Lesso 2.4: Agle Properties i Polygos, page 99 1. a) S(12) = 180 (12 2) S(12) = 180 (10) S(12) = 1800 A dodecago has 12 sides, so is 12. The sum of the iterior agles i a regular dodecago is 1800. S(12) = 1800 Show i part a). 1800 12 = 150 Each iterior agle i a regular dodecago is equal, so each agle must measure 1 of the sum of 12 the agles. The measure of each iterior agle of a regular dodecago is 150. 2. S(20) = 180 (20 2) S(20) = 180 (18) S(20) = 3240 A 20-sided covex polygo has 20 sides, so is 20. The sum of the iterior agles i a 20-sided covex polygo is 3240. 3. 3060 = 180 ( 2) 17 = 2 19 = Substitute the kow quatity. Divide both sides by 180. Add 2 to both sides. A 19-sided regular covex polygo has a sum of 3060 for its iterior agles. 4. e.g., The iterior agles of a hexago equal 120. Three hexagos will fit together sice the sum is 360 at the vertex where they are joied. 5. e.g., Yes. You ca alig parallel sides to create a tilig patter; the agles that meet are the four agles of the parallelogram, so their sum is 360. 6. A looie is a regular 11-sided covex polygo. S(11) = 180 (11 2) S(11) = 180 (9) S(11) = 1620 1620 = 147.272 11 A looie has 11 sides, so is 11. agles Each iterior agle i a regular 11-sided polygo is equal, so each agle must measure 1 of the sum of 11 the agles. The measure of each iterior agle of a regular 11- sided polygo is about 147.27. 7. a) agles Oe agle = S() Measure of oe iterior agle 180 ( 2) Oe agle = 180 ( 2) Substitute the = 140 kow quatity. 180 ( 2)= 140 180 360 = 140 40 = 360 = 9 A 9-sided regular covex polygo has a iterior agle measure of 140. 180 = exterior agle + iterior agle 180 = exterior agle + 140 40 = exterior agle Let S represet the sum of the exterior agles. S = umber of exterior agles measure of oe exterior agle S = 9(40 ) S = 360 Supplemetary agles Determie the sum of the measures of the exterior agles. The sum of the measures of the exterior agles of a regular 9-sided covex polygo is 360. 8. a) The sum of the measures of the exterior agles of a ay covex polygo is 360. Therefore, the sum of the measures of the exterior agles of a regular octago is 360. Let E represet the measure of a exterior agle. sum of exterior agles E = umber of sides E = 360º 8 E = 45 Foudatios of Mathematics 11 Solutios Maual 2-9
The measure of a exterior agle of a regular octago is 45. 180 = iterior agle + exterior agle 180 = iterior agle + 45 135 = iterior agle Supplemetary agles Substitute the kow quatity. The measure of a iterior agle of a regular octago is 135. c) Let S represet the sum of the iterior agles. S = umber of sides measure of oe iterior agle S = 8(135 ) S = 1080 The sum of the measures of the iterior agles of a regular octago is 1080. d) S(8) = 180 (8 2) S(8) = 180 (6) S(8) = 1080 A octago has 8 sides, so is 8. The aswers for parts c) ad d) are the same. 9. a) e.g., Agree. If you draw a regular hexago, you ca draw three rectagles usig opposite sides. The rectagles have opposite sides that are parallel. You caot do this for a regular polygo with a odd umber of sides. e.g., Opposite sides are parallel i a regular polygo that has a eve umber of sides. 10. a) LPO = 108 PL = OP OLP is isosceles. LOP = OLP LOP + OLP + LPO = 180 LOP + OLP + 108 = 180 LOP + OLP = 72 2 LOP = 72 LOP = 36 OLP = 36 LM = NM LMN is isosceles. LMN = NLM LMN = 108 LMNOP is a regular petago, so is 5. 