TJP TOP TIPS FOR IGCSE SHAPE & SPACE

Transcription

1 TJP TOP TIPS FOR IGCSE SHPE & SPCE Dr T J Prie, 2011

2 IGCSE SHPE & SPCE First, some importnt words; know wht they men (get someone to test you): ute ngle: less thn 90. Right ngle: extly 90. Otuse ngle: etween 90 nd 180. Reflex ngle: greter thn 180. Prllel: two or more lines going in the sme diretion (with the sme grdient). Perpendiulr: two lines t 90 to eh other. Polygon: losed 2-D shpe with stright sides. Regulr Polygon: polygon with ll ngles equl nd ll sides equl. Equilterl tringle: three equl ngles nd three equl sides. Isoseles tringle: two equl ngles nd two equl sides. Right-ngled tringle: one ngle of 90. Slene tringle: no speil ngles or sides. Pge 2

3 IGCSE SHPE & SPCE Qudrilterl: four-sided 2-D shpe. Cyli qudrilterl: four-sided shpe whose orners lie on irle. Trpezium: shpe with one pir of prllel sides. Prllelogrm: shpe with two pirs of prllel sides. Rhomus: prllelogrm whih hs ll sides the sme length ( dimond). Kite: shpe with line of symmetry pssing through two opposite orners. Interior ngle: n ngle inside orner of polygon. Exterior ngle: the ngle the line turns through t eh orner (the 'turtle' ngle). ngle of elevtion: the ngle ove the horizontl. ngle of depression: the ngle elow the horizontl. Pge 3

4 NGLES IGCSE SHPE & SPCE ngles re mesured in degrees ( ) so tht there re 360 degrees in full irle. Why 360? Well, it's proly to do with the numer of dys in yer, omined with the ft tht lots of numers go into 360 extly (it hs mny ftors). Here re some more importnt ngle fts: ngles on stright line ngles in tringle (interior ngles) dd up to 180. dd up to = = 180 Exterior ngle in tringle is the ngles in qudrilterl dd up to 360. sum of the other two interior ngles. d + = d = 360 Exterior ngles in polygon Interior ngles in n n-sided polygon dd up to 360, lwys. dd up to 180(n 2). If you wlk round the shpe one, You n divide n n-sided polygon you turn through totl ngle of 360. into n 2 tringles, eh 'worth' 180. Exterior ngle of regulr polygon with n sides is 360 n. Interior ngle of regulr polygon with n sides is 180(n 2) n. Pge 4

5 IGCSE SHPE & SPCE lternte ngles re equl. Corresponding ngles re equl. HB HELPFUL HINT: This is like n nd nd joined together, spelling lternte ngle. TJP TOP TIP: These re 'F'-ing Corresponding ngles. F Vertilly opposite ngles re equl. v v TJP TOP TIP: This is V nd V (for Vertil), Opposite one nother. VV IGCSE INSIDER INFO: You must use the proper nmes: 'Z ngle', 'F ngle' nd 'X ngle' will sore zero mrks. Bering is simply n ngle mesured lokwise from North. Berings re often used in nvigtion where ompss is used to find the diretion of North. The ering of from B mens we re t B, going towrds. The ering of B from is 180 different from the ering of from B. N N B = Pge 5

6 CONSTRUCTION IGCSE SHPE & SPCE You re expeted to e le to drw nd mesure: stright line to the nerest mm n ngle to the nerest degree so ring shrp penil, shrpener, n erser, ruler nd protrtor. You will lso need pir of ompsses for the following methods: SKILL: Find the perpendiulr isetor of line segment B (without mesuring). Set your ompsses to little over hlf wy etween the line ends nd B. Drw rs with the ompss point on nd then on B. Drw stright line through the points where these rs ross one nother. B SKILL: Biset n ngle (without protrtor). Set your ompsses to fixed rdius, just shorter thn the lines mking the ngle. Put the ompss point on the ngle orner O nd drw n r utting eh line t, B. Then drw rs with the ompss point on nd B in turn to mke rhomus OCB. Drw line through OC; this isets the ngle. C O B Pge 6

