11.1 Kick off with CAS Clculting res with CAS We cn use CAS to define formuls which llow us to quickly nd efficiently clculte the res of different shp

FS O PA G E PR O 11 O R R EC TE D Geometry: similrity nd mensurtion 11.1 Kick off with CAS U N C 11.2 Properties of ngles, tringles nd polygons 11.3 Are nd perimeter I 11.4 Are nd perimeter II 11.5 Gret circles 11.6 Totl surfce re 11.7 Volume of prisms, pyrmids nd spheres 11.8 Similr figures 11.9 Similr tringles 11.10 Tringultion similrity 11.11 Are nd volume scle fctors 11.12 Time zones 11.13 Review c11geometrysimilrityandmensurtion.indd 534 12/09/15 5:16 PM

11.1 Kick off with CAS Clculting res with CAS We cn use CAS to define formuls which llow us to quickly nd efficiently clculte the res of different shpes. 1 Use CAS to define nd sve the formul for clculting the re of trpezium O O FS 1 (A = ( + ) h). 2 2 Use your formul to clculte the re of trpeziums with the following lengths nd heights. = 7 cm, = 15 cm, h = 11 cm = 7.8 cm, = 9.4 cm, h = 4 cm PR 3 Use your formul to clculte the height of trpeziums with the following res E nd lengths. A = 63 cm2, = 6 cm, = 12 cm A = 116.25 mm2, = 13 mm, = 18 mm (A = πr2). PA G 4 Use CAS to define nd sve the formul for clculting the re of circle 5 Use your formul to clculte the re of circles with the following rdii, correct TE D to 2 deciml plces. Rdius = 4 cm Rdius = 7.5 cm 6 Use your formul to clculte the rdii of circles with the following res, correct U N C O R R EC to 1 deciml plce. Are = 181.5 cm2 Are = 47.8 m2 Plese refer to the Resources t in the Prelims section of your ebookplus for comprehensive step-y-step guide on how to use your CAS technology. c11geometrysimilrityandmensurtion.indd 535 12/09/15 5:16 PM

11.2 WORKED EXAMPLE 1 THINK Properties of ngles, tringles nd polygons Geometry Geometry is n importnt re of study. Mny professions nd tsks require nd use geometricl concepts nd techniques. Besides rchitects, surveyors nd nvigtors, ll of us use it in our dily lives for exmple, to descrie shpes of ojects, directions on cr trip nd spce or position of house. Much of this re of study is ssumed knowledge gined from previous yers of study. In this module, we will often encounter prolems in which some of the informtion we need is not clerly given. To solve the prolems, some missing informtion will need to e deduced using the mny common mon rules, definitions nd lws of geometry. Some of the more importnt rules re presented in this topic. Interior ngles of polygons For regulr polygon (ll sides nd ngles re equl) of n sides, the interior ngle is given y 180 360 n. For exmple, for squre the interior ngle is: 180 360 = 180 90 4 = 90. ED= The exterior ngle of regulr polygon is given y 360 n. Clculte te the interior nd exterior ngle of the regulr polygon shown. 80 1 UNCORRE RRECTED RECT ED ngp 1 This shpe is regulr pentgon, 5-sided figure. Sustitute n = 5 into the interior ngle formul. WRITE Interior ngle = 180 360 5 = 180 72 = 108 PROO ROOFS OFS Interior ngle Exterior ngle 2 Sustitute n = 5 into the exterior ngle formul. Exterior ngle = 360 5 = 72 3 Write your nswer. A regulr pentgon hs n interior ngle of 108 nd n exterior ngle of 72. 536 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

Geometry rules, definitions nd nottion rules The following geometry rules nd nottion will e most vlule in estlishing unknown vlues in the topics covered nd revised in this module. Definitions of common terms A ABC Between 90 nd 180 B Less thn 90 C Acute ngle Between 180 nd 360 Reflex ngle 90 Right ngle Some common nottions nd rules + + c = 180 No equl sides c Sclene tringle + = 90 Complementry ngles = c = d c d Alternte ngles 180 Stright ngle Otuse ngle A B A Line B A Ry B Prllel lines Perpendiculr lines Line segment Two equl sides nd ngles Isosceles tringle + = 180 All equl 60 sides nd ngles Verticlly opposite Supplementry ngles ngles 45 45 60 60 Right-ngled Equilterl tringle isosceles tringle = = c = d c d Corresponding ngles B + = d A C AB AB AB B D CD is perpendiculr isector of AB + d = 180 + c = 180 c d Co-interior ngles ED OFS c d + + c + d = 360 c d A C D BCD is n exterior ngle. Right ngle t the circumference in semicircle Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 537

WORKED EXAMPLE 2 Clculte the vlues of the pronumerls in the polygon shown. c d cm THINK 6 cm 1 This shpe is regulr hexgon. The ngles t the centre re ll equl. 2 The other two ngles in the tringle re equl. 60 3 The 6 tringles re equilterl tringles; therefore, ll sides re equl. WORKED EXAMPLE 3 WRITE = 360 6 = 60 + + c = 180 = c So: 60 + 2 = 180 = 60 c = 60 d cm = 6 cm Clculte the missing pronumerls in the digrm of rilings for the set of stirs shown. 35 THINK 1 Recognise tht the top nd ottom of the stir rils re prllel lines. c CTED WRITE/DRAW 35 35 = PR c 60P R 538 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

2 To find the unknown ngle, use the lternte ngle lw nd the given ngle. 3 Using the corresponding ngle lw nd the given right ngle, recognise tht the unknown ngle c is right ngle. Given ngle = 35 = 35 c 4 Use the stright ngle rule to find the unknown ngle. EXERCISE 11.2 PRACTISE CONSOLIDATE OLIDATE c = 90 + + c = 180 35 + + 90 = 180 = 180 125 = 55 Properties of ngles, tringles nd polygons 1 WE1 Clculte the interior nd exterior ngles for regulr nongon. 2 Clculte the interior nd exterior ngles for the following regulr polygon. For questions 3 6, clculte culte the vlue of the pronumerl in the following figures. 3 WE2 8 cm c 50 5 WE3 27 m n 4 6 81 m 32 140 7 Clculte the interior nd exterior ngles for ech of the following regulr polygons. Equilterl tringle Regulr qudrilterl c Hexgon d e Heptgon REC CT OFS Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 539

8 Clculte the vlue of the pronumerls in the following figures. 27 52 x y 130 c c 63 9 Clculte the vlue of the pronumerls in the following figures. 30 c t 62 x y 35 d d 0 15 c z 70 c d 10 Nme the regulr polygon tht hs the given ngle(s). Interior ngle of 108, exterior ngle of 72 Interior ngle of 150, exterior ngle of 30 c Interior ngle of 135, exterior ngle of 45 d Interior ngle of 120 e Exterior ngle of 120 11 In the figure, the vlue of is: A 30 B 75 C 90 150 D 120 E 150 12 An isosceles tringle hs known ngle of 50. The lrgest possile ngle for this tringle is: A 80 B 130 C 90 D 65 E 50 13 Find the vlues of the pronumerls in ech of the following figures. c UNCOR CORRECT RECTED ED 32 61 x y 145 t 70 540 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

ww 14 Find the vlues of the pronumerls in ech of the following figures. 70 d c c 38 MASTER NCOC 15 Find the vlues of the pronumerls in ech of the following figures. 28 m n x c 9 cm cm 110 70 110 y z FS30 16 Find the vlues of the pronumerls in ech of the following figures. 60 45 17 Clculte the unknown pronumerls. r x h 3.6 cm wworr CTE ROOFS OFS 30 15 110 y z 35 4.2 cm 18 86 40 d c Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 541

