Each side of square RSTV has length 8. Point W lies on Point Q is the center of regular hexagon ABCDEF. This

We now seek a general formula for the area of any regular polygon. AREA OF A REGULAR POLYGON In Chapter 7 and Chapter 8, we have laid the groundwork for determining the area of a regular polygon. In the proof of Theorem 8.3.1, the figure chosen is a regular pentagon; however, the proof applies to a regular polygon of any number of sides. It is also worth noting that the perimeter P of a regular polygon is the sum of its equal sides. If there are n sides and each has length s, the perimeter of the regular poly- gon is P ⫽ ns. The area A of a regular polygon whose apothem has length a and whose perimeter is P is given by A = 1 2 aP THEOREM 8.3.1 GIVEN: Regular polygon ABCDE in Figure 8.36a such that OF ⫽ a and the perimeter of ABCDE is P PROVE: PROOF: From center O, draw radii , , , , and . [See Figure 8.36b.] Now , , , , and are all by SSS. Where s represents the length of each of the congruent sides of the regular polygon and a is the length of an apothem, the area of each triangle is from . Therefore, the area of the pentagon is Because the sum or ns represents the perimeter P of the polygon, we have A ABCDE = 1 2 aP s + s + s + s + s = 1 2 a s + s + s + s + s A ABCDE = a 1 2 sa b + a 1 2 sa b + a 1 2 sa b + a 1 2 sa b + a 1 2 sa b A = 1 2 bh 1 2 sa ⬵ 䉭EOA 䉭DOE 䉭COD 䉭BOC 䉭AOB OE OD OC OB OA A ABCDE = 1 2 aP Exs. 9–11 EXAMPLE 4 Use to find the area of the square whose length of apothem is Solution For this repeat of Example 1, see Figure 8.33 as needed. When the length of apothem of a square is a ⫽ 2, the length of side is s ⫽ 4. In turn, the perimeter is Now becomes , so . NOTE: As expected, the answer from Example 1 is repeated in Example 4. A = 16 in 2 A = 1 2 2 16 A = 1 2 aP P = 16 in. a = 2 in. A = 1 2 aP D E B C F A O a D E B C F A O b Figure 8.36 쮿 쮿 For Examples 6 and 7, the measures of the line segments that represent the length of the apothem, the radius, or the side of a regular polygon depend upon relationships that are developed in the study of trigonometry. The methods used to find related measures will be developed in Chapter 11 but are not given attention at this time. Many of the meas- ures that are provided in the following examples and the exercise set for this section are actually only good approximations. Exs. 12–15 EXAMPLE 5 Use to find the area of the equilateral triangle whose apothem has the length 6 cm. Solution For this repeat of Example 3, refer to Figure 8.35 on page 374. Because the length of apothem is 6 cm, the length of . In turn, the length of side is . For the equilateral triangle, the perimeter is . Now becomes so . NOTE: As is necessary, the answer found in Example 3 is repeated in Example 5. 쮿 A = 108 13 cm 2 A = 1 2 6 36 13, A = 1 2 aP P = 36 13 cm 12 13 cm AB AD is 6 13 cm OD A = 1 2 aP EXAMPLE 6 In Figure 8.36a on page 375, find the area of the regular pentagon ABCDE with center O if OF ⫽ 4 and . Solution OF ⫽ a ⫽ 4 and . Therefore, or . Consequently, = 59 units 2 A ABCDE = 1 2 429.5 P = 29.5 P = 55.9 AB = 5.9 AB = 5.9 쮿 EXAMPLE 7 Find the area of the regular octagon shown in Figure 8.37. The center of PQRSTUVW is point O. The length of apothem is 12.1 cm, and the length of side is 10 cm. Solution If QR ⫽ 10 cm, then the perimeter of regular octagon PQRSTUVW is or 80 cm. With the length of apothem being , the area formula becomes , so . A = 484 cm 2 A = 1 2 12.1 80 A = 1 2 aP OX = 12.1 cm 8 10 cm QR OX S X T P W R U Q V O Figure 8.37 쮿 EXAMPLE 8 Find the exact area of equilateral triangle ABC in Figure 8.38 if each side measures 12 in. Use the formula . A = 1 2 aP Exs. 12–15 Solution In , the perimeter is or 36 in. To find the length a of an apothem, we draw the radius from center O to point A and the apothem from O to side . Because the radius bisects . Because apothem , . also bisects . Using the 30°-60°-90° relationship in , we see that . Thus a = 6 13 = 6 13 13 13 = 6 13 3 = 213 a 1 3 = 6 䉭OMA AB OM m ∠OMA = 90° OM ⬜ AB m ∠OAB = 30° ∠BAC, AB OM OA P = 3 12 䉭ABC C A B 12 12 12 Figure 8.38 C A M a 6 O 30° B NOTE: Using the calculator’s value for leads to an approximation of the area rather than to an exact area. 13 쮿 Discover TESSELLATIONS Tessellations are patterns composed strictly of interlocking and non-overlapping regular polygons. All of the regular polygons of a given number of sides will be congruent. Tessellations are commonly used in design, but especially in flooring tiles and vinyl sheets. A pure tessellation is one formed by using only one regular polygon in the pattern. An impure tessellation is one formed by using two different regular polygons. In the accompanying pure tessellation, only the regular hexagon appears. In nature, the beehive has compartments that are regular hexagons. Note that the adjacent angles’ measures must sum to 360°; in this case, . It would also be possible to form a pure tessellation of congruent squares because the sum of the adjacent angles’ measures would be . In the impure tessellation shown, the regular octagon and the square are used. In Champaign-Urbana, sidewalks found on the University of Illinois campus use this tessellation pattern. Again it is necessary that the sum of the adjacent angles’ measures be 360°; for this impure tessellation, . a Can congruent equilateral triangles be used to form a pure tessellation? b Can two regular hexagons and a square be used to build an impure tessellation? ANSWERS 135° + 135° + 90° = 360° 90° + 90° + 90° + 90° = 360° 120° + 120° + 120° = 360° a Yes, because b No, because 120 °+ 120° +90 °Z 360° 6 60° =360 ° Now becomes . A = 1 2 2 13 36 = 36 13 in 2 A = 1 2 aP Exercises 8.3