7 IGCSE SHPE & SPCE SKILL: Construt tringles nd other 2-D shpes (ompsses nd strightedge). Equilterl Tringle: Given seline B, set your ompsses to this distne. Drw two rs entred on points nd B so tht the rs ross t point C. BC then forms n equilterl tringle. C Regulr Hexgon: Set your ompsses to the required side length. Drw irle entre O. Use the ompsses to mrk off the side length ll round the irumferene. Join the mrks to onstrut the hexgon BCDEF. B C B D O E F Squre: Given seline B, onstrut perpendiulr line t (see previous pge). Use your ompsses to mrk off the distne B up this line to give C. Using this sme rdius, now drw rs entred on C nd B, rossing t D. C D B Pge 7

8 IGCSE SHPE & SPCE PRTS OF CIRCLES Lern ll the following words for prts of irles... DIMETER RDIUS TNGENT Just touhes the irle CIRCUMFERENCE Rdius < Dimeter < Cirumferene, just like the length of the words... CHORD Think guitr strings... MINOR SEGMENT Think Terry's hoolte ornge... MJOR SEGMENT RC Like n rh... MINOR SECTOR MJOR SECTOR Pge 8

9 IGCSE SHPE & SPCE CIRCLE THEOREMS O O θ θ Rdius meets tngent t 90. Isoseles tringle (with two orners on the irumferene nd one t the entre). θ O θ θ 2θ ngle t the entre is twie the ngle t the irumferene. ngles in the sme segment re equl. O θ Φ ngle in semiirle = 90. Opposite ngles in yli qudrilterl dd up to 180. θ+φ = 180. Pge 9

10 IGCSE SHPE & SPCE θ C D B θ lternte segment theorem. Mrked ngles re equl. Interseting Chords (I). B = C D HB HELPFUL HINT: The lines mke shpe, so we the lengths together. B B D C C Interseting Chords (II). B = C D Interseting Chords (III). B = C² IGCSE INSIDER INFO: You hve to give reson for eh step when finding the unknown ngles in question. Use the offiil nmes ove, not 'Str Trek', et. Pge 10

11 OPPOSITE FORMULE FOR TRINGLES IGCSE SHPE & SPCE Whih formul do I use to find ngles or sides? Here's flow hrt to help... STRT Is it RT (Right-ngled Tringle)? NO YES Does the question involve only sides (no ngles)? NO Do we know n opposite pir ( side nd n ngle)? NO YES YES Use PYTHGORS Use SOHCHTO Use SINE RULE Use COSINE RULE ll these formule (exept for one) re given inside the front over of your IGCSE pper. NTOMY OF RT Lern these speil nmes for the sides of right-ngled tringle. HYPOTENUSE θ DJCENT The Hypotenuse is lwys opposite the right ngle (Pythgors nd SOHCHTO) nd it is the longest side. The Opposite is opposite the given ngle (SOHCHTO). The djent is next to (djent to) the given ngle (SOHCHTO). Pge 11

12 PYTHGORS IGCSE SHPE & SPCE If we know two of the three sides of RT (Right-ngled Tringle), we n find the missing side using Pythgors' Theorem. SKILL: Find n unknown side of RT using Pythgors. Use 2 2 = 2 where 2 is the hypotenuse. TJP TOP TIP: Go hed nd relel the tringle if you like, to get in the right ple on the hypotenuse. Just rememer to hnge k to the orret letter t the end. Q: Find the length of side x in this tringle. 5 m x m : Use 2 2 = 2 sustituting to get = x = 169 = x 2 So x = 169 = 13 ; the side hs length 13 m. 12 m Q: Find the vlue of. 30 : = = = = 256 So = 256 = Q: Find the height of n equilterl tringle of side 5 m. : Hint: split it into two RTs! h 2 = h 2 = 25 h 2 = = So h = = The height is 4.33 m (3 sig figs). 5 m 5 m h m 2.5 m 2.5 m Pge 12