11.3 Are nd perimeter I Much of our world is descried y re (the mount of spce enclosed y closed figure) nd perimeter (the distnce round closed figure). Interctivity Are nd perimeter int-6474 17 Lot 603 645 m 2 Corner lock with expnsive 23.55 m frontge 13.05 $251 000 WORKED EXAMPLE 4 Some exmples re the re of house lock, the fencing of lock of lnd, the size of edroom nd the mount of pint required to cover n oject. In this section we will review the more common shpes. Perimeter Perimeter is the distnce round closed figure. Some common rules re: 1. For squres, the perimeter = 4l 2. For rectngles, the perimeter = 2(l + w) Squre l l l 36.56 14.07 l Find the perimeter of the closed figure shown. 8 cm 4 cm 12 cm THINK 1 To find the perimeter we need the lengths of ech side round the closed figure. So we need to find nd. 4.05 37.92 Corner lock with wide 17 m frontge $147 000 WRITE 8 cm w 10 cm 4 cm 32.18 Rectngle l 12 cm Lot 658 761m 2 23.55 UNCORREC RRECT RECTED l 10 cm 32.75 5.86 w 542 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

2 Find the length of. = 10 8 = 2 cm 3 Find the length of. = 12 4 = 8 cm 4 Clculte the perimeter y dding ll the lengths round the outside of the figure. Unit 4 AOS M3 Topic 1 Concept 6 Liner dimensions in 2D nd 3D Concept summry Prctice questions Are of common shpes The res of shpes commonly encountered re: 1. Are of squre: A = length 2 = l 2 2. Are of rectngle: A = length width = l w 3. Are of prllelogrm: Perimeter = 12 + 8 + 4 + 2 + 8 + 10 = 44 cm 5 Write your nswer. The perimeter of the closed figure is 44 cm. Squre l Rectngle Prllelogrm A = se height = P h h 4. Are of A = 1 Trpezium 2 ( + ) h trpezium: h of tringle: A = 1 Tringle 5. Are 2 h h ED PA = PA Are is mesured in mm 2, cm 2, m 2, km 2 nd hectres. 1 hectre = 100 m 100 m = 10 000 m 2 l l w WORKED EXAMPLE 5 Clculte the re of the grden ed given in the digrm (correct to the nerest squre metre). 5.7 m 2.4 m 7.5 m Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 543

THINK 1 The shpe of the grden is trpezium. Use the formul for re of trpezium. Rememer tht the lengths of the two prllel sides re nd, nd h is the perpendiculr distnce etween the two prllel sides. WRITE Are of trpezium = 1 ( + ) h 2 = 7.5 = 5.7 h = 2.4 2 Sustitute nd evlute. A = 1 (7.5 + 5.7) 2.4 2 = 1 13.2 2.4 2 = 15.84 m 2 3 Write your nswer. The re of the grden ed is pproximtely 16 squre metres. WORKED EXAMPLE 6 THINK Composite res Often closed figure cn e identified s comprising two or more different common figures. Such figures re clled composite figures.. The re of composite figure is the sum of the res of the individul common on figures. Are of composite figure = sum of the res of the individul common figures A composite = A 1 + A 2 + A 3 + A 4 +... Clculte the re of the composite shpe shown. WRITE 2PA A PAA 1 Identify the two shpes. The shpes re rectngle nd tringle. 2 Clculte the re of the rectngle. A Rectngle = l w = 12 7 = 84 3 Clculte the re of the tringle. A Tringle = 1 2 h UNCORRE RRECTED ED = 1 12 (10 7) 2 = 6 3 = 18 4 Sum the two res together. A Totl = A Rectngle + A Tringle = 84 + 18 = 102 5 Write your nswer. The totl re of the composite shpe is 102 m 2. 2 + OFS 7 m 12 m 10 m 544 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

Interctivity Conversion of units of re int-6269 Conversion of units of re Often the units of re need to e converted, for exmple, from cm 2 to m 2 nd vice vers. 1. To convert to smller units, for exmple, m 2 to cm 2, multiply ( ). 2. To convert to lrger units, for exmple, mm 2 to cm 2, divide ( ). WORKED EXAMPLE 7 EXERCISE 11.3 PRACTISE Some exmples re: () 1 cm 2 = 10 mm 10 mm = 100 mm 2 () 1 m 2 = 100 cm 100 cm = 10000 cm 2 (c) 1 km 2 = 1000 m 1000 m = 1000000 m 2 (d) 1 hectre = 10000 m 2 Convert 1.12 m 2 to squre centimetres (cm 2 ). Are nd perimeter I 10 2 100 2 1000 2 mm 2 cm 2 m 2 km 2 10 2 100 2 1000 2 THINK WRITE 1 To convert from m 2 to cm 2, multiply y 100 2 or 10000. 1.12 m 2 = 1.12 10 000 cm 2 = 11 200 cm 2 2 Write your nswer. 1.12 12 m 2 is equl to 11 200 squre centimetres (cm 2 ). WORKED EXAMPLE 8 Convert 156 000 metres 2 to: kilometres 2 hectres. THINK WRITE 1 To convert from metres es 2 to kilometres 2 divide y 1000 2 or 1 000 000. 156000 m 2 = 156000 1000000 km 2 = 0.156 km 2 2 Write the nswer in correct units. 156 000 m 2 = 0.156 squre kilometres (km 2 ) 1 10000 m 2 hectre. To convert 156000 m 2 156000 equls 1 = from hectres, divide y 10000. 10000 hectres m 2 to = 15.6 hectres 2 Write the nswer. 156 000 m 2 = 15.6 hectres RRECTEto: UNCO NCORRE RREC RECTED ED 1 WE4 Clculte the perimeter of the closed figure shown. 12 cm 4 cm 10 cm 21 cm Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 545

2 Clculte the perimeter of the closed figure shown. 21 m 8 m 3 m 16 m 32 m 3 WE5 Clculte the re of the ckyrd given in the digrm (correct to the nerest squre metre). 2.1 m 8.7 m 4.5 m 4 Clculte the re of the shpe shown (correct to the nerest squre metre). 3.8 m 5 WE6 Clculte the re of the composite shpe shown. 5 m 3 m 12 m 10 m 6 Clculte lte the re of the composite shpe shown. ED 15 cm 5 cm 7 WE7 Convert 2.59 m 2 to squre centimetres (cm 2 ). 8 Convert 34.56 cm 2 to squre millimetres (mm 2 ). 9 WE8 Convert 2 500 000 m 2 to kilometres 2 (km 2 ). 10 Convert 0.0378 m 2 to millimetres 2 (mm 2 ). 1.9 m 546 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