540 = 108 5 Give Defiitio of isosceles isosceles agles i Give Defiitio of isosceles isosceles Measure of iterior agle of regular petago LNM + NLM + LMN = 180 LNM + NLM + 108 = 180 LNM + NLM = 72 2 NLM = 72 NLM = 36 LNM = 36 MLP = 108 agles i Measure of iterior agle of regular petago MLP = OLN + OLP + MLN 108 = OLN + 36 + 36 36 = OLN NOP = ONM LOP = 36 ad LMN = 36 LOP = LMN NOP = LOP + LON 108 = 36 + LON 72 = LON LON = 72 ad LNO = 72 LON = LNO OLN is isosceles. Substitutio Measures of iterior agles i a regular polygo are equal. Determied i part a) Trasitive Substitutio Determied above Trasitive Defiitio of isosceles 11. The formula for the measure of a iterior 180 ( 2) agle of a regular polygo is S() =. I the formula, Sady wrote 1 istead of 2. 180 (10 2) S(10) = 10 S(10) = 180 (8) 10 S(10) = 144 12. a) e.g., A test could be a sigle lie draw aywhere through the polygo. For covex polygos, it itersects two sides oly. For ocovex polygos, it ca itersect i more tha two sides. If ay diagoal is exterior to the polygo, the polygo is o-covex. 13. a) Assume the hexagoal table top is i the shape of a regular hexago. Each trapezoidal piece of wood i a sectio forms a with the agle at the cetre vertex. Each i a sectio is similar to each other so their correspodig agles are equal. The correspodig agles i each trapezoid are equal. 2-10 Chapter 2: Properties of Agles ad Triagles
Each iterior agle of the hexago is 120. Each base agle of a sector is 60. x + 60 = 180 Each iterior agle of the octago is 135. Each base agle of a sector is 67.5. x + 67.5 = 180 x = 112.5 The table is a regular hexago, so is 6. S(6) = 180 (6 2) S(6) = 180 (4) S(6) = 720 720 = 120 6 Each diameter is a agle bisector of a iterior agle. 120 2 = 60 I each trapezoid, the sum of the iterior agles o the same side of a trasversal (diameter), are supplemetary. Let the other iterior agle be x. The table is a regular octago, so is 8. S(8) = 180 (8 2) S(8) = 180 (6) S(8) = 1080 1080 = 135 8 Each diameter is a agle bisector of a iterior agle. 135 2 = 67.5 I each trapezoid, the sum of the iterior agles o the same side of a trasversal (diameter), are supplemetary. Let the other iterior agle be x. 14. I drew a diagram to help me. v + 90 = 180 Supplemetary v = 90 agles w + 60 = 180 Supplemetary w = 120 agles x + 70 = 180 Supplemetary x = 110 agles y = z Give The polygo is a petago, so is 5. agles Let S represet the sum of the agles i the petago. S = y + z + v + w + x Substitutio S = 2y + 90 + 120 + 110 S = 2y + 320 Therefore, 2y + 320 = 540 2y = 220 y = 110 z = 110 Trasitive The measures of the iterior agles are 110, 120, 90, 110, ad 110. 15. The agles are exterior agles of a petago, so the sum of the measures of the exterior agles is 360. 16. a) Agle c is a iterior agle of a regular hexago. Foudatios of Mathematics 11 Solutios Maual 2-11
S(6) = 180 (6 2) S(6) = 180 (4) S(6) = 720 720º = 120 6 c = 120 d = 1 2 c d = 60 a + c = 180 a = 60 x + d = 180 z + x = 180 z = 60 b + z + a = 180 b + 60 + 60 = 180 b = 60 S(9) = 180 (9 2) S(9) = 180 (7) S(9) = 1260 1260 = 140 9 a = 140 z = 140 2b + z = 180 2b = 40 b = 20 x + b = 140 c + x = 180 c = 60 d + c + b = 140 d + 60 + 20 = 140 d = 60 The larger polygo is a hexago, so is 6. The measure of oe iterior agle The two trapezoids are cogruet so, the bisector of the hexago is a agle bisector of a iterior agle. Supplemetary agles Let x be the measure of the ukow agle that is a iterior agle o the same side of the trasversal as d. Let z be the measure of the ukow agle i the. agles i The polygo is a regular oago, so is 9. Measure of oe iterior agle Let z be the measure of the oago agle i the isosceles cotaiig b. Let x be the measure of the ukow agle adjacet to b. Iterior agles o the same side of a trasversal 17. There are two quadrilaterals. Quadrilateral 1 cotais agles: a, c, e, g Quadrilateral 2 cotais agles: b, d, f, h The polygo is a S(4) = 180 (4 2) quadrilateral, so is 4. S(4) = 180 (2) S(4) = 360 Measure of all iterior