hpter 0 ngles nd tringles 0.1 Introduction stright line which crosses two prllel lines is clled trnsversl (see MN in igure 0.1). Trigonometry is suject tht involves the mesurement of sides nd ngles of tringles nd their reltionship to ech other. This chpter involves the mesurement of ngles nd introduces types of tringle. P c d N Q 0. ngulr mesurement n ngle is the mount of rottion etween two stright lines. ngles my e mesured either in degrees or in rdins. If circle is divided into 360 equl prts, then ech prt is clled 1degreend is written s 1 R igure 0.1 M g h f e S i.e. or 1 revolution = 360 1degreeis 1 th of revolution 360 Some ngles re given specil nmes. ny ngle etween 0 nd 90 is clled n cute ngle. n ngle equl to 90 is clled right ngle. ny ngle etween 90 nd 180 is clled n otuse ngle. ny ngle greter thn 180 nd less thn 360 is clled reflex ngle. n ngle of 180 lies on stright line. If two ngles dd up to 90 they re clled complementry ngles. If two ngles dd up to 180 they re clled supplementry ngles. Prllel lines re stright lines which re in thesme plne nd never meet. Such lines re denoted y rrows, s in igure 0.1. With reference to igure 0.1, = c, = d, e = g nd f = h. Such pirs of ngles re clled verticlly opposite ngles. = e, = f, c = g nd d = h. Such pirs of ngles re clled corresponding ngles. (c) c = e nd = h. Such pirs of ngles re clled lternte ngles. (d) + e = 180 nd c + h = 180. Such pirs of ngles re clled interior ngles. 0..1 Minutes nd seconds One degree my e su-divided into 60 prts, clled minutes. i.e. 1degree= 60 minutes which is written s 1 = 60. OI: 10.1016/978-1-85617-697-.0000-X
166 sic ngineering Mthemtics 41 degrees nd 9 minutes is written s 41 9.41 9 is equivlent to 41 9 60 = 41.483 s deciml, correct to 3 deciml plces y clcultor. 1 minute further sudivides into 60 seconds, i.e. 1 minute = 60 seconds which is written s 1 = 60. (Notice tht for minutes, 1 dsh is used nd for seconds, dshes re used.) or exmple, 56 degrees, 36 minutes nd 13 seconds is written s 56 36 13. 0.. Rdins nd degrees One rdin is defined s the ngle sutended t the centre of circle y n rc equl in length to the rdius. (or more on circles, see hpter 6.) With reference to igure 0., for rc length s, θ rdins = s r Prolem 1. vlute 43 9 + 7 43 43 9 + 7 43 71 1 1 (i) 9 + 43 = 7 (ii) Since 60 = 1, 7 = 1 1 (iii) The 1 is plced in the minutes column nd 1 is crried in the degrees column. (iv) 43 + 7 + 1 (crried) = 71. Plce 71 in the degrees column. This nswer cn e otined using the clcultor s follows. 1. nter 43. Press 3. nter9 4. Press 5. Press+ 6. nter 7 7. Press 8. nter43 9. Press 10. Press = nswer = 71 1 Thus, 43 9 + 7 43 = 71 1. Prolem. vlute 84 13 56 39 igure 0. O r When s is the whole circumference, i.e. when s = πr, θ = s r = πr = π r In one revolution, θ = 360. Hence, the reltionship etween degrees nd rdins is i.e. 360 = π rdins or 180 = π rd 1rd= 180 π 57.30 Here re some worked exmples on ngulr mesurement. r S (i) (ii) 84 13 56 39 7 34 13 39 cnnot e done. 1 or 60 is orrowed from the degrees column, which leves 83 in tht column. (iii) (60 + 13 ) 39 = 34, which is plced in the minutes column. (iv) 83 56 = 7, which is plced in the degrees column. This nswer cn e otined using the clcultor s follows. 1. nter 84. Press 3. nter13 4. Press 5. Press 6. nter 56 7. Press 8. nter39 9. Press 10. Press = nswer = 7 34 Thus, 84 13 56 39 = 7 34. Prolem 3. vlute 19 51 47 + 63 7 34 19 51 47 + 63 7 34 83 19 1 1 1
ngles nd tringles 167 (i) 47 + 34 = 81 (ii) Since 60 = 1, 81 = 1 1 (iii) The 1 is plced in the seconds column nd 1 is crried in the minutes column. (iv) 51 + 7 + 1 = 79 (v) Since 60 = 1, 79 = 1 19 (vi) The 19 is plced in the minutes column nd 1 is crried in the degrees column. (vii) 19 + 63 + 1 (crried) = 83. Plce 83 in the degrees column. This nswer cn e otined using the clcultor s follows. 1. nter 19. Press 3. nter 51 4. Press 5. nter47 6. Press 7. Press + 8. nter 63 9. Press 10. nter 7 11. Press 1. nter 34 13. Press 14. Press = nswer = 83 19 1 Thus, 19 51 47 + 63 7 34 = 83 19 1. Prolem 4. form Hence, onvert 39 7 to degrees in deciml 39 7 = 39 7 60 7 60 = 0.45 y clcultor 39 7 = 39 7 60 = 39.45 This nswer cn e otined using the clcultor s follows. 1. nter 39. Press 3. nter7 4. Press 5. Press= 6. Press nswer = 39.45 Prolem 5. onvert 63 6 51 to degrees in deciml form, correct to 3 deciml plces 63 6 51 = 63 6 51 60 = 63 6.85 63 6.85 = 63 6.85 = 63.4475 60 Hence, 63 6 51 = 63.448 correct to 3 deciml plces. This nswer cn e otined using the clcultor s follows. 1. nter 63. Press 3. nter6 4. Press 5. nter51 6. Press 7. Press = 8. Press nswer= 63.4475 Prolem 6. nd seconds onvert 53.753 to degrees, minutes 0.753 = 0.753 60 = 45.18 0.18 = 0.18 60 = 11 to the nerest second Hence, 53.753 = 53 45 11 This nswer cn e otined using the clcultor s follows. 1. nter 53.753. Press = 3. Press nswer= 53 45 10.8 Now try the following Prctice xercise Prctice xercise 76 ngulr mesurement (nswers on pge 348) 1. vlute 5 39 + 9 48. vlute 76 31 48 37 3. vlute 77 + 41 36 67 47 4. vlute 41 37 16 + 58 9 36 5. vlute 54 37 4 38 53 5 6. vlute 79 6 19 45 58 56 +53 1 38 7. onvert 7 33 to degrees in deciml form. 8. onvert 7 45 15 to degrees correct to 3 deciml plces. 9. onvert 37.95 to degrees nd minutes. 10. onvert 58.381 to degrees, minutes nd seconds. Here re some further worked exmples on ngulr mesurement. Prolem 7. Stte the generl nme given to the following ngles: 157 49 (c) 90 (d) 45 ny ngle etween 90 nd 180 is clled n otuse ngle. Thus, 157 is n otuse ngle. ny ngle etween 0 nd 90 is clled n cute ngle.