CONSOLIDATE 11 Clculte the perimeters of the following figures (correct to the nerest whole units). 12 m 5 m 4 m 7 m c 23.7 cm d 120 m 15.9 cm 15.4 cm 83.2 m pfs 27.5 cm 210 m 12 Clculte the res of the closed figures in question 11. 13 Clculte the res of the following figures (correct to 1 deciml plce). 9 cm 8 cm 13 m 10 m 20 m 12 m 25 m 11.2 cm 21 cm 14 Clculte the perimeters of the closed figures in question 13. 16 cm 70 m 10 cm 15 Convert the following res to the units given in rckets. 20000 mm 2 (cm 2 ) 320000 cm 2 (m 2 ) c 0.035 m 2 (cm 2 ) d 0.035 m 2 (mm 2 ) e 2500000 m 2 (km 2 ) f 357000 m 2 (hectres) g 2750000 0 000 mm 2 (m 2 ) h 0.00006 km 2 (m 2 ) 16 Find the re of the regulr hexgon s shown in the digrm (correct to 2 deciml plces, in m 2 ). RECT) UNCORREC RECT ECTED ED the 1.20 m 2.08 m 17 The perimeter of the enclosed figure shown is 156.6 metres. The unknown length, x, is closest to: 20.5 m x A 20.5 m B 35.2 m 35.2 m C 40.2 m D 80.4 m E 90.6 m Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 547

18 Find the re of ech of the following figures. c 5 m 6 cm 10 cm 2 cm 4 m 4 m 4 m 8 m 10.8 m MASTER 11.4 1.4 19 Clculte the perimeter of ech of the shpes in question 18. Give your nswers correct to the nerest whole numer. 20 Convert the following res to the units given in the rckets. 45000 cm 2 (m 2 ) 0.71 m 2 (mm ) c 216000 m 2 (hectres) d 0.0737 km 2 (m ) 17 m 21 Find the perimeter, in millimetres, of the following shpe. Give your nswer correct to the nerest whole numer. 11 m 10 m 9 m 18 m 22 On the set of western movie, horse is tied to riling outside sloon r. The riling is 2 metres long; the reins re lso 2 metres long nd re tied t one of the ends of the riling. Drw digrm of this sitution. To how much re does the horse now hve ccess (correct to 1 deciml plce)? The reins re now tied to the centre of the riling. c Drw digrm of this sitution. d To how much re does the horse hve ccess (correct to 1 deciml plce)? Are nd perimeter II There re more shpes we cn clculte the perimeter nd re of. We cn lso clculte the perimeter of circle, known s the circumference, s well s the re of circle. RECT CTED Circumference (C) is the perimeter of circle. C = 2 π rdius = 2πr 1A OFS r Circumference of circle 548 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

WORKED EXAMPLE 9 Clculte the circumference of the following shpe correct to 2 deciml plces. 12 cm THINK 1 Use the formul for the circumference of circle with rdius in it. 2 Sustitute the rdius into the eqution, using the π key on your clcultor. Unit 4 AOS M3 Topic 2 Concept 1 Circle mensurtion Concept summry Prctice questions Interctivity Arc length int-6270 WORKED EXAMPLE 10 Arc length An rc is prt of the circumference of circle. The rc length of circle is clculted y finding the circumference of circle nd multiplying lying y the frction of the ngle tht it forms t the centre of the circle. Arc length = θ 360 2πr where r = rdius of the circle nd θ = the ngle the ends of the rc mkes with the centre of the circle. cle. Clculte the rc lengths for ech of the following shpes correct to 2 deciml plces. 7.2 cm 45 2.4 cm THINK 1 Determine the ngle the ends of the rcs mke with the centre. WRITE C = 2πr C = 2 π 12 = 75.398 223 69 75.40 cm 3 Write the nswer correct to 2 deciml plces. The circumference is 75.40 cm. =PA PA h WRITE θ = 360 90 = 270 2 Sustitute the vlues into the rc length eqution. Arc length = θ 360 2πr Arc length NKUNCORREC RRECTED ED PROO ROOFS = 270 360 2 π 2.4 = 11.3097 11.31 cm r θ r Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 549

3 Write the nswer. Arc length = 11.31 cm 1 Determine the ngle the ends of the rcs mke with the centre. WORKED EXAMPLE 11 THINK Perimeter of composite shpes The perimeter of composite shpe is the sme s the regulr perimeter; it is the totl distnce round the outside of closed figure. Clculte the perimeter of the closed figure given (correct to the nerest mm). 600 mm 1 The shpe is composed of semicircle nd three sides of rectngle. 2 Add together the three components of the perimeter. 300 mm WRITE θ = 360 45 = 315 2 Sustitute the vlues into the rc length eqution. Arc length = θ 360 2πr = 315 360 2 π 7.2 = 39.5841 39.58 cm 3 Write the nswer. Arc length = 39.58 cm Perimeter = 300 + 2 600 + 1 circumference where 2 1 circumference = 1 2 of 2 2πr = π 150 471.24 Perimeter = 300 + 2 600 + 471.24 = 1971.24 3 Write your nswer. The perimeter of the closed figure is 1971 mm, correct to the nerest millimetre. Are of circles, sectors nd segments Circles The re of circle is clculted y multiplying π y the rdius squred. This cn e used to clculte the re of sector, s shown. Are of circle: CTED orr Circle A = π rdius 2 = πr 2 r 550 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

Sectors A sector is prt of the re of circle, s shown y the shded re in the digrm. The ngle θ, t the centre of the circle, sutends the rc t AB. The re of sector is clculted using: A Sector = θ 360 πr2 r C θ r WORKED EXAMPLE 12 THINK A Segments A segment is formed y joining two points on the circumference with stright line, known s chord. This forms two res, with the smller re eing the minor segment nd the lrger re the mjor segment. This is shown y the lue (minor segment) nd pink (mjor segment) sections in the digrm. The re of the minor segment cn e clculted y: A Minor segment = A sector A tringle θ = πr2 2 r2 sinθ 360 r 2 1 = r 2 360 π 1 θ 2 sinθ From the circle shown, which hs rdius of 8 cm, clculte: the re of the minor sector the re of the minor segment of the circle. Give your nswers correct to 2 deciml plces. 1 Use the formul for the re of sector in degrees. C 8 cm RECT CTED 95º WRITE AGθ B A A Sector = θ 360 πr2 r = 8 θ = 95 A r B C θ Mjor segment Minor segment OFS nglpr r B Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 551

2 Sustitute the quntities into the formul nd simplify. 3 Write the nswer correct to 2 deciml plces. A Sector = θ 360 πr2 = 95 360 π1822 = 53.0580 A Sector = 53.06 cm 2 1 Use the formul for the re of segment in degrees. 2 Sustitute the quntities into the formul nd simplify. 3 Write the nswer correct to 2 deciml plces. WORKED EXAMPLE 13 Composite res As with composite res, the re is the sum of the individul common figures, including circles or frctions of circles. Clculte the re of the hotel foyer from the pln given (correct to the nerest squre metre). 25 m THINK 1 The shpe is composite nd needs to e seprted into two or more common shpes: in this cse, rectngle, tringle nd hlf of circle. 2 Find the re of ech shpe. (The width of the rectngle nd the se of the tringle is twice the rdius of the circle, tht is, 16 metres.) 20 m A Minor segment = r 2 θ 360 π 1 2 sin(θ) r = 8 θ = 95 20 mted A Minor segment = 8 2 95 360 π 1 2 sin 95 = 21.1798...... A Minor segment = 21.18 cm 2 8 m WRITE/DRAW 25 m A 2 A 1 16 m A 3 16 m 8 m 20 m Are of foyer = A 1 + A 2 + A 3 A 1 = re of tringle rectuncorrec RRECTED ED = 1 2 h = 1 16 20 2 = 160 m 2 A 2 = Are of rectngle = l w = 25 16 = 400 m 2 552 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

