pt)[13 cm])
  content((100.00pt, 16.00pt), anchor: "north", text(size: 9pt)[13 cm])
  content((158.86pt, 41.14pt), anchor: "west", text(size: 9pt)[13 cm])
  content((184.00pt, 100.00pt), anchor: "west", text(size: 9pt)[13 cm])
  line((100.00pt, 100.00pt), (100.00pt, 175.00pt), stroke: (paint: black, thickness: 0.6pt, dash: "dashed"))
  content((91.00pt, 137.50pt), anchor: "south", text(size: 9pt)[a = 18 cm])
  line((100.00pt, 175.00pt), (92.00pt, 175.00pt), (92.00pt, 167.00pt), (100.00pt, 167.00pt), close: true, stroke: 0.6pt + black, fill: none)
  circle((100.00pt, 100.00pt), radius: 1.8pt, fill: black, stroke: none)
})]),
  alt: "An octagon. Each side measures 13 centimeters. The apothem is marked a equals 18 centimeters.", caption: none,
)


      
      #solutionbox[
Using the formula, we have the perimeter #math.equation(block: false, alt: "p equals 8 open parenthesis 13 close parenthesis equals 104 cm .")[$p = 8 ( 13 ) = 104 #h(0.28em) "cm" .$] Then, the area is #math.equation(block: false, alt: "A equals the fraction 1 over 2 open parenthesis 10 close parenthesis open parenthesis 104 close parenthesis equals 936 cm squared .")[$A = frac(1, 2) ( 10 ) ( 104 ) = 936 #h(0.28em) "cm"^(2) .$]
]
]


    
    
      
==== Changing Units


      Often, we have the need to change the units of one or more items in a problem to find a solution. For example, suppose you are purchasing new carpet for a room measured in feet, but carpeting is sold in terms of yards. You will have to convert feet to yards to purchase the correct amount of carpeting. Or, you may need to convert centimeters to inches, or feet to meters. In each case, it is essential to use the correct equivalency.


      #examplebox("Example 10")[Changing Units][
Carpeting comes in units of square yards. Your living room measures 21 ft wide by 24 ft long. How much carpeting do you buy?


        
        #solutionbox[
We must convert feet to yards. As there are 3 ft in 1 yd, we have #math.equation(block: false, alt: "21 ft equals 7 yds")[$21 #h(0.28em) "ft" = 7 #h(0.28em) "yds"$] and #math.equation(block: false, alt: "24 ft equals 8 yds .")[$24 #h(0.28em) "ft" = 8 #h(0.28em) "yds" .$] Then, #math.equation(block: false, alt: "7 open parenthesis 8 close parenthesis equals 56 yd squared .")[$7 ( 8 ) = 56 #h(0.28em) "yd"^(2) .$]
]
]


      
    
  
  
    
=== Area of Circles


    Just as the circumference of a circle includes the number #math.equation(block: false, alt: "π ,")[$π ,$] so does the formula for the area of a circle. Recall that #math.equation(block: false, alt: "π")[$π$] is a non-terminating, non-repeating decimal number: #math.equation(block: false, alt: "π equals 3.14159 and so on")[$π = 3.14159 …$]. It represents the ratio of the circumference to the diameter, so it is a critical number in the calculation of circumference and area.


    #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
The area of a circle is given as #math.equation(block: false, alt: "A equals π r squared ,")[$A = π r^(2) ,$] where #math.equation(block: false, alt: "r")[$r$] is the radius.
]


    For example, to find the of the circle with radius equal to 3 cm, as shown in , is found using the formula #math.equation(block: false, alt: "A equals π r squared .")[$A = π r^(2) .$]


    #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 = 3 cm])
  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])
})]),
  alt: "A circle with its radius, r marked 3 centimeters.", caption: [Circle with Radius 3],
)


    We have


    #math.equation(block: true, alt: "A, equals, π r squared; equals, π open parenthesis 3 close parenthesis squared; equals, 9 π equals 28.27 cm squared")[$A & = & π r^(2) \  & = & π attach(( 3 ), t: 2) \  & = & 9 π = 28.27 #h(0.28em) "cm"^(2)$]
    #examplebox("Example 11")[Finding the Area of a Circle][
Find the area of a circle with diameter of 16 cm.