168 sic ngineering Mthemtics (c) Thus, 49 is n cute ngle. n ngle equl to 90 is clled right ngle. n ngle of 180 lies on stright line. Hence, from igure 0.3, (d) ny ngle greter thn 180 nd less thn 360 is clled reflex ngle. Thus, 45 is reflex ngle. from which, 180 = 53 + θ + 44 θ = 180 53 44 = 83 Prolem 8. 48 39 ind the ngle complementry to Prolem 11. vlute ngle θ in the digrm shown in igure 0.5 If two ngles dd up to 90 they re clled complementry ngles. Hence, the ngle complementry to 48 39 is 90 48 39 = 41 1 108 64 39 Prolem 9. 74 5 ind the ngle supplementry to 58 If two ngles dd up to 180 they re clled supplementry ngles. Hence, the ngle supplementry to 74 5 is 180 74 5 = 105 35 Prolem 10. vlute ngle θ in ech of the digrms shown in igure 0.3 igure 0.5 There re 360 in complete revolution of circle. Thus, 360 = 58 + 108 + 64 + 39 + θ from which, θ = 360 58 108 64 39 = 91 418 538 448 Prolem 1. Two stright lines nd intersect t 0. If O is 43,find O, O nd O igure 0.3 The symol shown in igure 0.4 is clled right ngle nd it equls 90. rom igure 0.6, O is supplementry to O. Hence, O = 180 43 = 137. When two stright lines intersect, the verticlly opposite ngles re equl. Hence, O = 43 nd O 137 438 0 igure 0.4 Hence, from igure 0.3, θ + 41 = 90 from which, θ = 90 41 = 49 igure 0.6
ngles nd tringles 169 Prolem 13. etermine ngle β in igure 0.7 igure 0.7 1338 α = 180 133 = 47 (i.e. supplementry ngles). α = β = 47 (corresponding ngles etween prllel lines). Prolem 14. igure 0.8 igure 0.8 etermine the vlue of ngle θ in 337 3549 Let stright line G e drwn through such tht G is prllel to nd. = (lternte ngles etween prllel lines nd G), hence = 3 37. = (lterntengles etween prllel lines G nd ), hence = 35 49. ngle θ = + = 3 37 + 35 49 = 59 6 G = = 46 (corresponding ngles etween prllel lines). lso, + c + 90 = 180 (ngles on stright line). Hence, 46 + c + 90 = 180, from which, c = 44. nd d re supplementry, hence d = 180 46 = 134. lterntively, 90 + c = d (verticlly opposite ngles). Prolem 16. onvert the following ngles to rdins, correct to 3 deciml plces. 73 5 37 lthough we my e more fmilir with degrees, rdins is the SI unit of ngulr mesurement in engineering (1 rdin 57.3 ). Since 180 = π rdthen1 = π 180 rd. Hence, 73 = 73 π rd = 1.74 rd. 180 5 37 = 5 37 60 = 5.616666... Hence, 5 37 = 5.616666... = 5.616666... π 180 rd Prolem 17. minutes = 0.447 rd. onvert 0.743 rd to degrees nd Since 180 = π rdthen1rd= 180 π Hence, 0.743 rd = 0.743 180 π = 4.57076... = 4 34 Since π rd = 180,then π rd = 90, π 3 rd = 60 nd π rd = 30 6 Now try the following Prctice xercise π 4 rd = 45, Prolem 15. igure 0.9 igure 0.9 etermine ngles c nd d in d 468 c Prctice xercise 77 urther ngulr mesurement (nswers on pge 348) 1. Stte the generl nme given to n ngle of 197.. Stte the generl nme given to n ngle of 136. 3. Stte the generl nme given to n ngle of 49.