EXERCISE 11.4 PRACTISE Are nd perimeter II 1 WE9 Clculte the circumference of the following correct to 2 deciml plces. 2.8 m 2 Clculte the circumference of the following correct to 2 deciml plces. 23.25 cm 3 WE10 Clculte the rc length of ech of the following shpes correct to 2 deciml plces. 120º º 2.45 m 5 cm 45º 4 Clculte the rc length of ech of the following shpes correct to 2 deciml plces. 14.25 cm A 3 = Are of hlf of circle = 1 2 π r2 = 1 2 π 82 = 100.53 m 2 3 Add together ll three res for the Are of foyer = A 1 + A 2 + A 3 composite shpe. = 160 + 400 + 100.53 = 660.53 m 2 4 Write your nswer. The re of the hotel foyer is pproximtely 661 m 2. RECTE OFS 110 23.48 mm 5 WE11 Clculte the perimeter of the following shpe, correct to 2 deciml plces. 12.8 cm 18.5 cm Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 553

6 Clculte the perimeter of the following shpe, correct to 1 deciml plce. 18 cm 12 cm 7 WE12 From the circle shown, which hs rdius of 12.5 cm, clculte: the re of the minor sector the re of the minor segment of the circle. Give your nswers correct to 2 deciml plces. B 150 C 12.5 cm 8 From the circle shown which hs rdius of 22 m, clculte: the re of the minor sector the re of the minor segment of the circle. Give your nswers s correct to 2 deciml plces. A C 34 22 m 9 WE13 Find the re of the following figure (correct to the nerest whole unit). CTED 5.5 mm A rd B AOFS 8 mm 554 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

10 Clculte the re of the shded shpe shown, correct to 2 deciml plces. CONSOLIDATE 16 km 11 Clculte the perimeter nd the re of the following figures (correct to the nerest whole unit). 13.5 mm 8 cm 12 Clculte the res of the following figures (where pproprite correct to 1 deciml plce). c 24 mm 3.5 m 2 m 17 m 12 m 13 Clculte the perimeter of the closed figures in question 12 correct to 1 deciml plce. 14 A cutting lde for crft knife hs the dimensions shown in the digrm. Wht is the re of steel in the lde (correct to the nerest mm 2 )? 15 The perimeter of the figure shown, in centimetres, is: OR3 A 34 B 24 + 5π C 24 + 2.5π D 29 + 5π E 29 + 2.5π RECT CTED 45.2 mm GE 125 mm 30 mm 20 mm 5 mm 40 mm 12 cm 16 A minor sector is formed y n ngle of 123 in circle of dimeter 9 cm. The re of the minor sector is closest to: A 86.94 cm 2 B 63.62 cm 2 C 41.88 cm 2 D 15.17 cm 2 E 21.74 cm 2 17 Clculte the difference in re etween sector of circle formed y n ngle of 50 nd sector of circle formed y n ngle of 70 in circle of rdius 3 cm. Give your nswer correct to 2 deciml plces. 7 cm 90 mm 2 cm 3 cm Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 555

MASTER 18 A church window is mde in the shpe of semicircle with three equl stined glss sections. Clculte the re of one of the stined glss sections. Give your nswer correct to 2 deciml plces. 19 A pool is to hve one row of cping tiles 500 mm wide round its perimeter. Clculte the inner perimeter of the tiles for the pool shpe shown. Give your nswer correct to 2 deciml plces. 20 Clculte the res of the shded region in the following shpes, s, correct to 2 deciml plces: A c 8 cm 60 12 cm C 100 60 60 60 20 cm 7.3 cm B 21 A circulr grden ed is to e lndscped with Grss Plnts grss, plnts nd mulch. Given the circulr grden is Mulch designed with the shown dimensions: s: 115 5.75 m clculte the re of the grssed re, correct to 2 deciml plces clculte the re of the mulched re, correct to 2 deciml plces c clculte the re of the plnted re, correct to 2 deciml plces. d If the cost of grss, plnts nd mulch re $40, $55 nd $15 per metre respectively, clculte the cost of lndscping the circulr grden correct to the nerest cent. 22 A 3-ring drtord hs dimensions s shown. (Give ll nswers to 1 deciml plce.) 40 cm 20 cm 6 cm RECT CTE ED 1 2 3 2 1 s:pa 10 m Wht is the totl re of the drtord? Wht is the re of the ullseye (inner circle)? c Wht is the re of the 2-point middle ring? d Express ech re of the three rings s percentge of the totl re (where pproprite correct to 2 deciml plces). 5 m 5 m OFS 556 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

11.5 Unit 4 AOS M3 Topic 2 Concept 2 Erth modelled y sphere Concept summry Prctice questions Interctivity Gret circles int-6271 4UN Unit 4 AOS M3 Topic 2 Concept 4 Distnce etween points on sphere Concept summry Prctice questions Gret circles A gret circle is circle tht divides sphere into two equl hemispheres, where the centre of the gret circle is the centre of the sphere. As with circle the circumference of gret circle cn e clculted y using C = 2πr. The gret circles tht we will focus on re those tht re relted to Erth. These re the ltitude nd longitude nd they re used to define loctions on the surfce of the Erth. The equtor nd the lines of longitude re exmples of gret circles. Equtor N Meridin of longitude PROMeridi Ltitude nd longitude Lines of ltitude re prllel nd run est to west. They mesure the ngle north or south of the equtor. The meridins, in or lines of longitude, mesure the ngle est or west of the prime meridin, which psses through Greenwich Oservtory (0 longitude) in London, Englnd. 30 Lines of ltitude 30 N Equtor 30 W S Lines of longitude 30 Prime meridin Mesuring distnces with gret circles The shortest distnce etween ny two points on the Erth with the sme longitude is long gret circle. To conduct clcultions in reltion to gret circles on Erth, the centre is the centre of the Erth, with the rdius of the Erth eing 6400 km. RECTED The shortest distnce etween two points on the Erth tht hve the sme longitude is the rc length etween those points. Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 557

WORKED EXAMPLE 14 Clculte the distnce trvelled in the following situtions correct to the nerest kilometre: n eroplne tht trvels long the equtor from point of longitude 48 W to 22 E ot tht trvels from point M: 36 S, 138 W to point N: 40 S, 138 W. THINK 1 Both points lie on the equtor, which is gret circle. The ngle etween the two points is mde of 2 prts, from the strting point to 0, then from 0 to the finishing point. 2 The distnce etween the two points is the rc length. Sustitute θ nd r into the rc length formul nd simplify. 3 Answer the question correct to the nerest kilometre. 1 Oserving tht M nd N lie on the sme longitude 138 W, pth MN is long gret circle. Since oth ltitudes re S, the ngle θ is formed y their difference. SUN 2 The distnce etween the two points is the rc length. Sustitute θ nd r into the rc length formul nd simplify, using the rdius of the Erth s 6400 km. UNCORRECT RECTE ECTED 3 Answer the question correct to the nerest kilometre. WRITE r 48 W 0 22 E r θ = 48 + 22 2 = 70 r = 6400 km rc length = θ 360 2πr = 70 2π 6400 360 = 7819.075... The eroplne trvelled 7819 km. r r M 36 N 40 θ = 40 36 = 4 rc length = θ 360 2πr = 4 2π 6400 360 = 446.804... The ot trvelled 447 km. 558 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

