Sect 8.3 Triangles and Hexagons

Transcription

1 13 Objective 1: Sect 8.3 Tringles nd Hexgons Understnding nd Clssifying Different Types of Polygons. A Polygon is closed two-dimensionl geometric figure consisting of t lest three line segments for its sides. If ll the sides re the sme length (the ngles will lso hve the sme mesure), then the figure is clled Regulr Polygon. The points where the two sides intersect re clled vertices. Tringle Qudrilterl Pentgon Hexgon Octgon 3 sides 4 sides 5 sides 6 sides 8 sides Regulr Polygons Tringle Qudrilterl Pentgon Hexgon Octgon Objective : Understnding nd Clssifying Different Types of Tringles. We cn clssify tringles by their sides: Equilterl Tringle Isosceles Tringle Sclene Tringle All sides re equl. Two sides re equl. No sides re equl. All ngles mesure 60. The two bse ngles re equl. No ngles re equl.

2 An equilterl tringle is lso clled n equingulr tringle. The third ngle in n isosceles tringle formed by the two equl sides is clled the vertex ngle. We cn lso clssify tringles by their ngles: 133 Acute Tringle Right Tringle Obtuse Tringle All ngles re cute. Hs on right ngle. Hs one obtuse ngle. In right tringle, the two sides tht intersect to form the right ngle re clled the legs of the right tringle while the third side (the longest side) of right tringle is clled the hypotenuse of the right tringle. Recll tht the sum of the mesures of the ngles of tringle is equl to 180. Determine wht type of tringle is picture below: Ex. 1 B Ex. X R 144 A C T Since the mesures of the Since TXR nd 144 re supplengles of tringle totl mentry ngles, then 180, then m TXR = = m B + 71 = 180 But, m TXR + m T = m B = 180 = 90. So, R = = = 109 Since the ngles re different, m B = 71. then none of the sides re equl. So, ABC is n Isosceles Thus, XRT is sclene nd nd n Acute Tringle. right tringle. 54

3 134 Objective 3: Right Tringles nd the Pythgoren Theorem To find the squre root of number on scientific clcultor, we will use the x key. Simplify the following (round to the nerest thousndth): Ex Ex. 3b 144 Ex. 3c 0 Ex. 3d Ex. 3e 15 Ex. 3f 3 7 ) 49 = 7. b) 144 = 1. c) 0 = 0. d) 65 = e) Use your clcultor: 15 = f) Use your clcultor. First, find 7 nd then multiply by 3: 3 7 = 3( ) = In right tringle, there is specil reltionship between the length of the legs ( nd b) nd the hypotenuse (c). This is known s the Pythgoren Theorem: Pythgoren Theorem In right tringle, the squre of the hypotenuse (c ) is equl to the sum of the squres of the legs ( + b ) c = + b Some lternte forms of the Pythgoren Theorem re: c b c = + b = c b b = c Keep in mind tht the hypotenuse is the longest side of the right tringle.

4 135 Find the length of the missing sides: Ex m Ex m 6.00 ft ft In this problem, we hve In this problem, we hve one leg the two legs of the tringle nd the hypotenuse nd we re nd we re looking for the looking for the other leg: hypotenuse: c = + b c = + b (15) = (6) + b c = (7.8) + (10.) 5 = 36 + b (solve for b ) c = = 36 c = = b To find c, tke the squre To find b, tke the squre root root* of : of 189: c = ± = ± b = ± 189 = ± c 1.8 m b 13.7 ft * - The eqution c = ctully hs two solutions 1.8 nd 1.8, but the lengths of tringles re positive so we ignore the negtive solution. Find the length of the digonl of the following rectngle: Ex. 6 A rectngle tht is 1 feet by 5 feet. First, drw picture. Since the ngles 5.0 feet of rectngle re right ngles, then the two sides re the legs of right 1.0 feet tringle while the digonl is the hypotenuse of the right tringle. Using the Pythgoren Theorem, we get: c = + b c = (1) + (5) c = = 169 c = ± 169 = ± 13. So, the digonl is 13 feet.