170 sic ngineering Mthemtics 4. Stte the generl nme given to n ngle of 90. 5. etermine the ngles complementry to the following. 69 7 37 (c) 41 3 43 6. etermine the ngles supplementry to 78 15 (c) 169 41 11 7. ind the vlues of ngle θ in digrms to (i) of igure 0.10. 108 708 8. With reference to igure 0.11, wht is the nme given to the line XY?Giveexmplesof ech of the following. x igure 0.11 1 4 3 5 6 7 8 verticlly opposite ngles supplementry ngles (c) corresponding ngles (d) lternte ngles 9. In igure 0.1, find ngle α. y 558 508 (c) (d) 378 1379 1649 igure 0.1 10. In igure 0.13, find ngles, nd c. (e) 708 1508 608 (f) 808 98 698 c 50.78 1118 578 688 458 igure 0.13 11. ind ngle β in igure 0.14. 133 (g) (h) 98 igure 0.14 igure 0.10 368 (i) 1. onvert 76 to rdins, correct to 3 deciml plces. 13. onvert 34 40 to rdins, correct to 3 deciml plces. 14. onvert 0.714 rd to degrees nd minutes.
ngles nd tringles 171 0.3 Tringles tringle is figure enclosed y three stright lines. The sum of the three ngles of tringle is equl to 180. 0.3.1 Types of tringle 308 438 n cute-ngled tringle is one in which ll the ngles re cute; i.e., ll the ngles re less thn 90.n exmple is shown in tringle in igure 0.15. right-ngled tringle is one which contins right ngle; i.e., one in which one of the ngles is 90.n exmple is shown in tringle in igure 0.15. 548 508 758 igure 0.17 758 G 88 558 678 598 igure 0.15 408 n otuse-ngled tringle is one which contins n otuse ngle; i.e., one ngle which lies etween 90 nd 180. n exmple is shown in tringle PQR in igure 0.16. n equilterl tringle is one in which ll the sides nd ll the ngles re equl; i.e., ech is 60. n exmple is shown in tringle in igure 0.16. P 8 608 igure 0.18 c With reference to igure 0.18, ngles, nd re clled interior ngles of the tringle. ngle θ is clled n exterior ngle of the tringle nd is equl to the sum of the two opposite interior ngles; i.e., θ = +. (c) + + c is clled the perimeter of the tringle. 1318 Q 78 R 608 608 igure 0.16 c n isosceles tringle is one in which two ngles nd two sides re equl. n exmple is shown in tringle G in igure 0.17. sclene tringle is one with unequl ngles nd therefore unequl sides. n exmple of n cute ngled sclene tringle is shown in tringle in igure 0.17. igure 0.19
17 sic ngineering Mthemtics right-ngled tringle is shown in igure 0.19. The point of intersection of two lines is clled vertex (plurl vertices); the three vertices of the tringle re lelled s, nd, respectively. The right ngle is ngle. The side opposite the right ngle is given the specil nme of the hypotenuse. The hypotenuse, length in igure 0.19, is lwys the longest side of right-ngled tringle. With reference to ngle, isthe opposite side nd isclled the djcent side. With reference to ngle, is the opposite side nd is the djcent side. Often sides of tringle re lelled with lower cse letters, eing the side opposite ngle, eing the side opposite ngle nd c eing the side opposite ngle. So, in the tringle, length = c, length = nd length =. Thus, c is the hypotenuse in the tringle. is the symol used for ngle. or exmple, in the tringle shown, = 90. nother wy of indicting n ngle is to use ll three letters. or exmple, ctully mens ; i.e., we tke the middle letter s the ngle. Similrly, mens nd mens. Here re some worked exmples to help us understnd more out tringles. (d) (e) Otuse-ngled sclene tringle (since one of the ngles lies etween 90 nd 180 ). Isosceles tringle (since two sides re equl). Prolem 19. In the tringle shown in igure 0.1, with reference to ngle θ, which side is the djcent? igure 0.1 The tringle is right-ngled; thus, side is the hypotenuse. With reference to ngle θ, the opposite side is. The remining side,, is the djcent side. Prolem 18. igure 0.0 Nme the types of tringle shown in Prolem 0. In the tringle shown in igure 0., determine ngle θ.6.8.1 398 (c) 518 568 1078 (d).5.5 (e).1 igure 0. (c) igure 0.0 quilterl tringle (since ll three sides re equl). cute-ngled sclene tringle (since ll the ngles re less thn 90 ). Right-ngled tringle (39 + 51 = 90 ; hence, the third ngle must e 90, since there re 180 in tringle). The sum of the three ngles of tringle is equl to 180. The tringle is right-ngled. Hence, 90 + 56 + θ = 180 from which, θ = 180 90 56 = 34. Prolem 1. igure 0.3 etermine the vlue of θ nd α in
