Onlin rsours Auto-mrk hptr pr-tst Vio monstrtions o ll work xmpls Intrtiv wigts Intrtiv wlkthroughs Downlol HOTshts Ass to ll HOTmths Austrlin Curriulum ourss Ass to th HOTmths gms lirry Lngth, r, sur 5r n volum Wht you will lrn 5A Lngth n primtr REVISION 5B Cirumrn n primtr o stors 5C Ar REVISION 5D Primtr n r o omposit shps 5E Sur r o prisms n pyrmis 5F Sur r o ylinrs 5G Volum o prisms 5H Volum o ylinrs REVISION Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
NSW syllus STRAND: MEASUREMENT AND GEOMETRY SUBSTRANDS: AREA AND SURFACE AREA (S4, 5.1, 5.2, 5.3) VOLUME (S4, 5.2) Outoms A stunt lults th rs o omposit shps, n th sur rs o rtngulr n tringulr prisms. (MA5.1 8MG) A stunt lults th sur rs o right prisms, ylinrs n rlt omposit solis. (MA5.2 11MG) Boy Sur Ar Boy Sur Ar (BSA) is msur in squr mtrs n is th totl sur r o prson s oy. It is us in miin to lult th orrt rug os n th volum o luis tht r to ministr intrvnously. Clulting th sur r o prisms n pyrmis is rltivly sy, s w us th r ormuls or h, n thn in th totl. Howvr, lulting th tul sur r o th irrgulr soli, suh s prson, is iiult tsk. To in th sur r o prson th ollowing ormul n us. oy wight (kg) BSA (m 2 ) = 3600 hight (m) A stunt pplis ormuls to in th sur rs o right pyrmis, right ons, sphrs n rlt omposit solis. (MA5.3 13MG) A stunt pplis ormuls to lult th volums o omposit solis ompos o right prisms n ylinrs. (MA5.2 12MG) A stunt pplis ormuls to in th volums o right pyrmis, right ons, sphrs n rlt omposits solis. (MA5.3 14MG) NESA, 2012 Not tht th vrg BSA or 12 13-yr-ol is 1.3 2 n 1.7 m 2 or th vrg ult. Pitriins n to know hil s BSA to us Clrk s ormul or hil s os to lult th orrt os rquir y h ptint. Clrk s ormul stts: Chil s os = Sur r o hil (m 2 ) Ault os Avrg sur r o ult (1.7 m 2 ) Dotors n nurss n to l to prorm ths lultions quikly n urtly in orr to sv livs. Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
274 Chptr 5 Lngth, r, sur r n volum Pr-tst 1 Clult th ollowing. 2.3 10 0.048 1000 270 100 52 134 10 000 0.0005 100 000 72 160 1000 2 For ths si shps in th: i numr o squrs insi th shp (r) ii istn roun th outsi o th shp (primtr) 2 m 2 m 2 m 3 Nm this soli. Wht is th nm o th shp (sh) on top o th soli? 4 Estimt th r (numr o squrs) in ths shps. 5 Count th numr o us in ths solis to in th volum. 8 m 6 Convrt th ollowing to th units shown in th rkts. (mm) 20 m (m) 1.6 km (m) 2m (m) 3167 m (km) 72 m (m) g 20000 mm (m) h 0.03 km (m) 7 Fin th r o ths si shps. 10 m 4 m 1. 4 m 2 m 8 m 8 Fin th irumrn (C = 2πr) n r (A = πr 2 ) o this irl, rouning to 2 iml pls. 1 m 7 km 10 m 4 m Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