13 OPPOSITE IGCSE SHPE & SPCE TRIGONOMETRY (SOHCHTO) Trigonometry gives the onnetion etween the sides nd the ngles of RT. sin = Opp Hyp os = dj Hyp tn = Opp dj HYPOTENUSE θ DJCENT Lern these tringles (they spell SOHCHTO) O O S H C H T Deide whih SOHCHTO tringle to use y seeing whih side does not feture in the question (s either numer or n unknown letter). Then over up the letter you re trying to find; the remining two letters give the formul you need. For exmple, to find H in the SOH tringle, we get O S. Two letters on the ottom line re multiplied; letter on top is divided y one underneth. SKILL: Find n unknown side of RT using SOHCHTO. Q: Find the length x. x : The djent side does not feture in this question, so use SOH (no dj). We re finding the Opposite side, so from the SOH tringle, Opp = Sin Hyp. x = sin = 4.99 (3 sig figs). Q: Find the length y. y 31 8 : The Hypotenuse does not feture in this question, so use TO (no Hyp). We re finding the djent side, so from the TO tringle, dj = Opp Tn y = 8 tn 31 = 13.3 (3 sig figs). Pge 13

14 IGCSE SHPE & SPCE SKILL: Find n unknown ngle of RT using SOHCHTO. TJP TOP TIP: lwys use the SHIFT utton to find n ngle. This gives the inverse sin, os or tn, written s sin 1, os 1, tn 1. If you get error on your lultor, you hve got your numers the wrong wy round. Q: Find the ngle θ in this tringle. 57 : The Opposite does not feture in this question, so use CH (no Opp). We re finding Cos, so from the CH tringle, Cos = dj Hyp = os = 30.7 (1 dp). θ 49 THREE-DIMENSIONL PROBLEMS We my get 3-D question whih we hve to rek down into 2-D RTs in order to find mystery length or ngle. SKILL: Use Pythgors nd Trigonometry to solve 3-D prolem. Q: Find the long digonl D of this uoid, nd lso ngle DC. D B C : Find digonl C first, using this tringle: C 2 = = 436 C 6 Then find D using this tringle: D 2 = = 485 D = 485 = 22.0 (3 sig figs) 436 C To find DC (the ngle t ), we ould use SOH, CH or TO; we know ll the sides. Here we'll use TO. DC = tn = 18.5 (1 dp). 20 B D 7 Pge 14

15 SINE RULE IGCSE SHPE & SPCE If we know the vlues of n opposite pir ( side nd the ngle opposite it) in non right-ngled tringle (non-rt), we use the Sine Rule. TJP TOP TIP: OppoSite pir Sine Rule. sin = sin B = sin C IGCSE INSIDER INFO: You n flip the printed formul upside-down to get the unknown quntity on the top line this mkes life muh esier. Note: little letters re sides, CPITL letters re ngles, nd is opposite, opposite B, et. B C SKILL: Use the Sine Rule to find n unknown side. Q: Find the vlue of side in this tringle : First, lel the sides nd ngles with, nd B (we don't need nd C here). 17 sin 67 = sin sin 35 = sin 67 = 10.6 (3 sig figs) B=35 =17 =67 SKILL: Use the Sine Rule to find n unknown ngle. Q: Find the vlue of ngle C in this tringle C : Lel the other sides nd ngles involved in the question. Then use the flipped version of the formul to get the ngle on the top. [Note: we ould hve used nd B insted of nd.] sin 19 = sin C 8 13 sin = sin C 8 sin C = C = sin = 31.9 (1 dp). =19 =13 C =8 Pge 15

16 COSINE RULE IGCSE SHPE & SPCE If we don't know the vlues of n opposite pir ( side nd the ngle opposite it) in non right-ngled tringle (non-rt), we use the Cosine Rule. TJP TOP TIP: No Sine Rule Cosine Rule. 2 = os Be reful with BIDMS when working out this formul. Don't press '=' until the end... nd rememer to squre root! B C SKILL: Use the Cosine Rule to find n unknown side. TJP TOP TIP: Relel the tringle so tht the side you re finding is lled. 'Cos tht's wht is on the left hnd side of the formul... Q: Find side y. 2.6 y : Relel the tringle with = y, nd put ngle opposite it. It doesn't mtter where you put nd, y the wy = os35 = os35 2 = = y = 4.33 (3 sig figs). = = =6.2 =y SKILL: Use the Cosine Rule to find n unknown ngle. IGCSE INSIDER INFO: LERN THIS FORMUL! os = It's not given in the front of the exm pper. 2 Q: Find ngle. : os = os = = =4 =13 =12 = os = Pge 16