ngles nd tringles 173 6 igure 0.3 15 In tringle, + + = 180 (the ngles in tringle dd up to 180 ). Hence, = 180 90 6 = 8. Thus, = 8 (verticlly opposite ngles). θ = + (the exterior ngle of tringle is equl to the sum of the two opposite interior ngles). Hence, θ = 8 + 15 = 43. α nd re supplementry; thus, α = 180 15 = 165. Prolem. is n isosceles tringle in which the unequl ngle is 56. is extended to s shown in igure 0.4. ind, for the tringle, nd. lso, clculte 568 Prolem 3. igure 0.5 igure 0.5 ind ngles,, c, d nd e in e 558 68 d c = 6 nd c = 55 (lternte ngles etween prllel lines). 55 + + 6 = 180 (ngles in tringle dd up to 180 ); hence, = 180 55 6 = 63. = d = 63 (lternte ngles etween prllel lines). e + 55 + 63 = 180 (ngles in tringle dd up to 180 ); hence, e = 180 55 63 = 6. heck: e = = 6 (corresponding ngles etween prllel lines). Now try the following Prctice xercise Prctice xercise 78 pge 348) Tringles (nswers on 1. Nme the types of tringle shown in digrms to (f) in igure 0.6. 48 66 81 15 igure 0.4 39 Since tringle is isosceles, two sides i.e. nd re equl nd two ngles i.e. nd re equl. The sum of the three ngles of tringle is equl to 180. Hence, + = 180 56 = 14. Since = then = = 14 = 6. n ngle of 180 lies on stright line; hence, + = 180 from which, = 180 = 180 6 = 118. lterntively, = + (exterior ngle equls sum of two interior opposite ngles), i.e. = 56 + 6 = 118. 45 (c) 60 5 (e) igure 0.6 45 5 97 (d) 53 37 (f)
174 sic ngineering Mthemtics. ind the ngles to f in igure 0.7. O 658 38 578 N 838 c 114 M 1058 P d igure 0.9 f 6. etermine φ nd x in igure 0.30. e 1058 (c) 588 igure 0.7 3. In the tringle of igure 0.8, which side is the hypotenuse? With reference to ngle, which side is the djcent? x 198 igure 0.30 7. In igure 0.31 nd, find ngles w,x, y nd z. Wht is the nme given to the types of tringle shown in nd? igure 0.8 388 1108 x y 1108 cm z 708 cm 4. In tringle of igure 0.8, determine ngle. 5. MNO is n isosceles tringle in which the unequl ngle is 65 s shown in igure 0.9. lculte ngle θ. igure 0.31 8. ind the vlues of ngles to g in igure 0.3 nd.
ngles nd tringles 175 56899 148419 688 c d f 1318 igure 0.3 9. ind the unknown ngles to k in igure 0.33. e g (c) (d) the three sides of one re equl to the three sides of the other (SSS), two sides of one re equl to two sides of the other nd the ngles included y these sides re equl (SS), two ngles of the one re equl to two ngles of the other nd ny side of the first is equl to the corresponding side of the other (S), or their hypotenuses re equl nd one other side of one is equl to the corresponding side of the other (RHS). f Prolem 4. Stte which of the pirs of tringles shown in igure 0.35 re congruent nd nme their sequence 8 igure 0.33 158 e g h d j i c k 998 H G I J K L 10. Tringle hs right ngle t nd is 34. is produced to. Ifthe isectors of nd meet t, determine. 11. If in igure 0.34 tringle is equilterl, find the interior ngles of tringle. M N P (c) O Q R S V X W (d) T U (e) igure 0.35 978 ongruent, (ngle, side, ngle; i.e., S). igure 0.34 (c) ongruent GIH, JLK (side, ngle, side; i.e., SS). ongruent MNO, RQP (right ngle, hypotenuse, side; i.e., RHS). 0.4 ongruent tringles Two tringles re sid to e congruent if they re equl in ll respects; i.e., three ngles nd three sides in one tringle re equl to three ngles nd three sides in the other tringle. Two tringles re congruent if (d) (e) Not necessrily congruent. It is not indicted tht ny side coincides. ongruent, (side, side, side; i.e., SSS).
176 sic ngineering Mthemtics Prolem 5. In igure 0.36, tringle PQR is isosceles with Z, the mid-point of PQ. Prove tht tringles PXZ nd QYZ re congruent nd tht tringles RXZ nd RYZ re congruent. etermine the vlues of ngles RPZ nd RXZ X 678 R Y L G K I H O N J P (c) V M P igure 0.36 88 Z 88 Q Q R S U X W Y Since tringle PQR is isosceles, PR = RQ nd thus QPR = RQP. RXZ = QPR + 8 nd RYZ = RQP + 8 (exterior ngles of tringle equl the sum of the two interior opposite ngles). Hence, RXZ = RYZ. PXZ = 180 RXZ nd QYZ = 180 RYZ. Thus, PXZ = QYZ. Tringles PXZ nd QYZ re congruent since XPZ = YQZ, PZ = ZQ nd XZP = YZQ (S). Hence, XZ = YZ. Tringles PRZ nd QRZ re congruent since PR = RQ, RPZ = RQZ nd PZ = ZQ (SS). Hence, RZX = RZY. Tringles RXZ nd RYZ re congruent since RXZ = RYZ, XZ = YZ nd RZX = RZY (S). QRZ = 67 nd thus PRQ = 67 + 67 = 134. Hence, RPZ = RQZ = 180 134 = 3. RXZ = 3 + 8 = 51 (externl ngle of tringle equls the sum of the two interior opposite ngles). Now try the following Prctice xercise Prctice xercise 79 ongruent tringles (nswers on pge 349) 1. Stte which of the pirs of tringles in igure 0.37 re congruent nd nme their sequences. T (d) igure 0.37. In tringle, = nd nd re points on nd, respectively, such tht =. Show tht tringles nd re congruent. 0.5 Similr tringles Two tringles re sid to e similr if the ngles of one tringle re equl to the ngles of the other tringle. With reference to igure 0.38, tringles nd PQR re similr nd the corresponding sides re in proportion to ech other, i.e. igure 0.38 c Prolem 6. side 578 658 588 p = q = r c r (e) P 578 Z q 658 588 Q p R In igure 0.39, find the length of