5A Lngth n primtr 275 5A Lngth n primtr REVISION Lngth is t th ountion o msurmnt rom whih th onpts o primtr, irumrn, r n volum r vlop. From th us o th royl uit (istn rom tip o mil ingr to th low) us y th nint Egyptins to th lultion o pi ( π ) y morn omputrs, units o lngth hv hlp to rt th worl in whih w liv. Units o lngth r ssntil in msuring istn, r n volum. Lt s strt: Not nough inormtion? All th ngls t h vrtx in this shp r 90 n th two givn lngths r 10 m n 1. Is thr nough inormtion to in th primtr o th shp? I thr is nough inormtion, in th primtr n isuss your mtho. I not, thn sy wht inormtion ns to provi. To onvrt twn mtri units o lngth, multiply or ivi y th pproprit powr o 10. km 1000 1000 m 100 100 Primtr is th istn roun los shp. Sis with th sm mrkings r o qul lngth. m 10 10 mm P = 2 5 + 3 = 1 1 10 m Stg 5.3# 5.3 5.3 5.2 5.2 5.1 4 Ky is Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
276 Chptr 5 Lngth, r, sur r n volum Exmpl 1 Fining primtrs o simpl shps Fin th primtr o h o th ollowing shps. SOLUTION 18 m 12 m Primtr = 2 12 + 2 18 = 24 + 36 = 60 m Primtr = (2 5) + 6 + 3 + 2 + 1 = 22 m Altrntiv mtho: Primtr = 2 6 + 2 5 EXPLANATION 6 m Two lngths o 12 m n two lngths o 18 m. Missing sis r: = 2 m 6 m = 1 m 6 m Exmpl 2 Fining missing sis givn primtr 2 m 1 m Th horizontl sis r h. Th vrtil sis r 6 m h. Fin th unknown si lngth in this shp with th givn primtr. SOLUTION 2 5 + 4 + x = 16 14 + x = 16 x = 2 Th missing lngth is 2 m x m 4 m Primtr = 16 m EXPLANATION A ll th lngths n st qul to th givn primtr. Solv or th unknown. Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
5A Lngth n primtr 277 Exmpl 1 Exmpl 1 Exris 5A REVISION UNDERSTANDING AND FLUENCY 1 Convrt th ollowing lngth msurmnts into th units givn in th rkts. (mm) 2.8 m (m) 521 mm (m) 83.7 m (m) 4.6 km (m) 2170 m (km) 2 A stl m is 8.2 long n 22.m wi. Writ own th imnsions o th m in ntimtrs. 3 Writ own th vlus o th pronumrls in ths shps. 10 m m m 7 m m m 4 Fin th primtr o h o th ollowing shps. 2 m 10 m 6 km 4 m 3 km 7 m 2. 8 m 12 m 5 Fin th primtr o h o th ollowing omposit shps. 1 1 8 m 18 m 6 m 6 m 12.57 m 4 m 10.7 m 10 m 1 7 m 7.4 m 1 7 3, 4 5(½), 6, 7 8(½) 5(½), 6(), 7 8(½) 3.1 m m 3.9 m m 30 m 8 mm 12 m 2.3 km 40 m 2 6 mm 30 m 1.8 km 5.9 m m 7 m 40 m Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
278 Chptr 5 Lngth, r, sur r n volum Exmpl 2 6 Fin th primtr o h o ths shps. You will n to onvrt th msurmnts to th sm units. Giv your nswrs in th units givn in r. 9 mm 0.4 950 m 1.8 m 400 m 1.315 km 201 mm 39 m 7 Convrt th ollowing msurmnts into th units givn in th rkts. 8 m (mm) 110000 mm (m) 0.00001 km (m) 0.02 m (mm) 28400 m (km) 62743000 mm (km) 8 Fin th unknown si lngth in ths shps with th givn primtrs. 2.1 m x m P = 14 m x km 7 km 2 km P = 25.5 km PROBLEM-SOLVING AND REASONING P = 6.4 m x m x m 10.1 m P = 36. 7.2 m 7 mm P = 3m x mm x m P = 33.6 m 9 A lion nlosur is m up o iv stright n stions. Thr stions r 20 m in lngth n th othr two stions r 15. n 32.. Fin th primtr o th nlosur. 10 Fin th primtr o ths shps. Assum ll ngls r right ngls. 20 m 1 4.1 m 2.7 m 11 Fin th vlu o x in ths shps with th givn primtr. 2x m x m x m 7 m P = 12.6 m P = 36 m 9, 10, 13 9 11, 13, 14 10 12, 14, 15 3x km 4.7 m 9. 5x km 4x km P = 84 km Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