17 RE OF TRINGLE IGCSE SHPE & SPCE The re of tringle is ½ se height, ut if we hve non right-ngled tringle the height my not e ovious. Fortuntely, we re given formul to work out the re. re = = 1 2 sin C TJP TOP TIP: Relel the tringle so tht the ngle is lled C, sndwihed etween sides nd. SKILL: Find the re of non right-ngled tringle. Q: Clulte the re of this tringle. 66 m m 59 : sneky question; we first hve to fill in the missing ngle t the top, whih is sndwihed etween the two given sides. This ngle is = 88. = sin 88 2 = So the re is 1350 m² (3 sig fig). 33 =66 m C=88 59 =41 m Q: If this tringle hs n re of 132 m², find the vlue of x. =x C=42 =2x : Sustitute into the re formul: = 1 x 2 x sin 42 = x 2 sin 42 = 132 x 2 = 132 sin 42 = x = 14.0 m (3 sig figs). Pge 17

18 PERIMETER ND RE IGCSE SHPE & SPCE The perimeter of shpe is the distne ll the wy round it (imgine wlking round it). irle's perimeter is lled the irumferene; it gets its own speil nme. Tringle Perimeter = sum of the sides re = ½ se vertil height Retngle Perimeter = 2 se + 2 height re = se height Prllelogrm Perimeter = 2 se + 2 slope height re = se vertil height Trpezium Perimeter = sum of the sides re = men width vertil height Cirle Cirumferene = 2 π r re = π r² TJP TOP TIP: re is squrier! SKILL: Find the perimeter of setor of irle. Hint: length of r = frtion of irle irumferene Perimeter = rdius + rdius + r = (130/360) 2 π 5 = = 21.3 m (3 sig figs) m SKILL: Find the re of setor of irle. Hint: re of setor = frtion of irle re of irle re = (130/360) π 5² = = 28.4 m² (3 sig figs) m Pge 18

19 SURFCE RE ND VOLUME IGCSE SHPE & SPCE Fe flt side of 3-D shpe. Edge line joining two orners of 3-D shpe. Vertex orner of 3-D shpe. The surfe re is the totl re of the outside of 3-D shpe. If you re sked for the urved surfe re, just leve out ny flt fes. [The formule you re given inlude only the urved surfe re for ones/ylinders.] Cuoid Surfe re = 2 Volume = Prism Surfe re = sum of retngles + 2 ends Volume = re of end length Cylinder Surfe re = 2 r h2 r 2 Volume = r 2 h Note: h my e the length if the ylinder is lying on its side. r h Cone Surfe re = r l r 2 Volume = 1 3 r 2 h Note: l, h, r oey l 2 = h 2 r 2 where l is the slnt height nd h is the vertil height. h r l Sphere Surfe re = 4 r 2 Volume = 4 3 r3 r To find the volume or urved surfe re of hemisphere ( hlf-sphere), simply work it out for whole sphere nd then hlve your nswer. Pge 19

20 SIMILRITY ND ENLRGEMENT IGCSE SHPE & SPCE Shpes re similr if one is n enlrgement of the other (so tht ll its lengths re multiplied y the sme mount). The sle ftor = new length old length. Shpes re ongruent if they re the sme shpe nd size (so tht you ould fit one on top of the other). If we enlrge shpe on speil 3-D photoopier, wht hppens to the lengths, the res nd the volumes? Here is lue: set the enlrgement to doule ll lengths, nd then see wht hppens. The re of eh fe is multiplied y 4 = 2² (re is squrier) The volume of the ue is multiplied y 8 = 2³ (volume is 3-D) But the ngles in the orners sty t 90 (don't hnge the ngles!) In generl use LV to work out sle ftors. L: N (This stnds for Length re Volume) : N² lwys strt y finding the sle ftor N. V: N³ SKILL: Solve prolem involving similr shpes. Q: Two shpes re similr; find the missing lengths nd m m 2 m 4 m 10 m : Use two mthing numers to get the sle ftor = 10 4 = 2.5. Now use it to find = = 5 m nd = = 5 m. m Q: ylinder is enlrged y sle ftor 3. Find the new length, re nd volume. L = 7m = 40m² V = 100m³ : New L = 7 3 = 21 m New = 40 3² = 40 9 = 360 m² New V = 100 3³ = = 2700 m³ Pge 20