ngles nd tringles 177 c 5 1.0 cm igure 0.39 708 508 f 5 5.0 cm 508 608 d 5 4.4 cm In tringle, 50 + 70 + = 180, from which = 60. In tringle, = 180 50 60 = 70. Hence, tringles nd re similr, since their ngles re the sme. Since corresponding sides re in proportion to ech other, d = c f i.e. 4.4 = 1.0 5.0 Hence, side, = 1.0 (4.4) = 10.61cm. 5.0 Prolem 7. In igure 0.40, find the dimensions mrked r nd p P q 5 6.8 cm 358 R igure 0.40 r p Q X z 5 1.97 cm Y 558 y 5 10.63 cm x 5 7.44 cm In tringle PQR, Q = 180 90 35 = 55. In tringle XYZ, X = 180 90 55 = 35. Hence, tringles PQR nd ZYX re similr since their ngles re the sme. The tringles my he redrwn s shown in igure 0.41. 558 r P Q p 358 q 5 6.8 cm R x 5 7.44 cm Y Z 558 z 5 1.97 cm Z 358 y 5 10.63 cm X y proportion: p z = r x = q y i.e. p 1.97 = r 7.44 = 6.8 10.63 from which, ( ) 6.8 r = 7.44 = 4.77cm 10.63 y proportion: p z = q y i.e. p 1.97 = 6.8 10.63 ( ) 6.8 Hence, p = 1.97 = 8.3cm 10.63 Prolem 8. In igure 0.4, show tht tringles nd re similr nd hence find the length of nd igure 0.4 10 6 1 Since is prllel to then = nd = (corresponding ngles etween prllel lines). lso, is common to tringles nd. Since the ngles in tringle re the sme s in tringle, the tringles re similr. Hence, y proportion: i.e. lso, = ( 9 = 9 6 + 9 =, from which 1 ( ) 9 = 1 = 7.cm 15 ) 9 15 =, from which 10 ( ) 9 = 10 = 6cm 15 igure 0.41
178 sic ngineering Mthemtics Prolem 9. rectngulr shed m wide nd 3m high stnds ginst perpendiculr uilding of height 5.5 m. ldder is used to gin ccess to the roof of the uilding. etermine the minimum distnce etween the ottom of the ldder nd the shed 14.58 mm 111 5.69 mm side view is shown in igure 0.43, where is the minimum length of the ldder. Since nd re prllel, = (corresponding ngles etween prllel lines). Hence, tringles nd re similr since their ngles re the sme. = = = 5.5 3 =.5m y proportion: = i.e..5 3 = ( ) 3 Hence, = =.4m = minimum distnce.5 from ottom of ldder to the shed. x 3 igure 0.44 4.74 mm y 3 37 7.36 mm. PQR is n equilterl tringle of side 4cm. When PQ nd PR re produced to S nd T, respectively, ST is found to e prllel with QR. IfPS is 9cm, find the length of ST. X is point on ST etween S nd T such tht the line PX is the isector of SPT.indthe length of PX. 3. In igure 0.45, find the length of when = 6cm, = 8cm nd = 3cm, the length of when = cm, = 5cm nd = 10cm. m 5.5 m 3m Shed igure 0.45 igure 0.43 4. In igure 0.46, = 8m, = 5m nd = 3m. ind the length of. Now try the following Prctice xercise Prctice xercise 80 Similr tringles (nswers on pge 349) 1. In igure 0.44, find the lengths x nd y. igure 0.46
hpter 5 res of common shpes 5.1 Introduction re is mesure of the size or extent of plne surfce. re is mesured in squre units such s mm,cm nd m. This chpter dels with finding the res of common shpes. In engineering it is often importnt to e le to clculte simple res of vrious shpes. In everydy life its importnt to e le to mesure re to, sy, ly crpet, order sufficient pint for decorting jo or order sufficient ricks for new wll. On completing this chpter you will e le to recognize common shpes nd e le to find the res of rectngles, squres, prllelogrms, tringles, trpeziums nd circles. 5. ommon shpes 5..1 Polygons polygon is closed plne figure ounded y stright lines. polygon which hs 3 sides is clled tringle see igure 5.1 4 sides is clled qudrilterl see igure 5.1 5 sides is clled pentgon see igure 5.1(c) 6 sides is clled hexgon see igure 5.1(d) 7 sides is clled heptgon see igure 5.1(e) 8 sides is clled n octgon see igure 5.1(f) 5.. Qudrilterls There re five types of qudrilterl, these eing rectngle, squre, prllelogrm, rhomus nd trpezium. If the opposite corners of ny qudrilterl re joined y stright line, two tringles re produced. Since the sum of the ngles of tringle is 180, the sum of the ngles of qudrilterl is 360. Rectngle In the rectngle shown in igure 5., ll four ngles re right ngles, the opposite sides re prllel nd equl in length, nd (c) digonls nd re equl in length nd isect one nother. Squre In the squre PQRS shown in igure 5.3, ll four ngles re right ngles, the opposite sides re prllel, (c) ll four sides re equl in length, nd (d) digonls PR nd QS re equl in length nd isect one nother t right ngles. Prllelogrm In the prllelogrm WXYZ shown in igure 5.4, opposite ngles re equl, opposite sides re prllel nd equl in length, nd (c) digonls WY nd XZ isect one nother. Rhomus In the rhomus shown in igure 5.5, opposite ngles re equl, opposite ngles re isected y digonl, (c) opposite sides re prllel, (d) ll four sides re equl in length, nd (e) digonls nd isect one nother t right ngles. OI: 10.1016/978-1-85617-697-.0005-9