5A Lngth n primtr 279 12 A photo 12 m wi n 20 m long is surroun with pitur rm wi. Fin th outsi primtr o th rm pitur. 13 Giv th rul or th primtr o ths shps using th givn pronumrls,.g. P = 3 + 2. x y 14 Explin why you o not n ny mor inormtion to in th primtr o this shp, ssuming ll ngls r right ngls. x 15 A pi o string 1 m long is ivi into th givn rtios. Fin th lngth o h prt. 1 : 3 2 : 3 5 : 3 1 : 2 : 3 : 4 ENRICHMENT Pitur rming 16 A squr pitur o si lngth 20 m is surroun y rm o with x m. Fin th primtr o th rm pitur i: i x = 2 ii x = 3 iii x = 5 Writ rul or th primtr P o th rm pitur in trms o x. Us your rul to in th primtr i: i x = 3.7 ii x = 7.05 Us your rul to in th vlu o x i th primtr is: i 90 m ii 102 m Is thr vlu o x or whih th primtr is 7? Explin. 10 m x 3 1 16 20 m x m Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
280 Chptr 5 Lngth, r, sur r n volum 5B Cirumrn n primtr o stors A portion o irl nlos y two rii n n r is ll stor. Th primtr o stor is m up o thr omponnts: two rii o th sm lngth n th irulr r. Givn n ngl, it is possil to in th lngth o th r using th rul or th irumrn o irl C = 2πr or C = π. Lt s strt: Primtr o stor A stor is orm y iviing irl with two rius uts. Th ngl twn th two rii trmins th siz o th stor. Th primtr will thror pn on oth th rius lngth n th ngl twn thm. Complt this tl to s i you n trmin rul or th primtr o stor. Rmmr tht th irumrn C o irl is givn y C = 2πr whr r is th rius lngth. 6 60 Shp Frtion o ull irl Working n nswr 3 4 5 q r 90 360 = 1 4 270 360 = P = REVISION P = 2 3 + 1 2π 3 10.71 4 P = P = P = Mjor stor Minor θ stor Stg 5.3# 5.3 Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400 5.3 5.2 5.2 5.1 4
5B Cirumrn n primtr o stors 281 Cirumrn o irl C = 2πr or C = π. 22 Us or 3.14 to pproximt π or us thnology or mor 7 pris lultions. A stor is portion o irl nlos y two rii n n r. Spil stors inlu th ollowing. A hl irl is ll smiirl. A qurtr irl is ll qurnt. Th primtr o stor is givn y P = 2r + 360 2πr. Th symol or pi (π) n us to writ n nswr xtly. For xmpl: C = 2πr = 2π 3 = 6π Exmpl 3 Fining th irumrn o irl n primtr o stor Fin th irumrn o this irl n primtr o this stor orrt to 2 iml pls. SOLUTION C = 2πr = 2 π 3 = (6π) m = 18.8 (to 2.p.) P = 2r + 360 2πr = 2 7 + 100 360 2 π 7 = 14 + 35π ( 9 ) m = 26.22 m (to 2.p.) 100 EXPLANATION 7 m r q 3 r r = 2r Us th ormul C = 2πr or C = π n sustitut r = 3 (or = 6 ). 6π woul th xt nswr n 18.85 is th roun nswr. Writ th ormul. Th rtion o th irl is 100 360 (or 5 18 ). 14 + 35π is th xt nswr. 9 Us lultor n roun to 2 iml pls. Ky is Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