EXERCISE 11.5 PRACTISE CONSOLIDATE Gret circles 1 WE14 Clculte the distnce trvelled in the following situtions correct to the nerest kilometre: n eroplne tht trvels long the equtor from point of longitude 23 W to 35 E ot tht trvels from point M: 12 S, 100 W to point N: 58 S, 100 W. 2 Clculte the distnce trvelled in the following situtions correct to the nerest kilometre: plne tht trvels long the equtor from point of longitude 67 W to 15 E ycht tht trvels from point P: 36 S, 98 E to point Q: 81 S, 98 E. 3 Clculte the distnce trvelled y n eroplne trvelling long the equtor from point of longitude 36 W to 123 E. Give your nswer correct FS E. FSE. to the nerest km. 4 Clculte the distnce trvelled y cruise ship tht trvels from point P: 18 N, 25 W to point Q: 30 S, 25 W. Give your nswer correct to the nerest km. 5 Clculte the gret circle distnces shown in these figures. Give your nswers correct to the nerest kilometre. N N P 30 N 134 E 72 N 58 W Y 21 N 58 W Q W E W E c W 34 W D N S 38 E F E d S X 72 S 134 E S S 6 Clculte the shortest distnce, correct to the nerest kilometre, etween two points with the sme longitude tht form the following ngles t the centre of the Erth. 12 3 c 0.5 d 0.08 7 Clculte the distnce etween two points on the equtor tht form the following ngles t the centre of the Erth. Give your nswers correct to the nerest kilometre. 15.75 10.31 CTED W OFS N 128 W L M 70 E E Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 559

MASTER 11.6 Unit 4 AOS M3 Topic 1 Concept 7 Surfce re nd volume Concept summry Prctice questions 8 Use meridins to clculte the shortest distnce, correct to the nerest kilometre, etween the following cities nd the North Pole. London (51.51 N, 0.13 W) Rome (41.90 N, 12.50 E) c Dlls (32.78 N, 96.80 W) 9 Melourne (37.83 S, 145 E) nd Port Dougls (16.48 S, 145 E) re on the sme meridin. Wht is the shortest distnce (correct to the nerest kilometre) etween the two towns? 10 The cities of Wshington in the USA nd Lim in Peru re 5625 km prt. They oth lie on the sme meridin. Wht is the ngle formed t the centre of the Erth y the two cities? Give your nswer correct to 2 deciml plces. 11 The cities of Port Hedlnd in Austrli nd Nnjing in Chin re 3795 km prt. They oth lie on the sme line of longitude. Wht is the ngle formed t the centre of the Erth y the two cities? Give your nswer correct to 2 deciml plces. 12 Use meridins to clculte the shortest distnce, correct to the nerest kilometre, etween the following cities nd the equtor. Bnglore (12.97 N, 77.57 E) São Pulo (23.55 S, 46.63 W) c Tokyo (35.68 N, 139.68 E) 13 Melourne nd Hokkido Islnd in Jpn lie on the sme line of longitude nd re 9094 km prt. If Melourne hs ltitude of 37.83 S, find the ltitude of Hokkido Islnd, which is north of Melourne. Give your nswer correct to 2 deciml plces. 14 The cities of Perth nd Beijing lie on the sme line of longitude nd re 7985 km prt. If Perth hs ltitude of 31.95 S, find the ltitude of Beijing, which is north of Perth. Give your nswer correct to 2 deciml plces. Totl surfce re The totl surfce re (TSA) of solid oject is the sum of the res of the surfces. In some cses we cn use estlished formuls of very common everydy ojects. In other situtions we will need to derive formul y using the net of n oject. Totl surfce re formuls of common ojects RECT CTED Cue l Cuoid h w l OFS Cylinder Cues: Cuoids: Cylinders: TSA = 6l 2 TSA = 2(lw + lh + wh) TSA = 2πr (r + h) r r h 560 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

Interctivity Surfce re int-6477 WORKED EXAMPLE 15 Cone h r Slnt height s Sphere Cones: Spheres: TSA = πr(r + s), TSA = 4πr 2 where s is the slnt height r Squre pyrmid: TSA = 2 + 2s, where s is the FSs slnt height Clculte the totl surfce re of poster tue with length of 1.13 metres nd rdius of 5 cm. Give your nswer correct to the nerest 100 cm 2. 5 cm THINK 1 A poster tue is cylinder. Express ll dimensions in centimetres. Rememer 1 metre equls 100 centimetres. 1.13 m WRITE TSA of cylinder = 2πr(r + h) Rdius, r = 5 cm Height, h = 1.13 m = 113 cm 2 Sustitute nd evlute. Rememer BODMAS. TSA = 2 π 5(5 + 113) = 2 π 5 118 3707.08 3 Write your nswer. The totl surfce re of poster tue is pproximtely 3700 cm 2. UNCORR ORRECTED RECT ED WORKED EXAMPLE 16 Clculte the totl surfce re of size 7 sketll with dimeter of 25 cm. Give your nswer correct to the nerest 10 cm 2. OFS h 25 cm s Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 561

THINK 1 Use the formul for the totl surfce re of sphere. Use the dimeter to find the rdius of the sketll nd sustitute into the formul. WORKED EXAMPLE 17 THINK A die used in ord gme hs totl surfce re of 1350 mm 2. Clculte the liner dimensions of the die (correct to the nerest millimetre). 1 A die is cue. We cn sustitute into the totl surfce re of cue to determine the dimension of the cue. Divide oth sides y 6. 2 Tke the squre root of oth sides to find l. Totl surfce re using net If the oject is not common oject or vrition of one, such s n open cylinder, then it is esier to generte the formul from first principles y constructing net of the oject. A net of n oject is plne figure tht represents the surfce of 3-dimensionl oject. Squre pyrmid WRITE TSA of sphere = 4πr 2 Dimeter = 25 cm Rdius = 12.5 cm TSA = 4 π 12.5 2 1963.495 2 Write your nswer. Totl surfce re of the ll is pproximtely 1960 cm 2. WRITE TSA = 6 l 2 = 1350 mm 2 1350 = 6 l 2 l 2 = 1350 6 = 225 l =!225 2 = 15 mm 3 Write your nswer. The dimensions of the die re 15 mm 15 mm 15 mm. UNCORRECT RECTED ED Slnt height OFS Net 562 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