0 sic ngineering Mthemtics (c) (d) (e) (f) igure 5.1 P Q Prolem 1. Stte the types of qudrilterl shown in igure 5.7 nd determine the ngles mrked to l igure 5. W S igure 5.3 X R d x x J 408 308 x c H G (c) M e f L x K 1158 S Z igure 5.4 Y N O g h i 658 Q j (d) 58 P R l 358 U k 758 (e) T igure 5.7 igure 5.5 H igure 5.6 G is squre The digonls of squre isect ech of the right ngles, hence Trpezium In the trpezium GH shown in igure 5.6, only one pir of sides is prllel. GH is rectngle = 90 = 45
res of common shpes 1 (c) (d) (e) In tringle GH, 40 + 90 + = 180, since the ngles in tringle dd up to 180, from which = 50.lso,c = 40 (lternte ngles etween prllel lines nd HG). (lterntively, nd c re complementry; i.e., dd up to 90.) d = 90 + c (externl ngle of tringle equls the sum of the interior opposite ngles), hence d = 90 + 40 = 130 (or H = 50 nd d = 180 50 = 130 ). JKLM is rhomus The digonls of rhomus isect the interior ngles nd the opposite internl ngles re equl. Thus, JKM = MKL = JMK = LMK = 30, hence, e = 30. In tringle KLM, 30 + KLM + 30 = 180 (the ngles in tringle dd up to 180 ), hence, KLM = 10. The digonl JL isects KLM, hence, f = 10 = 60. NOPQ is prllelogrm g = 5 since the opposite interior ngles of prllelogrm re equl. In tringle NOQ, g + h + 65 = 180 (the ngles in tringle dd up to 180 ), from which h = 180 65 5 = 63. i = 65 (lternte ngles etween prllel lines NQ nd OP). j = 5 + i = 5 + 65 = 117 (the externl ngle of tringle equls the sum of the interior opposite ngles). (lterntively, PQO = h = 63 ; hence, j = 180 63 = 117.) RSTU is trpezium 35 + k = 75 (externl ngle of tringle equls the sum of the interior opposite ngles), hence, k = 40. STR = 35 (lternte ngles etween prllel lines RU nd ST). l + 35 = 115 (externl ngle of tringle equls the sum of the interior opposite ngles), hence, l = 115 35 = 80. Now try the following Prctice xercise 408 p q 758 478 s 388 r 158 68 958 (c) igure 5.8 5.3 res of common shpes t 578 The formule for the res of common shpes re shown in Tle 5.1. Here re some worked prolems to demonstrte how the formule re used to determine the re of common shpes. Prolem. lculte the re nd length of the perimeter of the squre shown in igure 5.9 igure 5.9 4.0 cm 4.0 cm re of squre = x = (4.0) = 4.0cm 4.0cm = 16.0cm (Note the unit of re is cm cm = cm ; i.e., squre centimetres or centimetres squred.) Perimeter of squre = 4.0cm+ 4.0cm+ 4.0cm + 4.0cm = 16.0cm Prolem 3. lculte the re nd length of the perimeter of the rectngle shown in igure 5.10 7.0 cm Prctice xercise 96 ommon shpes (nswers on pge 351) 1. ind the ngles p nd q in igure 5.8.. ind the ngles r nd s in igure 5.8. 3. ind the ngle t in igure 5.8(c). igure 5.10 4.5 cm
sic ngineering Mthemtics Tle 5.1 ormule for the res of common shpes re of plne figures Squre x re = x x Rectngle re = l I Prllelogrm h re = h Tringle h re = 1 h Trpezium re = 1 ( + )h h ircle r re = πr or πd 4 ircumference = πr Rdin mesure: π rdins = 360 degrees Sector of circle s re = θ 360 (πr ) r
res of common shpes 3 re of rectngle = l = 7.0 4.5 = 31.5cm Perimeter of rectngle = 7.0cm+ 4.5cm + 7.0cm+ 4.5cm = 3.0cm Prolem 4. lculte the re of the prllelogrm shown in igure 5.11 Prolem 5. lculte the re of the tringle shown in igure 5.13 J 1.9 cm I 5.68 cm 9mm igure 5.13 K re of tringle IJK = 1 se perpendiculr height H 16 mm G = 1 IJ JK igure 5.11 1 mm re of prllelogrm = se perpendiculr height The perpendiculr height h is not shown in igure 5.11 ut my e found using Pythgors theorem (see hpter 1). rom igure 5.1, 9 = 5 + h, from which h = 9 5 = 81 5 = 56. Hence, perpendiculr height, h = 56 = 7.48mm. To find JK, Pythgors theorem is used; i.e., 5.68 = 1.9 + JK, from which JK = 5.68 1.9 = 5.346cm Hence, re of tringle IJK = 1 1.9 5.346 = 5.13cm. Prolem 6. lculte the re of the trpezium shown in igure 5.14 igure 5.14 5.5 mm 7.4 mm 8.6 mm 9mm h re of trpezium = 1 (sum of prllel sides) H 16 mm 5 mm G igure 5.1 Hence, re of prllelogrm GH = 16mm 7.48mm = 10mm. Hence, re of trpezium LMNO (perpendiculr distnce etween the prllel sides) = 1 (7.4 + 8.6) 5.5 = 1 36 5.5 = 99mm