5 Solve the following: Ex. 7 Find the height of AXT if ech side hs length of 17.0 in. The ltitude (height) of the X tringle forms right ngle with the bse of the tringle nd 17.0 in 17.0 in h A 136 divides the bse into two equl prts. Thus, the distnce from point A nd this point of intersection is 17 in = 8.5 in. The right hlf of the tringle together with the ltitude form right tringle with hypotenuse of 17 in, one leg equl to 8.5 in, nd the other leg equl to the height of the equilterl tringle being the length of the other leg. Using the Pythgoren Theorem, we cn find the height: 17 = (8.5) + h 17 in 17.0 in T 89 = h ( 7.5 from both sides) h = h (solve for h) h = ± = ± So, the height is bout 14.7 inches. 8.5 in In generl, the formul for the height of n equilterl tringle with side of length is h = 3. Objective 4: Understnding Heron's Formul. Recll the re of tringle is equl to one-hlf the bse times the height. This formul works well if we hve the bse nd the height of the tringle. But, if we only know the lengths of the sides of the tringle, we cn use Heron's formul to find the re. The key to using Heron's Formul is to first find the perimeter of the tringle nd then divide the rest by two to get the semi-perimeter. We then plug the vlues into the formul. Heron's Formul Let, b, nd c be the lengths of the tringle nd s be the semi-perimeter of the tringle. Then the re of the tringle is A = s(s )(s b)(s c) where s = +b+c b c

6 137 Find the re of the following tringles: 5.00 m Ex. 8 Ex. 8b Ex. 8c 4.8 cm 9. cm 7.3 cm 5.00 m 5.00 m ) First, clculte the semi-perimeter: s = ( )/ = 1.3/ = cm Now, plug into Heron's Formul: A = 10.65( )( )( ) (subtrct) = 10.65(5.85)(3.35)(1.45) (multiply) = (squre root) = (round to two significnt digits) 17 cm b) First, clculte the semi-perimeter: s = ( )/ = 15/ = 7.5 cm Now, plug into Heron's Formul: A = 7.5(7.5 5)(7.5 5)(7.5 5) (subtrct) = 7.5(.5)(.5)(.5) (multiply) = (squre root) = (round to three significnt digit) 10.8 m c) First, clculte the semi-perimeter: s = ( )/ = 10/ = 5 in Now, plug into Heron's Formul: A = 5(5 3.95)(5 3.95)(5.1) (subtrct) = 5(1.05)(1.05)(.9) (multiply) = (squre root) = (round to two significnt digits) 4.0 in 3.95 in 3.95 in.1 in In generl, we cn use Heron's Formul to derive the formuls for the re of specil types of tringles. For instnce, suppose we wnt to find the

7 formul for the re of n equilterl tringle. Let = the length of one side of n equilterl tringle. Then the semi-perimeter is: s = ( + + )/ = 3 nd the re is: 3 A = ( 3 )( 3 )( 3 ) (subtrct) = = = ( 1 )( 1 )( 1 ) (multiply) (squre root) 138 Objective 5: Understnding Properties of Regulr Hexgons. In regulr hexgon, ll the sides re equl nd ll the ngles re equl. To find the mesure of the ngles of regulr hexgon, we cn split the regulr hexgon into series four of tringles s shown. This mens tht the sum of the mesures of the ngles of hexgon is equl to four times the sum of the mesures of the ngles of tringle or = 70. Since the ngles of regulr hexgon re equl, then ech ngle is equl to 70 6 = 10. To find the re of regulr hexgon, first drw the digonls of the regulr hexgon. Ech digonl bisects the ngle of the hexgon into two 60 ngles. Thus, the regulr hexgon consists of six equilterl tringles. Since the length of of ech side is equl to, then the re of regulr hexgon is six times the re of n equilterl tringle or = 3 3 This lso implies tht the length of the digonl of regulr hexgon, d, is equl to. This is lso referred to s the distnce cross the corners..