20 cm 20 cm 20 cm 20 cm Trpezoidl prism Net WORKED EXAMPLE 18 THINK Cylinder Clculte the totl surfce re of the tringulr prism shown in the digrm. 20 cm 10 cm 8 cm 1 Form net of the tringulr prism, trnsferring ll the dimensions to ech of the sides of the surfces. 2 Identify the different-sized common figures nd set up sum of the surfce res. The two tringles re the sme. CTED 6 cm WRITE/DRAW 10 cm A 1 10 cm A 4 Net O 10 cm A 4 6 cm 8 cm 6 cm A 2 A 3 8 cm 10 cm 6 cm 6 cm TSA = A 1 + A 2 + A 3 + 2 A 4 A 1 = l 1 w 1 = 20 10 = 200 cm 2 A 2 = l 2 w 2 = 20 8 = 160 cm 2 A 3 = l 3 w 3 = 20 6 = 120 cm 2 A 4 = 1 2 h = 1 2 8 6 = 24 cm 2 Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 563

3 Sum the res. TSA = A 1 + A 2 + A 3 + 2 A 4 = 200 + 160 + 120 + 2 24 = 528 cm2 4 Write your nswer. The totl surfce re of the tringulr prism is 528 cm 2. WORKED EXAMPLE 19 THINK Totl surfce re of composite solids When clculting the totl surfce re of composite solids, e creful to include only the surfces tht form the outer prt of the solid. 1 Identify the components of the composite solid. Clculte the totl surfce re of the following solid. 2 Use the formuls to clculte the surfce res of the component prts. Rememer to exclude internl surfces. 3 Add the res together ther to otin the totl surfce re.. WRITE The composite solid consists of cue of length 6 cm nd squre pyrmid with se length 6 cm nd slnt height 4 cm. TSA of cue (minus top) = 5l 2 = 5 6 2 = 5 36 = 180 cm 2 TSA of pyrmid (minus ottom) = 2s = 2 6 4 = 48 cm 2 180 + 48 = 228 cm 2 4 Write your nswer. The totl surfce re of the solid is 228 cm 2. EXERCISE 11.6 PRACTISE Totl surfce re CT 1 WE15 Clculte the totl surfce re of closed cylinder with rdius of 1.2 cm nd height of 6 cm. Give your nswer correct to 1 deciml plce. 2 Clculte the totl surfce re of closed cone with rdius of 7 cm nd slnt height of 11 cm. Give your nswer correct to 1 deciml plce. 6 cm 3 WE16 Find the totl surfce re of sphere with rdius of 0.8 m. Give your nswer correct to 1 deciml plce. 4 Find the totl surfce re of cue with side length 110 cm. 4 cm OF 564 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

CONSOLIDATE 5 WE17 Find the unknown length of cue, given tht the totl surfce re is 255 m 2. Give your nswer correct to 1 deciml plce. 6 Find the unknown rdius of sphere, given tht the totl surfce re is 785.12 cm 2. Give your nswer correct to 1 deciml plce. 7 WE18 Clculte the totl surfce re of this prism. 22 cm 40 cm 20 cm 40 cm 17 cm 8 Clculte the surfce re of n open cylindricl cn tht is 12 cm high nd 8 cm in dimeter (correct to 1 deciml plce) y completing the following steps. Form net of the open cylinder, trnsferring ll the dimensions to ech of the surfces. Identify the different-sized common figures nd set up sum of the surfce res. The length of the rectngle is the circumference of the circle. c Add the res. d Write your nswer. 9 WE19 Clculte the totl surfce re of the following solid 4 cm 7 cm 10 Clculte the totl surfce re of the following solid, correct to 2 deciml plces. 5.5 cm ED 18 cm 11 Clculte the totl surfce re for ech of the solids to d from the following informtion. Give nswers correct to 1 deciml plce where pproprite. A cue with side lengths of 135 mm A cuoid with dimensions of 12 m 5 m 8 m (l w h) c A sphere with rdius of 7.1 cm d An opened cylinder with dimeter of 100 mm nd height of 30 mm 8 cm 12 cm Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 565

12 Clculte the totl surfce re of the ojects given in the digrms. Give nswers correct to 1 deciml plce. Length = 1.5 m c 410 mm 14 cm Dimeter = 43 cm 7 cm 4 cm 13 Clculte the unknown dimensions, given the totl surfce re of the ojects. Give nswers correct to 1 deciml plce. Length of cue with totl surfce re of 24 m 2 The rdius of sphere with totl surfce re of 633.5 cm 2 c Length of cuoid with width of 12 mm, height of 6 cm nd totl surfce re of 468 cm 2 d Dimeter of plying ll with totl surfce re of 157630 cm 2 14 Clculte the totl surfce res for the ojects given in the digrms. Give nswers correct to 1 deciml plce. 5 cm 15 cm 10 cm c Are = 22 cm 2 13 cm 8 cm 4 cm 6.06 cm 30 cm 7 cm d 7.5 mm 6 mm 4.2 mm 15 Clculte the totl surfce re of ech of the ojects in the digrms. Give nswers correct to 1 deciml plce. Ruish in UNCORRECT RECTED 10 cm 13.5 cm 12.3 cm 10.5 cm 250 mm 20 cm c 2 cm 250 mm 4.5 cm 7 cm 3 cm 566 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

MASTER 11.7 16 A concrete swimming pool is cuoid with the following dimensions: length of 6 metres, width of 4 metres nd depth of 1.3 metres. Wht surfce re of tiles is needed to line the inside of the pool? (Give your nswer in m 2 nd cm 2.) 17 Wht is the totl re of cnvs needed for the tent (including the se) shown in the digrm? 1.0 m 2.5 m Give your nswer correct to 2 deciml plces. 1.5 m 18 The totl surfce re of 48 mm dimeter ll used in gme of pool is closest to: A 1810 mm 2 B 2300 mm 2 C 7240 mm 2 D 28950 mm 2 E 115800 mm 2 19 The totl surfce re of squre-sed pyrmid with se length of 3.55 cm nd slnt height of 6.71 cm is closest to: A 107.88 cm 2 B 60.24 cm 2 C 60.25 cm 2 D 92.66 cm 2 E 92.67 cm 2 20 The formul for the totl surfce re for the oject shown is: A 1 2 h B 2 1 + + 2 h 2 h h C 3 1 h + 2 1 D h + 3 2 E h + 3 21 Clculte the totl surfce re of the following solid, correct to 2 deciml plces. 2.5 cm 9 cm 18 cm 7.5 cm 4.5 m 22 A ker is investigting the est shpe for lof of red. The shpe with the smllest surfce re stys freshest. The ker hs come up with two shpes: rectngulr prism with 12 cm squre se nd cylinder with round end tht hs 14 cm dimeter. Which shpe stys fresher if they hve the sme overll length of 32 cm? Wht is the difference etween the totl surfce res of the two loves of red? h 1 PRO 2 2PR PR m Volume of prisms, pyrmids nd spheres 6.5 m The most common volumes considered in the rel world re the volumes of prisms, pyrmids, spheres nd ojects tht re comintion of these. For exmple, people who rely on tnk wter need to know the cpcity (volume) of wter tht the tnk is holding. Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 567