agles of oe quadrilateral 2(360 ) = 720 Measure of all iterior agles of two quadrilaterals 18. AB, BC, CD, DE, ad EA regular are equal. petago EO = DO Give DO = CO Give EOD DOC Three pairs of correspodig sides are equal. ODE = ODC EOD ad DOC are ODE = OED cogruet, isosceles s. 540 = 108 5 ODE + ODC = 108 2 ODE = 108 ODE = 54 ADE is isosceles. EAD = EDA ADE + EAD + DEA = 180 2 ADE + 108 = 180 2 ADE = 72 ADE = 36 EFD + EDF + FED = 180 EFD + 36 + 54 = 180 EFD = 90 The larger polygo is a regular petago, so is 5. Measure of oe iterior agle Trasitive ADE is isosceles because AE ad DE are equal. isosceles. agles i ( ADE ad EDA are the same agle.) ADE ad EDF are the same agle. FED ad OED are the same agle. agles i EFD is a right. 19. e.g., If a polygo is divided ito s by joiig oe vertex to each of the other vertices, there are always two fewer s tha the origial umber of sides. Every has a agle sum of 180. 2-12 Chapter 2: Properties of Agles ad Triagles
20. e.g., 3a + 2(90 ) = 540 3a + 180 = 540 3a = 360 a = 120 The polygo is a petago, so is 5. agles Yes. A tilig patter ca be created by puttig four 90 agles together or three 120 agles together for a sum of 360 at the commo vertex. 21. 5x + x = 180 6x = 180 x = 30 180 ( 2) 150 = 150 = 180 360 30 = 360 = 12 Let x be the measure of a exterior agle. The 5x is the measure of a correspodig iterior agle. Supplemetary agles Measure of oe iterior agle The regular polygo has 12 sides. It is called a regular dodecago. Math i Actio, page 100 e.g., Number of Sides Sum of Measures of Agles: 180 ( 2) 12 1800 150 18 2880 160 24 3960 165 Measure of Each Agle I used software to draw several regular polygos. I oticed that the measure of each iterior agle gets closer ad closer to 180 as the umber of sides icreases, so there is less of a bed goig from oe side to aother. I other words, the agle is ot obvious ad it seems to be smoothed out. Practical limitatios o the umber of sides of a buildig could iclude the followig: The availability of materials i a coveiet size to build the walls. If the wall is too arrow, the framig for the wall would be early solid. Isulatio could ot be placed betwee the framig. It would be difficult to fiish the isides of the walls. There would be too may seams i the drywall. If the sides were very short, there would ot be eough space for electrical outlets to be istalled. The optimal umber of sides for a home depeds o the square footage of the home. I decided to make my home about 1000 square feet. Usig the formula A = r 2, I determied that the radius should be about 17.8 feet. I approximated the perimeter of the home by usig the formula for the circumferece of a circle: C = 2 r. The perimeter should be approximately 112 feet. If I built a 18- sided house, I could use 6 foot paels, givig a perimeter of 108 feet. Drywall for iterior walls ca be purchased i 12 foot legths, so I could cut the drywall i half ad ot have ay waste. This seems reasoable. History Coectio, page 103 e.g., A. The buckyball was created with regular petagos ad regular hexagos. B. First, I determied the measures of the iterior agles i both shapes. Shape Sum of Measures of Measure of Agles Oe Agle regular 180 (5 2) = 540 108 petago regular 180 (6 2) = 720 120 hexago At each vertex, there are two agles from the hexago ad oe agle from the petago. The sum of the measures of these agles is: 2(120 ) + 108 = 348 C. This value makes sese. If the sum were 360, the the three shapes would lie flat. To get them to form a covex shape, the agle must be less tha 360. I the diagram I drew, you ca see that the petagos must be bet to be sew to the hexagos. Foudatios of Mathematics 11 Solutios Maual 2-13