8 139 One finl mesurement of regulr hexgon we will need is the distnce cross the flts, f. Notice tht f is twice the height of n equilterl tringle formed by the digonls. As noted in exmple #7, the formul for finding the height of n equilterl tringle is h = Hence, f = h = 3 3. = 3. Let's summrize our findings: Properties of Regulr Hexgons with side of length : d f 1) Ech ngle mesures 10. ) Distnce cross the corners (digonl): d = 3) Distnce cross the flts: f = 3 4) The re: A = 3 3 Complete the following tble: Ex. 9 ) 1 in d f A b) 5 8 cm c) 15.3 mm ) Since = 1 in, then d = = 1 = 4 in f = 3 = 1 3 = in A = 3 3 = 3(1) 3 = in

9 140 b) Since d = 5 cm = 0.65 cm, then 8 d = implies 0.65 = (divide by ) = cm f = 3 = = cm A = 3 3 = 3(0.315) 3 = cm c) Since f = 15.3 mm, then f = 3 implies 15.3 = 3 (divide by 3 ) = mm d = = = mm A = 3 3 = 3( ) 3 = mm d f A ) 1 in 4 in 1 in 370 in b) cm 5 8 cm cm 0.54 cm c) mm mm 15.3 mm 03.3 mm Objective 6: Understnding Composite Figures A Composite Figure is geometric figure tht consists of two or more bsic geometric figures. We wnt to find the perimeter nd re of composite figures. The perimeter of ny figure is the totl length round the outside of the figure. Think of person wlking long the outside of the figure; how fr does the person hve to trvel to get bck to the beginning?

10 141 In terms of the re of the composite figure, think of how much pint is needed to pint the inside of the figure. The key to solving composite figure problem is to split it prt into set of bsic figures. Let's look t some exmples. Ex.10 ) Find the perimeter of the following. b) Find the re of the following m 9.00 m 4.4 m 4.4 m ) We begin by splitting the figure prt: 6.00 m 9.00 m 9.00 m 4.4 m 4.4 m The perimeter of this figure is the sum of the lengths of the three sides of the rectngle plus the two sides of the tringle. Thus, the perimeter is: P = ( ) meters = 3.48 meters

11 14 b) We begin by splitting the figure prt: 6.00 m 6.00 m 4.4 m 4.4 m 9.00 m The re of this figure is the sum of the re of the rectngle nd the re of the tringle. The re of the rectngle is L w = 6 9 = 54 m. To find the re of the tringle, we will use Heron's Formul: s = ( ) = = 7.4 m A = 7.4(7.4 6)( )( ) (subtrct) = 7.4(1.4)(3)(3) (multiply) = (squre root) = m Thus, the totl re = = m. Ex. 11 Find the re of the following figure: 13 m 14 m 4 m 5 m 9 m First, we will need to split the figure prt nd determine if there ny dimensions we will need to find:

12 m 14 m 4 m 4 m 5 m 9 m We need to find the height of the tringles nd the width of the rectngle. Notice tht the width of the rectngle is 4 m less thn the height of the tringles. Consider the tringle on the left. We hve the hypotenuse nd leg of right tringle. We cn use the Pythgoren Theorem to solve for the unknown side: + 5 = = m = m = ± 144, so = 1 m since > 0. Thus, the heights of the two tringles re 1 m. Now, we cn find the width of the rectngle. 5 m 1 m 4 m The dshed side corresponds to the width of the rectngle in the middle. Since the height of the tringle is 1 m nd the length of side from the bse of the tringle to the rectngle is 4 m, then the width of the rectngle is 1 4 = 8 m. We cn find the re of the composite figure by dding the res of the tringle nd the re of the rectngle:

13 144 8 m 1 m 1 m 14 m 5 m 9 m A = 1 bh + Lw + 1 bh = [ 1 (5)(1) + 14(8) + 1 (9)(1)] m = [ ] m = 196 m 00 m. Ex. 1 A wooden door hs 9 rectngulr pieces of glss embedded in it. If ech piece of glss is 1/3 ft by 1/ ft, find the re of the wood. Round to the nerest hlf-foot. (See figure) 3 ft 7 ft The re of the wood is equl to the re of the door minus the re of the nine pieces of glss: A = Lw 9Lw A = 3(7) 9(1/3)(1/) = = 19 1 ft