Interctivity Volume int-6476 Volume is the mount of spce occupied y 3-dimensionl oject. The units of volume re mm 3 (cuic millimetres), cm 3 (cuic centimetres or cc) nd m 3 (cuic metres). 1000 mm 3 = 1 cm 3 1 000 000 cm 3 = 1 m 3 Interctivity Conversion of units of volume int-6272 WORKED EXAMPLE 20 Another mesure of volume is the litre, which is used primrily for quntities of liquids ut lso for cpcity, such s the cpcity of refrigertor or the size of motor cr engines. 1 litre = 1000 cm 3 1000 litres = 1 m 3 Conversion of units of volume Often the units of volume need to e converted, for exmple, from cm 3 to m 3 nd vice vers. Convert 1.12 cm 3 to mm 3. THINK 1 To convert from cm 3 to mm 3 multiply y 10 3 or 1000. (Tht is, 1 cm 3 equls 1000 mm 3.) WRITE 1.12 cm 3 = 1.12 1000 mm 3 = 1120 mm 3 2 Write the nswer in the correct units. 1.12 cm 3 is equl to 1120 mm 3. WORKED EXAMPLE 21 Convert 156 000 0 cm 3 to: m 3 litres. THINK 1 To convert from cm 3 to m 3 divide y 100 3 or 1000000. (Tht is, 1000 000 cm 3 equls 1 m 3.) 10 3 100 3 mm 3 cm 3 m 3 10 3 100 3 WRITE 156000 cm 3 = 156000 1000000 m 3 = 0.156 m 3 2 Write the nswer in correct units. 156000 cm 3 = 0.156 cuic metres (m 3 ) 1 1000 0 cm 3 is equivlent to 1 litre; therefore, to convert from cm 3 to litres, divide y 1000. 156000 cm 3 156000 = 1000 litres = 156 litres 2 Write the nswer. 156000 cm 3 = 156 litres UNCOR CORRE RRECTE RECT CTED 1.1 PRO 1.1 AGE Volume of prisms A prism is polyhedron with uniform cross-section. It is nmed in ccordnce with its uniform crosssectionl re. Uniform cross-section Height 568 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

Tringulr prism Trpezoidl prism Hexgonl prism Rectngulr prism To find the volume of prism we need to determine the re of the uniform cross-section (or se) nd multiply y the height. This is the sme for ll prisms. Volume of prism, V prism, cn e generlised y the formul: V prism = re of uniform cross-section height V = AH For exmple: V rect. prism = A rect. H V tringulr prism = A tringle H Note: Although cylinders re not prisms, they hve uniform cross-section (which is circle) therefore, the sme formul cn e pplied to find volume of cylinder. Tht is, V cylinder = A circle H = πr 2 H In fct, the formul V = A H cn e pplied to ny solid with uniform cross-section of re A.. Cylinder WORKED Clculte the the oject shown. Give your nswer EXAMPLE 22 volume of correct to the nerest cm 3. 15 cm THINK 1 The oject hs circle s uniform cross-section. It is cylinder. The re of the se is: re of circle = πrr 2. Volume is cross-sectionl re times height. WRITE V cylinder = A H, where A circle = πr 2 = π r 2 H = π 15 2 20 = 4500 π 14137.1669cm 3 2 Write your nswer. The volume of the cylinder is pproximtely 14 137 cm 3. UNCO NCORRE RRECTED ED UNC. op OFS 20 cm WORKED Clculte (correct to the nerest mm 3 ) the volume EXAMPLE 23 of the slice of red with uniform cross-sectionl re of 8000 mm 2 nd thickness of 17 mm. Are = 8000 mm 2 17 mm Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 569

THINK WRITE 1 The slice of red hs uniform cross-section. V = A H The cross-section is not common figure ut where A = 8000 mm 2 its re hs een given. V = 8000 mm 2 17 mm = 136000 mm 3 2 Write your nswer. The volume of the slice of red is 136 000 mm 3. Given the volume of n oject, we cn use the volume formul to find n unknown dimension of the oject y trnsposing the formul. WORKED Clculte the height of the tringle (correct EXAMPLE 24 Volume of prism = 6.6 m 3 to 1 deciml plce) from the informtion provided in the digrm. h THINK 1 The volume of the oject is given, long with the width of the tringulr cross-section nd the height of the prism. WRITE V = 6.6 m 3, H = 1.1 m, = 2 m V = A H, where A = 1 2 h V = 1 2 h H 2 m 1.1 m 2 Sustitute the vlues into the 6.6 1 = formul nd solve for h. 2 h 1.1 2 = 1.1 h TEh h = 6.6 1.1 = 6 3 Write your nswer. The height of the tringle in the given prism is 6.0 metres. Volume of pyrmids A pyrmid is polyhedron, where the se is ny polygon nd ll other fces re tringles meeting t the vertex. The nme of the pyrmid is relted to the shpe of the polygon t the se. Vertex Th UNCORRE RREC RECTED ED PA 2PAGH 1 OOFS6.6 m mpro 2 OFS Tringulr pyrmid Squre-sed pyrmid Hexgonl pyrmid The shpe of the cross-section of the pyrmid remins unchnged, ut its size reduces s it pproches the vertex. Similrly, for cones, the shpe of the cross-section is lwys the sme ( circle), ut its size reduces s we move from se towrds the vertex. The volume of pyrmid is lwys one-third of the volume of prism with the sme se nd sme height, H. This holds for ll pyrmids. Cone Vertex 570 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

Volume of pyrmid, V pyrmid, cn e found y using the formul: V pyrmid = 1 re of the se height 3 = 1 AH 3 H V cone = 1 3 πr 2 H A The height of pyrmid, H, is sometimes clled the ltitude. WORKED Clculte the volume of the pyrmid shown EXAMPLE 25 Height of pyrmid = 40 m (correct to the nerest m 3 ). 30 m 30 m THINK WRITE 1 The pyrmid hs squre se. The re of the V pyrmid = 1 se is: Are of squre = l 2. 3 A PR, H, where A squre = l 2 = 1 l2 3 l HPH = 1 302 40 3 = 12000 m 3 2 Write your nswer. The volume of the pyrmid is 12 000 m 3. Volume of spheres nd composite ojects Volume of sphere Spheres re unique (ut common) ojects tht deserve specil ttention. The formul for the volume of sphere is: V sphere = 4 πr 3 3 where r is the rdius of the sphere. Volume RRwh of composite ojects Often the oject cn e identified s comprising two or more different common prisms, pyrmids or spheres. Such figures re clled composite ojects. The volume of composite oject is found y dding the volumes of the individul common figures or deducting volumes. For exmple, ech of the structures on the right cn e roughly modelled s the sum of cylinder nd cone. Volume of composite oject = the sum of the volumes of the individul components. RECT CTED mfs 30 PROO ROOFS OFS V composite = V 1 + V 2 + V 3 +... (or V composite = V 1 V 2 ) r Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 571

WORKED EXAMPLE 26 Clculte the cpcity of the continer shown. (Give your nswer correct to the nerest litre.) 12 cm 20 cm THINK 1 The oject consists of cylinder nd squre-sed prism. 2 The volume of the composite oject is the sum of the volumes of the cylinder nd the prism. 3 Convert to litres using the conversion of 1000 cm 3 equls 1 litre. EXERCISE 11.7 PRACTISE WRITE/DRAW H = 20 cm Volume of prisms, pyrmids nd spheres 1 WE20 Convert 0.35 cm 3 to mm 3. 2 Convert 4800 cm 3 to m 3. r = 6 cm 3 WE21 Convert 1.6 m 3 to litres. 4 Convert 250000 mm 3 to cm 3. 18 cm 5 WE22 Clculte the volume of the following solid correct to the nerest whole unit. 30 mm 20 mm 25 cm 18 cm 18 cm V composite = volume of cylinder + volume of squre-sed prism = A circle H cylinder + A squre prism = (π r 2 H c ) + (l 2 H H = (π 6 2 20) p) + (18 2 25) = 2261.946711 1 + 8100 = 10361.946 6711 711cm 3 10 362 cm 3 = 10.362 litres 4 Write your nswer. The cpcity of the continer is 10 litres, correct to the nerest est litre. UNCORRECTE RECT ECTED 25 cm PR HPR 34 mm 6 Clculte the volume of the following solid correct to the nerest whole unit. 14 mm 22 mm 57 mm 572 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