4 sic ngineering Mthemtics Prolem 7. rectngulr try is 80mm long nd 400mm wide. ind its re in mm cm (c) m re of try = length width = 80 400 = 38000mm (c) Since 1 cm = 10mm, 1cm = 1cm 1cm = 10mm 10mm = 100mm, or 1mm = 1 100 cm = 0.01cm Hence, 38000 mm = 38000 0.01cm = 380cm. Since 1 m = 100cm, 1m = 1m 1m = 100cm 100cm = 10000cm, or 1cm = 1 10000 m = 0.0001m Hence, 380cm = 380 0.0001m = 0.380m. Prolem 8. The outside mesurements of picture frme re 100cm y 50cm. If the frme is 4cm wide, find the re of the wood used to mke the frme sketch of the frme is shown shded in igure 5.15. 100 cm 9 cm Prolem 9. ind the cross-sectionl re of the girder shown in igure 5.16 igure 5.16 8mm 5mm 50 mm 70 mm 6mm The girder my e divided into three seprte rectngles, s shown. re of rectngle = 50 5 = 50mm 75 mm re of rectngle = (75 8 5) 6 = 6 6 = 37mm re of rectngle = 70 8 = 560mm Totl re of girder = 50 + 37 + 560 = 118mm or 11.8cm Prolem 10. igure 5.17 shows the gle end of uilding. etermine the re of rickwork in the gle end 6m 5m 5m 50 cm 4 cm igure 5.17 8m The shpe is tht of rectngle nd tringle. igure 5.15 re of wood = re of lrge rectngle re of smll rectngle = (100 50) (9 4) = 5000 3864 = 1136cm re of rectngle = 6 8 = 48m re of tringle = 1 se height = 4m nd = 5 m, hence = 3m (since it is 3, 4, 5 tringle or y Pythgors). Hence, re of tringle = 1 8 3 = 1m.
res of common shpes 5 Totl re of rickwork = 48 + 1 = 60m. Now try the following Prctice xercise 7cm 30 mm Prctice xercise 97 res of common shpes (nswers on pge 351) 1. Nme the types of qudrilterl shown in igure 5.18(i) to (iv) nd determine for ech the re nd the perimeter. cm 1cm 1cm igure 5.19 cm 5 mm 8mm 10 mm 50 mm 6mm 4 cm 5.94 cm (i) 3.5 cm 6 mm 30 mm 38 mm (ii) 10 mm (iii) 30 mm 8. igure 5.0 shows 4m wide pth round the outside of 41m y 37m grden. lculte the re of the pth. 41 6 cm 4 37 10 cm igure 5.18 1 cm 16 cm (iv). rectngulr plte is 85mm long nd 4mm wide. ind its re in squre centimetres. 3. rectngulr field hs n re of 1.hectres nd length of 150 m. If 1 hectre = 10000m, find the field s width nd the length of digonl. 4. ind the re of tringle whose se is 8.5cm nd perpendiculr height is 6.4cm. 5. squre hs n re of 16cm. etermine the length of digonl. 6. rectngulr picture hs n re of 0.96m. If one of the sides hs length of 800mm, clculte, in millimetres, the length of the other side. 7. etermine the re of ech of the ngle iron sections shown in igure 5.19. 5 igure 5.0 9. The re of trpezium is 13.5cm nd the perpendiculr distnce etween its prllel sides is 3cm. If the length of one of the prllel sides is 5.6cm, find the length of the other prllel side. 10. lculte the re of the steel plte shown in igure 5.1. 60 igure 5.1 imensions in mm 140 5 5 100
6 sic ngineering Mthemtics 11. etermine the re of n equilterl tringle of side 10.0cm. 1. If pving sls re produced in 50mm y 50mm squres, determine the numer of slsrequiredtocovernreofm. Here re some further worked prolems on finding the res of common shpes. Prolem 11. ind the re of circle hving rdius of 5cm re of circle = πr = π(5) = 5π = 78.54cm If dimeter = 80mm then rdius, r = 40mm, nd re of sector = 107 4 107 4 360 (π40 ) = 60 360 (π40 ) = 107.7 360 (π40 ) = 1504mm or 15.04cm Prolem 16. hollow shft hs n outside dimeter of 5.45 cm nd n inside dimeter of.5 cm. lculte the cross-sectionl re of the shft The cross-sectionl re of the shft is shown y the shded prt in igure 5. (often clled n nnulus). Prolem 1. ind the re of circle hving dimeter of 15 mm re of circle = πd 4 = π(15) = 5π = 176.7mm 4 4 Prolem 13. ind the re of circle hving circumference of 70mm igure 5. d 5.5 cm 5 5.45 cm ircumference, c = πr, hence rdius,r = c π = 70 π = 35 π mm re of circle = πr = π ( 35 π ) = 35 π = 389.9mm or 3.899cm Prolem 14. lculte the re of the sector of circle hving rdius 6cm with ngle sutended t centre 50 re of sector = θ 360 (πr ) = 50 360 (π6 ) = 50 π 36 360 = 15.71cm Prolem 15. lculte the re of the sector of circle hving dimeter 80mm with ngle sutended t centre 107 4 re of shded prt = re of lrge circle re of smll circle = π 4 πd 4 = π 4 ( d ) = π 4 (5.45.5 ) = 19.35cm Now try the following Prctice xercise Prctice xercise 98 res of common shpes (nswers on pge 351) 1. rectngulr grden mesures 40m y 15m. 1m flower order is mde round the two shorter sides nd one long side. circulr swimming pool of dimeter 8m is constructed