      
      #solutionbox[
The formula for the area of a circle is given in terms of the radius, so we cut the diameter in half. Then, the area is


        #math.equation(block: true, alt: "A equals π open parenthesis 8 close parenthesis squared equals 201.1 cm squared .")[$A = π attach(( 8 ), t: 2) = 201.1 #h(0.28em) "cm"^(2) "."$]
]
]


    
    #examplebox("Example 12")[Determining the Better Value for Pizza][
You decide to order a pizza to share with your friend for dinner. The price for an 8-inch diameter pizza is \$7.99. The price for 16-inch diameter pizza is \$13.99. Which one do you think is the better value?


      
      #solutionbox[
The area of the 8-inch diameter pizza is #math.equation(block: false, alt: "A equals π open parenthesis 4 close parenthesis squared equals 50.3 in squared .")[$A = π attach(( 4 ), t: 2) = 50.3 #h(0.28em) "in"^(2) .$] The area of the 16-inch diameter pizza is #math.equation(block: false, alt: "A equals π open parenthesis 8 close parenthesis squared equals 201.1 in squared .")[$A = π attach(( 8 ), t: 2) = 201.1 #h(0.28em) "in"^(2) .$] Next, we divide the cost of each pizza by its area in square inches. Thus, #math.equation(block: false, alt: "the fraction 13.99 over 201.1 equals $ 0.07")[$frac(13.99, 201.1) = "$" 0.07$] per square inch and #math.equation(block: false, alt: "the fraction 7.99 over 50.3 equals $ 0.16")[$frac(7.99, 50.3) = "$" 0.16$] per square inch. So clearly, the 16-inch pizza is the better value.
]
]


    
    #examplebox("Example 13")[Applying Area to the Real World][
You want to purchase a tinted film, sold by the square foot, for the window in . (This problem should look familiar as we saw it earlier when calculating circumference.) The bottom part of the window is a rectangle, and the top part is a semicircle. Find the area and calculate the amount of film to purchase.


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((16.67pt, 16.67pt), (183.33pt, 16.67pt), (183.33pt, 100.00pt), (16.67pt, 100.00pt), close: true, fill: none, stroke: 0.7pt + black)
  content((100.00pt, 7.67pt), anchor: "north", text(size: 9pt)[10 ft])
  content((192.33pt, 58.33pt), anchor: "west", text(size: 9pt)[5 ft])
  line((58.33pt, 100.00pt), (75.00pt, 136.67pt), (100.00pt, 141.67pt), (125.00pt, 136.67pt), (141.67pt, 100.00pt), close: true, fill: none, stroke: 0.7pt + black)
  content((100.00pt, 153.33pt), anchor: "center", text(size: 9pt)[d = 5 ft])
})]),
  alt: "A figure shows a semicircle placed on top of a rectangle. The length and width of the rectangle are 10 feet and 5 feet. The diameter of the semicircle is 5 feet.", caption: none,
)


      
      #solutionbox[
First, the rectangular portion has #math.equation(block: false, alt: "A equals 5 open parenthesis 10 close parenthesis equals 50 ft squared")[$A = 5 ( 10 ) = 50 #h(0.28em) "ft"^(2)$]. For the top part, we have a semicircle with a diameter of 5 ft, so the radius is 2.5 ft. We want one half of the area of a circle with radius 2.5 ft, so the area of the top semicircle part is #math.equation(block: false, alt: "A equals the fraction 1 over 2 π open parenthesis 2.5 close parenthesis squared equals 9.8 ft squared .")[$A = frac(1, 2) π attach(( 2.5 ), t: 2) = 9.8 #h(0.28em) "ft"^(2) .$] Add the area of the rectangle to the area of the semicircle. Then, the total area to be covered with the window film is #math.equation(block: false, alt: "A equals 50 plus 9.8 equals 59.8 ft squared .")[$A = 50 + 9.8 = 59.8 #h(0.28em) "ft"^(2) .$]
]
]


    
  
  
    
=== Area within Area


    Suppose you want to install a round hot tub on your backyard patio. How would you calculate the space needed for the hot tub? Or, let’s say that you want to purchase a new dining room table, but you are not sure if you have enough space for it. These are common issues people face every day. So, let’s take a look at how we solve these problems.


    #examplebox("Example 14")[Finding the Area within an Area][
The patio in your backyard measures 20 ft by 10 ft . On one-half of the patio, you have a 4-foot diameter table with six chairs taking up an area of approximately 36 sq feet. On the other half of the patio, you want to install a hot tub measuring 6 ft in diameter. How much room will the table with six chairs and the hot tub take up? How much area is left over?