7 WE23 Clculte the volume of the following oject (correct to 2 deciml plces). Are = 120 mm 2 14.5 mm 8 Clculte the volume of the following oject. Are = 400 cm 2 20 cm 9 WE24 Clculte the mesurement of the unknown dimension dpd (correct to 1 deciml plce). Volume of tringulr prism = 1316.1 cm 3 15.0 cm x 21.4 cm 10 Clculte the mesurement of the unknown dimension (correct to 1 deciml plce). Volume of cylinder = 150 796.4 mm 3 x 120 mm 11 WE25 Clculte the volume of the oject (correct to the nerest whole unit). CTED VO = 10 cm V O 12 cm 11 cm 12 Clculte the volume of the oject (correct to the nerest whole unit). VO = 17 m V 8 m 12 m O Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 573

13 WE26 Clculte the volume of the figure elow. Give your nswer in m 3 correct to 3 significnt figures. (m 3 ) 18 km CONSOLIDATE 12 km 14 Clculte the volume of the figure. Give your nswer in m 3 correct to 4 significnt figures. 100 mm 15 Convert the volumes to the units specified. 530 cm 3 to m 3 56000 cm 3 to litres c 15 litres to cm 3 d 72.1 m to litres e 0.0023 cm 3 to mm 3 f 0.00057 m 3 to cm 3 g 140000 mm 3 to litres h 16000 mm 3 to cm 3 16 Clculte the volume of the following solids correct to the nerest whole unit. 7 cm c 75 mm 4000 mm 51.2 cm 15 cm 4 cm 23 cm 104.8 cm 17 Clculte the volume of the following ojects (correct to 2 deciml plces). Are = 4.2 m 2 Are = 15 cm 2 2.9 m RECT CTED d 2.1 m 6.4 m Are = 32 cm 2 4.8 m 8.5 cm 18 Clculte the mesurement of the unknown dimension (correct to 1 deciml plce). Volume of Volume of prism = 10 1 litres cue = 1.728 m 3 8 x 10 mm 3PR PR m x 3x 574 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

19 Clculte the volume of these ojects (correct to the nerest whole unit). 12 mm c V 35 cm 11 cm VO = 15 cm Altitude of squre-sed pyrmid = 18 mm Bse of pyrmid O 6 cm 10 cm 20 Clculte the volume of these ojects (correct to the nerest whole unit). 4 cm c 3 m 7 cm 8 cm 5 m 10 cm d 10 cm r = 8 cm 10 cm 15 cm e 1 m 2.1 m 2.5 m 20 cm 21 Clculte the volume of cue with sides 4.5 cm long. Clculte the volume of room, 3.5 m y 3 m y 2.1 m high. c Clculte the rdius of sell tht hs volume of 125 cm 3. d Clculte the height of cylinder tht is 20 cm in dimeter with volume of 2.5 litres (correct to the nerest unit). e Clculte the height of tringulr prism with se re of 128 mm 2 nd volume of 1024 mm 3. f Clculte the depth of wter in swimming pool tht hs cpcity of 56000 litres. The pool hs rectngulr dimensions of 8 metres y 5.25 metres. 22 The rtio of the volume of sphere to tht of cylinder of r similr dimensions, s shown in the digrm, is est expressed s: A 4 B 2 4 C 3 r 3 3 r h RECT CTED D 3 4 E 3 2 6 m f 19 m 42 m 4 m 60 m OFS 6 cm 42 m 23 If the volume of the squre-sed pyrmid shown is 6000 m 3, then the perimeter of the se is closest to: A 900 m B 20 m C 30 m D 80 m E 120 m V O VO = 20 m Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 575

MASTER 24 A tin of fruit is 13 cm high nd 10 cm in dimeter. Its volume, correct to 1 deciml plce, is: A 1021.0 cm 3 B 510.5 cm 3 C 1021.4 cm 3 D 1020.1 cm 3 E 4084.1 cm 3 25 The medicine cup shown hs the shpe of cone with dimeter 4 cm of 4 cm nd height of 5 cm (not including the cup s se). Clculte the volume of the cone correct to the nerest millilitre, where 1 cm 3 = 1 ml. 5 cm 11.8 Unit 4 AOS M3 Topic 1 Concept 5 Similrity Concept summry Prctice questions 26 Tennis lls hve dimeter of 6.5 cm nd re pckged in cylinder tht cn hold four tennis lls. Assuming the lls just fit inside ide the cylinder, clculte: the height of the cylindricl cn the volume of the cn (correct to 1 deciml plce) c the volume of the four tennis lls (correct to 1 deciml plce) d the volume of the cn occupied y ir e the frction of the cn s volume occupied y the lls. Similr figures Ojects tht hve the sme shpe ut different ent size re sid to e similr. For two figures to e similr, they must hve the following properties: 1. The rtios of the corresponding sides must e equl. A B AB = B C BC = C D CD = A D = common rtio AD 2. The corresponding ngles must e equl. A = A B = B C = C D = D OFS Scle fctor, k A mesure of the reltive size of the two similr figures is the scle fctor. The scle fctor is the common rtio of the corresponding sides nd quntifies the mount of 4 A' B' 2 B' 125 85 A' 2 D' 6 D' C' C' 60 B B 2 A 85 A 1 125 1 D D 3 C C 60 576 MATHS QUEST 12 FURTHER MATHEMATICS VCE Units 3 nd 4

KEU WORKED EXAMPLE 27 enlrgement or reduction one figure undergoes to trnsform into the other figure. The strting shpe is commonly referred to s the originl nd the trnsformed shpe s the imge. Scle fctor, k, is the mount of enlrgement or reduction nd is expressed s integers, frction or scle rtios. For exmple, k = 2, k = 1 or 1 : 10000. 12 B' length of imge Scle fctor, k = length of originl = A B AB = B C BC = C A B 9 9 CA where for enlrgements k is greter thn 1 nd for 3 3 reductions k is etween 0 nd 1. PROA A 1 C A' 3 C' For k = 1, the figures re exctly the sme shpe nd size nd re referred to s congruent. Enlrgements nd reductions re importnt in mny spects of photogrphy, mp mking nd modelling. Often, photogrphs re incresed in size (enlrged) to exmine fine detil without distortion, while house plns re n exmple of reduction to scle; for exmple, 1 : 25. RECT For the similr shpes shown: find the scle fctor for the reduction of the shpe find the unknown length in the smller shpe. 20 cm Originl 45 cm 10 cm Imge x Topic 11 GEOMETRY: SIMILARITY AND MENSURATION 577