1. Find the area of a square with

a sides of length 3.5 cm each. b apothem of length 4.7 in.

2. Find the area of a square with

a a perimeter of 14.8 cm. b radius of length

3. Find the area of an equilateral triangle with

a sides of length 2.5 m each. b apothem of length 3 in.

4. Find the area of an equiangular triangle with

a a perimeter of 24.6 cm. b radius of length 4 in.

5. In a regular polygon, each central angle measures 30°. If

each side of the regular polygon measures 5.7 in., find the perimeter of the polygon.

6. In a regular polygon, each interior angle measures 135°. If

each side of the regular polygon measures 4.2 cm, find the perimeter of the polygon.

7. For a regular hexagon, the length of the apothem is 10 cm.

Find the length of the radius for the circumscribed circle for this hexagon.

8. For a regular hexagon, the length of the radius is 12 in.

Find the length of the radius for the inscribed circle for this hexagon.

9. In a particular type of regular polygon, the length of the

radius is exactly the same as the length of a side of the polygon. What type of regular polygon is it?

10. In a particular type of regular polygon, the length of the

apothem is exactly one-half the length of a side. What type of regular polygon is it?

11. In one type of regular polygon, the measure of each

interior angle is equal to the measure of each central angle. What type of regular polygon is it?

12. If the area and the perimeter of a regular

polygon are numerically equal, find the length of the apothem of the regular polygon.

13. Find the area of a square with apothem and

perimeter .

14. Find the area of an equilateral triangle with apothem

and perimeter .

15. Find the area of an equiangular triangle with apothem

and perimeter

16. Find the area of a square with apothem and

perimeter . In Exercises 17 to 30, use the formula to find the area of the regular polygon described.

17. Find the area of a regular pentagon with an apothem of

length and each side of length .

18. Find the area of a regular pentagon with an apothem of

length and each side of length s = 9.4 in. a = 6.5 in. s = 7.5 cm a = 5.2 cm A = 1 2 aP P = 65.6 ft a = 8.2 ft P = 27.6 13 in. a = 4.6 in. P = 19.2 13 cm a = 3.2 cm P = 25.6 cm a = 3.2 cm A = 1 2 aP AI = n - 2180° n B 4 12 in.

19. Find the area of a regular octagon with an apothem of

length and each side of length

20. Find the area of a regular octagon with an apothem of

length and each side of length .

21. Find the area of a regular hexagon whose sides have

length 6 cm. 22. Find the area of a square whose apothem measures 5 cm. 23. Find the area of an equilateral triangle whose radius measures 10 in.

24. Find the approximate area of a regular pentagon whose

apothem measures 6 in. and each of whose sides measures approximately 8.9 in.

25. In a regular octagon, the approximate ratio of the length

of an apothem to the length of a side is 6:5. For a regular octagon with an apothem of length 15 cm, find the approximate area.

26. In a regular dodecagon 12 sides, the approximate ratio

of the length of an apothem to the length of a side is 15:8. For a regular dodecagon with a side of length 12 ft, find the approximate area.

27. In a regular dodecagon 12 sides, the approximate ratio

of the length of an apothem to the length of a side is 15:8. For a regular dodecagon with an apothem of length 12 ft, find the approximate area.

28. In a regular octagon, the approximate ratio of the length of an

apothem to the length of a side is 6:5. For a regular octagon with a side of length 15 ft, find the approximate area.

29. In a regular polygon of 12 sides, the measure of each side

is 2 in., and the measure of an apothem is exactly Find the exact area of this regular polygon.