res of common shpes 7 in the middle of the grden. ind, correct to the nerest squre metre, the re remining.. etermine the re of circles hving rdius of 4cm dimeter of 30mm (c) circumference of 00mm. 3. n nnulus hs n outside dimeter of 60mm nd n inside dimeter of 0mm. etermine its re. 4. If the re of circle is 30mm, find its dimeter nd its circumference. 5. lculte the res of the following sectors of circles. rdius 9cm, ngle sutended t centre 75. dimeter 35mm, ngle sutended t centre 48 37. 6. etermine the shded re of the templte shown in igure 5.3. 10 mm igure 5.3 90 mm 80 mm rdius 7. n rchwy consists of rectngulr opening topped y semi-circulr rch, s shown in igure 5.4. etermine the re of the opening if the width is 1m nd the gretest height is m. Prolem 17. lculte the re of regulr octgon if ech side is 5 cm nd the width cross the flts is 1cm n octgon is n 8-sided polygon. If rdii re drwn from the centre of the polygon to the vertices then 8 equl tringles re produced, s shown in igure 5.5. igure 5.5 1 cm 5m re of one tringle = 1 se height = 1 5 1 = 15cm re of octgon = 8 15 = 10cm Prolem 18. etermine the re of regulr hexgon which hs sides 8cm long hexgon is 6-sided polygon which my e divided into 6 equl tringles s shown in igure 5.6. The ngle sutended t the centre of ech tringle is 360 6 = 60. The other two ngles in the tringle dd up to 10 nd re equl to ech other. Hence, ech of the tringles is equilterl with ech ngle 60 nd ech side 8 cm. 4cm h 8cm m 608 igure 5.4 1m 8cm igure 5.6 re of one tringle = 1 se height Here re some further worked prolems of common shpes. = 1 8 h
15 m 8 sic ngineering Mthemtics h is clculted using Pythgors theorem: 8 = h + 4 from which h = 8 4 = 6.98cm Hence, re of one tringle = 1 8 6.98 = 7.71cm re of hexgon = 6 7.71 = 166.3cm Prolem 19. igure 5.7 shows pln of floor of uilding which is to e crpeted. lculte the re of the floor in squre metres. lculte the cost, correct to the nerest pound, of crpeting the floor with crpet costing 16.80perm, ssuming 30% extr crpet is required due to wstge in fitting m 0.6 m L 0.6 m H m 0.8 m igure 5.7 G K J I m 0.8 m.5 m 3m M 4m 3m 30 60 3m re of floor pln = re of tringle + re of semicircle + re of rectngle GLM + re of rectngle re of trpezium HIJK Tringle is equilterl since = = 3m nd, hence, ngle = 60. re of semicircle = 1 πr = 1 π(.5) = 9.817m re of GLM = 5 7 = 35m re of = 0.8 3 =.4m re of HIJK = 1 (KH + IJ)(0.8) Since M = 7m thenlg = 7 m, hence JI = 7 5. = 1.8 m. Hence, re of HIJK = 1 (3 + 1.8)(0.8) = 1.9m Totl floor re = 3.897 + 9.817 + 35 +.4 1.9 = 49.194m To llow for 30% wstge, mount of crpet required = 1.3 49.194 = 63.95m ost of crpet t 16.80 per m = 63.95 16.80 = 1074, correct to the nerest pound. Now try the following Prctice xercise Prctice xercise 99 res of common shpes (nswers on pge 351) 1. lculte the re of regulr octgon if ech side is 0mm nd the width cross the flts is 48.3mm.. etermine the re of regulr hexgon which hs sides 5mm. 3. plot of lnd is in the shpe shown in igure 5.8. etermine 0 m 0 m 0 m 30 m 0 m 10 m 0 m 30 m i.e. sin = /3 = 3sin60 =.598m. re of tringle = 1 ()( ) = 1 (3)(.598) = 3.897m 15 m 40 m igure 5.8 0 m
res of common shpes 9 its re in hectres (1 h = 10 4 m ). the length of fencing required, to the nerest metre, to completely enclose the plot of lnd. 5.4 res of similr shpes igure 5.9 shows two squres, one of which hs sides three times s long s the other. re of grge on the pln = 10mm 0mm = 00mm Since the res of similr shpes re proportionl to the squres of corresponding dimensions, True re of grge = 00 (50) = 1.5 10 6 mm = 1.5 106 10 6 m since 1 m = 10 6 mm = 1.5m x igure 5.9 x 3x 3x re of igure 5.9 = (x)(x) = x re of igure 5.9 = (3x)(3x) = 9x Hence, igure 5.9 hs n re (3) ; i.e., 9 times the re of igure 5.9. In summry, the res of similr shpes re proportionl to the squres of corresponding liner dimensions. Prolem 0. rectngulr grge is shown on uilding pln hving dimensions 10mm y 0mm. If the pln is drwn to scle of 1 to 50, determine the true re of the grge in squre metres Now try the following Prctice xercise Prctice xercise 100 res of similr shpes (nswers on pge 351) 1. The re of prk on mp is 500mm.If the scle of the mp is 1 to 40000, determine the true re of the prk in hectres (1 hectre = 10 4 m ).. model of oiler is mde hving n overll height of 75mm corresponding to n overll height of the ctul oiler of 6m. If the re of metl required for the model is 1500mm, determine, in squre metres, the re of metl required for the ctul oiler. 3. The scle of n Ordnnce Survey mp is 1:500. circulr sports field hs dimeter of 8 cm on the mp. lculte its re in hectres, giving your nswer correct to 3 significnt figures. (1 hectre = 10 4 m.)