282 Chptr 5 Lngth, r, sur r n volum Exmpl 4 Using xt vlus Giv th xt irumrn or primtr o ths shps. 2 4 SOLUTION C = 2πr = 2 π 4 = 8π P = 2 r + 360 2πr = 2 2 + 270 360 2π 2 = 4 + 3 4 4π = 4 + 3π Exris 5B REVISION UNDERSTANDING AND FLUENCY EXPLANATION Writ th ormul with r = 4. 2 π 4 = 2 4 π 1 Wht is th rius o irl with imtr 5.6 m? Wht is th imtr o irl with rius 48 mm? Writ th nswr xtly in trms o π. Us th ormul or th primtr o stor. Th ngl insi th stor is 270 so th rtion is 270 360 = 3 4. 4 + 3π nnot simplii urthr. 2 Simpliy ths xprssions to giv n xt nswr. Do not vlut with lultor or roun o. Th irst on is on or you. 2 3 π = 6π 6 2π 7 2.5 π 3 + 1 2 4π 2 6 + 1 4 12π 2 5 + 2 5 2 π 5 g 2 4 + 90 360 2 π 4 h 3 + 270 360 π i 7 + 30 360 π 3 Dtrmin th rtion o irl shown in ths stors. Writ th rtion in simplst orm. 1, 2 7(½) 2(½), 4 8(½) 5 8(½) 225 150 60 Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
5B Cirumrn n primtr o stors 283 Exmpl 3 Exmpl 3 4 Fin th irumrn o ths irls orrt to 2 iml pls. Us lultor or th vlu o π. 8 m m 4 km 5 Fin th primtr o ths stors orrt to 2 iml pls. 60 265 (Smiirl) 3.9 m 14 m 3. 130 (Qurnt) 2.8 m 15.9 km 6 Fin th irumrn o ths irls without lultor, using th givn pproximtion o π. 10 m 2 m π 3.14 π 3.14 7 mm 14 m π 22 π 22 Unorrt 3r smpl 7 pgs Cmrig Univrsity Prss Plmr, t l. 2018 7 978-1-108-46816-9 Ph 03 8671 1400
284 Chptr 5 Lngth, r, sur r n volum Exmpl 4 Exmpl 4 7 Giv th xt irumrn o ths irls. 10 7.5 8 Giv th xt primtr o ths stors. 60 PROBLEM-SOLVING AND REASONING 4 6 20 1.5 300 9 Fin th istn roun th outsi o irulr pool o rius 4., orrt to 2 iml pls. 10 Fin th lngth o string rquir to surroun th irulr trunk o tr tht hs imtr o 1., orrt to 1 iml pl. 11 Th n o ylinr hs rius o. Fin th irumrn o th n o th ylinr, orrt to 2 iml pls. 3 2 225 12 A whl o rius 30 m is roll in stright lin. Fin th irumrn o th whl orrt to 2 iml pls. How r, orrt to 2 iml pls, hs th whl roll tr omplting: i 2 rottions? ii 10.5 rottions? Cn you in how mny rottions woul rquir to ovr t lst 1 km in lngth? Roun to th nrst whol numr. 11 20 4 20.5 9, 10, 13 9 11, 13 10 12, 13(½), 14 30 m Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
5B Cirumrn n primtr o stors 285 13 Giv xt nswrs or th primtr o ths shps. Exprss nswrs using rtions. 11. 1.2 m 135 2.5 km 20 mm 7 m 11. 14 W know tht th rul or th irumrn o irl is C = 2πr. Fin rul or r in trms o C. Fin th rius o irl to 1 iml pl i its irumrn is: i 10 m ii 2 Giv th rul or th imtr o irl in trms o its irumrn C. Atr 1000 rottions whl hs trvll 2.12 km. Fin its imtr to th nrst ntimtr. ENRICHMENT Th rris whl 15 A lrg rris whl hs rius o 21 m. Roun to 2 iml pls or ths qustions. Fin th istn prson will trvl on on rottion o th whl. A ri inlus 6 rottions o th whl. Wht istn is trvll in on ri? How mny rottions woul rquir to ri istn o: i 500 m? ii 2 km? A rris whl hs sign tht rs, On ri o 10 rottions will ovr 2 km. Wht must th imtr o th whl? 15 W n in th istn trvll uring on rottion on rris whl y lulting th irumrn. Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