30. In a regular octagon, the measure of each apothem is

4 cm, and each side measures exactly . Find the exact area of this regular polygon.

31. Find the ratio of the area of a square circumscribed about

a circle to the area of a square inscribed in the circle.

32. Given regular hexagon ABCDEF with each side of length

6 and diagonal , find the area of pentagon ACDEF. AC 8 12 - 1 cm 2 + 13 in. s = 6.5 ft a = 7.9 ft s = 8.1 in. a = 9.8 in. A B C F E D G

33. Given regular octagon RSTUVWXY with each side of

length 4 and diagonal , find the area of hexagon RYXWVU . RU

34. Regular octagon

ABCDEFGH is inscribed in a circle whose radius is . Considering that the area of the octagon is less than the area of the circle and greater than the area of the square ACEG, find the two integers between which the area of the octagon must lie. NOTE: For the circle, use with .

35. Given regular pentagon

RSTVQ and equilateral triangle PQR, the length of an apothem not shown of RSTVQ is 12, while the length of each side of the equilateral triangle is 10. If , find the approximate area of kite VPST. PV L 8.2 ␲ L 22 7 A = ␲r 2 7 2 12 cm A B C D E F G H T V S R Q P

36. Consider regular pentagon RSTVQ not shown. Given

that diagonals and intersect at point F, show that .

37. Consider a regular hexagon ABCDEF not shown. By

joining midpoints of consecutive sides, a smaller regular hexagon MNPQRS is formed. Show that the ratio of areas is A MNPQRS A ABCDEF = 3 4 VF FR = TF FQ VR QT Circumference of a Circle ␲ Pi Length of an Arc Limit Area of a Circle Circumference and Area of a Circle 8.4 KEY CONCEPTS In geometry, any two figures that have the same shape are described as similar. For this reason, we say that all circles are similar to each other. Just as a proportionality exists among the sides of similar triangles, experimentation shows that there is a proportion- ality among the circumferences distances around and diameters distances across of circles; see the Discover activity on page 380. Representing the circumferences of the circles in Figure 8.39 by C 1 , C 2 , and C 3 and their respective lengths of diameters by d 1 , d 2 , and d 3 , we claim that where k is the constant of proportionality. C 1 d 1 = C 2 d 2 = C 3 d 3 = k C 1 d 1 C 2 d 2 C 3 d 3 Figure 8.39 The ratio of the circumference of a circle to the length of its diameter is a unique positive constant. POSTULATE 22 The constant of proportionality k described in the opening paragraph of this section, in Postulate 22, and in the Discover activity is represented by the Greek letter . ␲ pi GIVEN: Circle O with length of diameter d and length of radius r. See Figure 8.40. PROVE: PROOF: By Postulate 22, . Multiplying each side of the equation by d, we have . Because d 2r the diameter’s length is twice that of the radius, the formula for the circumference can be written , or . VALUE OF ␲ In calculating the circumference of a circle, we generally leave the symbol ␲ in the an- swer in order to state an exact result. However, the value of ␲ is irrational and cannot be represented exactly by a common fraction or by a terminating decimal. When an approximation is needed for ␲, we use a calculator. Approximations of ␲ that have been commonly used throughout history include , , and . Although these approximate values have been used for centuries, your calculator provides greater accuracy. A calculator will show that . ␲ L 3.141592654 ␲ L 3.1416 ␲ L 3.14 ␲ L 22 7 C = 2␲r C = ␲ 2r = C = ␲d ␲ = C d C = 2␲r Exs. 1–2 Discover Find an object of circular shape, such as the lid of a jar. Using a flexible tape measure such as a seamstress or carpenter might use, measure both the distance around circumference and the distance across length of diameter the circle. Now divide the circumference C by the diameter length d. What is your result? ANSWER The ratio should be slightly larger than 3. Technology Exploration Use computer software if available. 1. Draw a circle with center O. 2. Through O, draw diameter . 3. Measure the circumference C and length d of diameter . 4. Show that . C d L 3.14 AB AB ␲ is the ratio between the circumference C and the diameter length d of any circle; thus, in any circle. ␲ = C d DEFINITION The circumference of a circle is given by the formula C = ␲d or C = 2␲r THEOREM 8.4.1 r O Figure 8.40 EXAMPLE 1 In in Figure 8.41, OA ⫽ 7 cm. Using , a find the approximate circumference C of . b find the approximate length of the minor arc . Solution a b Because the degree of measure of is 90°, we have or of the circumference for length of AB ¬ = 90 360 44 = 1 4 44 = 11 cm 1 4 90 360 AB ¬ = 44 cm = 2 22 7 7 C = 2␲r AB ¬ }O ␲ L 22 7 }O A B O Figure 8.41 쮿 the arc length. Then In the following theorem, the lengths of the diameter and radius of the circle are rep- resented by d and r respectively. 쮿