        #figure(figph[A patio is divided into two equal parts. The length and width of the patio measure 20 feet and 10 feet. The left part of the patio has a table with six chairs. The right part of the patio has a hot tub whose diameter is marked 6 feet.], alt: "A patio is divided into two equal parts. The length and width of the patio measure 20 feet and 10 feet. The left part of the patio has a table with six chairs. The right part of the patio has a hot tub whose diameter is marked 6 feet.", caption: none)


      
      #solutionbox[
The hot tub has a radius of 3 ft. That area is then #math.equation(block: false, alt: "A equals π open parenthesis 3 close parenthesis squared equals 9 π equals 28.27 ft squared .")[$A = π attach(( 3 ), t: 2) = 9 π = 28.27 #h(0.28em) "ft"^(2) .$] The total square feet taken up with the table and chairs and the hot tub is #math.equation(block: false, alt: "36 plus 28.27 equals 64.27 ft squared .")[$36 + 28.27 = 64.27 #h(0.28em) "ft"^(2) .$] The area left over is equal to the total area of the patio, #math.equation(block: false, alt: "200 ft squared")[$200 #h(0.28em) "ft"^(2)$] minus the area for the table and chairs and the hot tub. Thus, the area left over is #math.equation(block: false, alt: "200 minus 64.27 equals 135.7 ft squared .")[$200 − 64.27 = 135.7 #h(0.28em) "ft"^(2) .$]
]
]


    
    #examplebox("Example 15")[Finding the Cost of Fertilizing an Area][
A sod farmer wants to fertilize a rectangular plot of land 150 ft by 240 ft. A bag of fertilizer covers #math.equation(block: false, alt: "5,000 ft squared")[$5,000 #h(0.28em) "ft"^(2)$] and costs \$200. How much will it cost to fertilize the entire plot of land?


      
      #solutionbox[
The plot of land is #math.equation(block: false, alt: "36 , 000 ft squared .")[$36 , 000 #h(0.28em) "ft"^(2) .$] It will take 7.2 bags of fertilizer to cover the land area. Therefore, the farmer will have to purchase 8 bags of fertilizer at \$200 a bag, which comes to \$1,600.
]
]


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


      #figure(figph[A sketch depicts Heron of Alexandria.], alt: "A sketch depicts Heron of Alexandria.", caption: [Heron of Alexandria])


      Heron of Alexandria, born around 20 A.D., was an inventor, a scientist, and an engineer. He is credited with the invention of the Aeolipile, one of the first steam engines centuries before the industrial revolution. Heron was the father of the vending machine. He talked about the idea of inserting a coin into a machine for it to perform a specific action in his book, #emph[Mechanics and Optics]. His contribution to the field of mathematics was enormous. #emph[Metrica], a series of three books, was devoted to methods of finding the area and volume of three-dimensional objects like pyramids, cylinders, cones, and prisms. He also discovered and developed the procedures for finding square roots and cubic roots. However, he is probably best known for Heron’s formula, which is used for finding the area of a triangle based on the lengths of its sides. Given a triangle #math.equation(block: false, alt: "A B C")[$A B C$] ,


      #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  content((100.00pt, 180.67pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[triangle ABC])
  line((33.33pt, 33.33pt), (166.67pt, 33.33pt), (100.00pt, 133.33pt), close: true, fill: none, stroke: 0.7pt + black)
  content((100.00pt, 24.33pt), anchor: "north", text(size: 9pt)[c])
  content((141.38pt, 87.36pt), anchor: "west", text(size: 9pt)[a])
  content((58.62pt, 87.36pt), anchor: "east", text(size: 9pt)[b])
  content((37.33pt, 37.33pt), anchor: "south-west", text(size: 9pt)[A])
  content((162.67pt, 37.33pt), anchor: "south-east", text(size: 9pt)[B])
  content((100.00pt, 129.33pt), anchor: "north", text(size: 9pt)[C])
})]),
  alt: "A triangle, A B C. A B measures c. A C measures b. C B measures a.", caption: none,
)


      Heron’s formula is #math.equation(block: false, alt: "A equals the square root of s open parenthesis s minus a close parenthesis open parenthesis s minus b close parenthesis open parenthesis s minus c close parenthesis ,")[$A = sqrt(s ( s − a ) ( s − b ) ( s − c )) ,$] where #math.equation(block: false, alt: "s")[$s$] is the semi-perimeter calculated as #math.equation(block: false, alt: "s equals the fraction a plus b plus c over 2 .")[$s = frac(a + b + c, 2) .$]
]


  
  
  
  