286 Chptr 5 Lngth, r, sur r n volum 5C Ar REVISION Th numr o squr ntimtrs in this rtngl is 6 ; thror th r is 6 m 2. A quikr wy to in th numr o squrs is to not tht thr r two rows o thr squrs n hn th r is 2 3 = 6 m 2. This ls to th ormul A = l or th r o rtngl. For mny ommon shps, suh s th prlllogrm n trpzium, th ruls or thir r n vlop through onsirtion o simpl rtngls n tringls. Shps tht involv irls or stors rly on lultions involving pi (π). Lt s strt: Formul or th r o stor W know tht th r o irl with rius r is givn y th rul A = πr 2. Complt this tl o vlus to vlop th rul or th r o stor. Shp Frtion o ull irl Working n nswr 5 21 A = π 2 2 12.57 units 2 4 3 180 360 = A = 1 2 A = 120 A = θ A = r How mny tils ovr this r? Stg 5.3# 5.3 2 m 5.3 5.2 5.2 5.1 4 Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
5C Ar 287 Convrsion o r units 1000 2 km 2 m 2 1000 2 100 2 100 2 m 2 10 2 10 2 mm 2 10 2 = 10 10 = 100 100 2 = 100 100 = 10000 1000 2 = 1000 1000 = 1000000 Th r o two-imnsionl shp is msur o th sp nlos within its ounris. Squr A = s 2 s Prlllogrm/Rhomus Kit x 1 A = 2 xy h A = h y Exmpl 5 Convrting units o r Rtngl l A = l Trpzium h 1 A = 2 h( + ) Cirl r A = πr 2 Convrt th ollowing r msurmnts into th units givn in th rkts. 859 mm 2 (m 2 ) 2.37 m 2 (m 2 ) SOLUTION EXPLANATION 859 mm 2 = 859 10 2 m 2 m 2 mm 2 = 8.59 m 2 10 2 = 100 Tringl h 1 A = 2 h y Rhomus x 1 A = 2 xy Stor θ r r θ A = 360 πr 2 859. 2.37 m 2 = 2.37 100 2 m 2 = 23700 m 2 100 2 = 10 000 m 2 m 2 2.3700 Ky is Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
288 Chptr 5 Lngth, r, sur r n volum Exmpl 6 Fining rs o rtngls, tringls n prlllogrms Fin th r o h o th ollowing pln igurs. 5 km 1. 7 m SOLUTION A = l = 7 3 = 21 m 2 A = 1 2 h = 1 2 11 5 = 27.5 km 2 A = h = 3 1.5 = 4. 2 11 km EXPLANATION Us th r ormul or rtngl. Sustitut l = 7 n = 3. Inlu th orrt units. Us th r ormul or tringl. Sustitut = 11 n h = 5. Inlu th orrt units. Us th r ormul or prlllogrm. Multiply th s lngth y th prpniulr hight. Exmpl 7 Fining rs o rhomuss n trpziums Fin th r o h o th ollowing pln igurs. 10 mm SOLUTION A = 1 2 x y A = 1 2 = 1 2 10 9 = 4m 2 9 mm h ( + ) = 1 3 (4 + 6) 2 = 1 2 EXPLANATION 4 m 6 m Us th r ormul or rhomus. Sustitut x = 10 n y = 9. Inlu th orrt units. Us th r ormul or trpzium. Sustitut = 4, = 6 n h = 3. Inlu th orrt units. Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
5C Ar 289 Exmpl 8 Fining rs o irls n stors Fin th r o this irl n stor orrt to 2 iml pls. SOLUTION 5.1 m A = πr 2 = π (5.1) 2 = 81.71 m 2 (to 2.p.) A = 360 πr2 = 260 360 π 42 = 13 18 π 16 = 36.30 m 2 (to 2.p.) Exris 5C REVISION UNDERSTANDING AND FLUENCY EXPLANATION 260 4 m Writ th rul n sustitut r = 5.1. 81.7128... rouns to 81.71 sin th thir iml pl is 2. Us th stor ormul. Th rtion o th ull irl is 260 360 = 13 18, so multiply this y πr 2 to gt th stor r. 104π m 2 woul th xt nswr 9 1 Fin th r o ths shps. Eh squr in h shp rprsnts on squr unit. 2 1 4 6 1 3, 4 9(½) 2, 4 9(½) 4 9(½) 17 9 Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
290 Chptr 5 Lngth, r, sur r n volum Exmpl 5 Exmpl 6 2 Nm th shp tht hs th givn r ormul. A = l A = πr 2 A = 1 xy (2 shps) 2 A = 360 πr2 A = 1 2 h A = 1 h( + ) 2 g A = h h A = l 2 A = 1 2 πr2 3 Wht rtion o ull irl is shown y ths stors? Simpliy your rtion. 120 60 225 4 Convrt th ollowing r msurmnts into th units givn in th rkts. 2 m 2 (mm 2 ) 500 mm 2 (m 2 ) 2.1 m 2 (m 2 ) 210 000 m 2 (m 2 ) 0.001 km 2 (m 2 ) 3200000 m 2 (km 2 ) 5 Fin th r o h o th ollowing pln igurs. 4 m 7 m 6 m 1. 12 m 4.2 m 8 km 100 5 km 9.2 m 4 m Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