21 METRIC UNITS IGCSE SHPE & SPCE You need to know ll these ommon units of length nd volume. 10 mm = 1 m 1000 ml = 1 litre 1000 mm = 1 metre 1 ml = 1 m³ 100 m = 1 metre 1 litre = 1000 m³ 1000 m = 1 km milli = 1 thousndth enti = 1 hundredth kilo = one thousnd SKILL: Convert etween re/volume mesures. Q: Convert 3 m² to m². : Eh squre metre hs = 100² squre entimetres in it (not just 100). 100 squres So the nswer is 3 100² = 30,000 m². 100 squres Q: How mny ui millimetres re there in one ui kilometre? : long one edge of this gint ue, there re = 10 6 millimetres in kilometre. 1 million So there re = mm³ in one km³. 1 million 1 million Q: How mny litres of fruit juie would it tke to fill swimming pool 10m y 50m y 3m? : The pool hs volume of = 1500 m³. This onverts to ³ = 1,500,000,000 m³ Sine 1 litre = 1000 m³, the pool will hold 1,500,000 litres of juie. Pge 21

22 VECTORS IGCSE SHPE & SPCE slr hs size (n ordinry numer), e.g. time, mss, speed, distne. vetor hs size nd diretion, e.g. fore, weight, veloity, displement. The size (or length) of vetor is lled its mgnitude or modulus. vetor quntity is shown y old letter (in print), or underlined if hndwritten. The sme letter '' written normlly mens the modulus or length of the vetor. If we hve vetor going from to B, we n write this s B. We n show the diretion of vetor in severl different wys: y drwing digrm: y giving n ngle or ering: 10km on ering of 327. y giving the x nd y mounts in rkets ( olumn vetor): = 8 6. If you multiply vetor y slr (ordinry numer) you just mke the vetor longer or shorter. For exmple, douling vetor mkes it twie s long (in the sme diretion). The resultnt is the result of dding or sutrting two or more vetors. Use Pythgors to find the modulus of olumn vetor. SKILL: Solve prolems involving olumn vetors. Q: If = 2 5 nd = 7 7, find (i) 3 (ii) (iii) the modulus of. : (i) 3 = = = 15 6 (ii) = (iii) modulus of = = 29 = 5.39 (3 sig figs) SKILL: Solve geometril vetor questions. Q: Use the grid elow to write vetor expressions for (i) C (ii) HG (iii) KF (iv) DI Hint: you n only move in the nd diretions, nd kwrds is negtive. B C D E F G H I J K L : (i) C = 2 (ii) HG = (iii) KF = (iv) DI = 3 2 Pge 22

23 IGCSE SHPE & SPCE TJP TOP TIP: Two vetors re prllel if one is multiple of the other. SKILL: Use vetors to prove geometril results. Q: If M is the midpoint of B nd N is the midpoint of C, find MN nd BC nd hene stte two fts out these vetors. N M B C : MN = nd BC = 2 2. Therefore BC = 2MN, so: (i) BC is twie s long s MN, (ii) BC is prllel to MN. Pge 23

24 IGCSE SHPE & SPCE CONTENTS Pge Topi 2-3 ngle nd Polygon Words 4-5 ngle nd Polygon Fts 6-7 Constrution 8 Prts of Cirles 9-10 Cirle Theorems 11 Flowhrt for Tringle Formule 12 Pythgors Trigonometry (SOHCHTO) 14 3-D Prolems 15 Sine Rule 16 Cosine Rule 17 re of Tringle 18 Perimeter nd re 19 Surfe re nd Volume 20 Similrity nd Enlrgement 21 Metri Units Vetors Pge 24