    
=== Key Terms


    - triangle
- square
- rectangle
- rhombus
- apothem
- radius
- circle


  
  
    
=== Key Concepts


    - The area #math.equation(block: false, alt: "A")[$A$] of a triangle is found with the formula #math.equation(block: false, alt: "A equals the fraction 1 over 2 b h ,")[$A = frac(1, 2) b h ,$] where #math.equation(block: false, alt: "b")[$b$] is the base and #math.equation(block: false, alt: "h")[$h$] is the height.
- The area of a parallelogram is found using the formula #math.equation(block: false, alt: "A equals b h ,")[$A = b h ,$] where #math.equation(block: false, alt: "b")[$b$] is the base and #math.equation(block: false, alt: "h")[$h$] is the height.
- The area of a rectangle is found using the formula #math.equation(block: false, alt: "A equals l w ,")[$A = l w ,$] where #math.equation(block: false, alt: "l")[$l$] is the length and #math.equation(block: false, alt: "w")[$w$] is the width.
- The area of a trapezoid is found using the formula #math.equation(block: false, alt: "A equals the fraction 1 over 2 h open parenthesis b sub 1 plus b sub 2 close parenthesis ,")[$A = frac(1, 2) h ( b_(1) + b_(2) ) ,$] where #math.equation(block: false, alt: "h")[$h$] is the height, #math.equation(block: false, alt: "b sub 1")[$b_(1)$] is the length of one base, and #math.equation(block: false, alt: "b sub 2")[$b_(2)$] is the length of the other base.
- The area of a rhombus is found using the formula #math.equation(block: false, alt: "A equals the fraction d sub 1 d sub 2 over 2 ,")[$A = frac(d_(1) d_(2), 2) ,$] where #math.equation(block: false, alt: "d sub 1")[$d_(1)$] is the length of one diagonal and #math.equation(block: false, alt: "d sub 2")[$d_(2)$] is the length of the other diagonal.
- The area of a regular polygon is found using the formula #math.equation(block: false, alt: "A equals the fraction 1 over 2 a p ,")[$A = frac(1, 2) a p ,$] where #math.equation(block: false, alt: "a")[$a$] is the apothem and #math.equation(block: false, alt: "p")[$p$] is the perimeter.
- The area of a circle is found using the formula #math.equation(block: false, alt: "A equals π r squared ,")[$A = π r^(2) ,$] where #math.equation(block: false, alt: "r")[$r$] is the radius.


  
  
    
=== Formulas


    The area of a triangle is given as #math.equation(block: false, alt: "A equals the fraction 1 over 2 b h ,")[$A = frac(1, 2) b h ,$] where #math.equation(block: false, alt: "b")[$b$] represents the base and #math.equation(block: false, alt: "h")[$h$] represents the height.


    The formula for the area of a square is #math.equation(block: false, alt: "A equals s times s")[$A = s ⋅ s$] or #math.equation(block: false, alt: "A equals s squared .")[$A = s^(2) .$]


    The area of a rectangle is given as #math.equation(block: false, alt: "A equals l w .")[$A = l w .$]


    The area of a parallelogram is #math.equation(block: false, alt: "A equals b h .")[$A = b h .$]


    The formula for the area of a trapezoid is given as #math.equation(block: false, alt: "A equals the fraction 1 over 2 h open parenthesis a plus b close parenthesis .")[$A = frac(1, 2) h ( a + b ) .$]


    The area of a rhombus is found using one of these formulas:


    - #math.equation(block: false, alt: "A equals the fraction d sub 1 d sub 2 over 2 ,")[$A = frac(d_(1) d_(2), 2) ,$] where #math.equation(block: false, alt: "d sub 1")[$d_(1)$] and #math.equation(block: false, alt: "d sub 2")[$d_(2)$] are the diagonals.
- #math.equation(block: false, alt: "A equals the fraction 1 over 2 b h ,")[$A = frac(1, 2) b h ,$] where #math.equation(block: false, alt: "b")[$b$] is the base and #math.equation(block: false, alt: "h")[$h$] is the height.


    The area of a regular polygon is found with the formula #math.equation(block: false, alt: "A equals the fraction 1 over 2 a p ,")[$A = frac(1, 2) a p ,$] where #math.equation(block: false, alt: "a")[$a$] is the apothem and #math.equation(block: false, alt: "p")[$p$] is the perimeter.


    The area of a circle is given as #math.equation(block: false, alt: "A equals π r squared ,")[$A = π r^(2) ,$] where #math.equation(block: false, alt: "r")[$r$] is the radius.