5C Ar 291 Exmpl 7 Exmpl 8 6 Fin th r o h o th ollowing pln igurs. 7 mm 4 m 6 mm 11 m 9 m 1.7 m 4.2 m 6 m 3.1 m 7 Fin th r o h o th ollowing mix-pln igurs. 0. 6 m 3. 0.6 m 1.6 m 2 mm 2.m 4 mm 1.51 m 1.6 1.9 m 2 m 7 m 10 m 8 Fin th r o ths irls using th givn vlu or pi (π). Roun to 2 iml pls. π = 3.14 π = 22 7 4.9 km 6.2 km 8.3 km π (rom lultor) 2.6 m 7.9 m Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
292 Chptr 5 Lngth, r, sur r n volum Exmpl 8 9 Fin th r o ths stors, rouning to 2 iml pls. 10 m 45 7 m 6 m PROBLEM-SOLVING AND REASONING 4 m 190 147 10 Convrt th ollowing msurmnts into th units givn in th rkts. 1.5 km 2 (m 2 ) 0.00000 2 (mm 2 ) 75000 mm 2 (m 2 ) 11 A vlur tlls you tht your pi o ln hs n r o hl squr kilomtr (0.5 km 2 ). How mny squr mtrs (m 2 ) o you own? 12 A rtngulr prk ovrs n r o 175 000 m 2. Giv th r o th prk in km 2. 13 An ol pitur rm tht ws on squr now lns to on si to orm rhomus. I th istns twn pirs o opposit ornrs r 8 n 1.2 m, in th r nlos within th rm in m 2. 14 A pizz shop is onsiring inrsing th imtr o its mily pizz try rom 32 m to 34 m. Fin th prntg inrs in r, orrt to 2 iml pls, rom th 32 m try to th 34 m try. 15 A tnnis ourt r is illumint y our ornr lights. Th illumintion o th stor r los to h light is onsir to goo (G) whil th rmining r is onsir to lit stistorily (S). Wht prntg o th r is onsir goo? Roun to th nrst pr nt. 16 Th rul or th r o irl is givn y A = πr 2. Rrrng this rul to in rul or r in trms o A. Fin th rius o irl with th givn rs. Roun to 1 iml pl. i 2 ii 6.9 m 2 iii 20 km 2 7 m 10 12, 16 10, 11, 13, 14, 16 10, 13 17 G G 30 m S G G 12 m Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
5C Ar 293 17 A stor hs rius o. Fin th ngl, orrt to th nrst gr, i its r is: i 2 ii 2 2 Explin why th r o th stor oul not 30 m 2. ENRICHMENT Winows 18 Six squr winows o si lngth 2 m r to pl into 12 m wi y 8. high wll s shown. Th winows r to position so tht th vrtil sping twn th winows n th wll gs r qul. Similrly, th horizontl spings r lso qul. 2 m 2 m 12 m i Fin th horizontl istn twn th winows. ii Fin th vrtil istn twn th winows. Fin th r o th wll not inluing th winow sps. I th wll inlu 3 rows o 4 winows (inst o 2 rows o 3 ), invstigt i it woul possil to sp ll th winows so tht th horizontl n vrtil spings r uniorm (lthough not nssrily qul to h othr). 19 A rtngulr winow is wip y wipr l orming th givn stor shp. Wht prntg r is ln y th wipr l? Roun to 1 iml pl. Wip r 30 m 100 10 m 70 m θ 18, 19 8. 40 m Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
294 Chptr 5 Lngth, r, sur r n volum 5D Primtr n r o omposit shps A omposit shp n thought o s omintion o mor simplisti shps suh s tringls n rtngls. Fining primtrs n rs o suh shps is mttr o intiying th mor si shps thy onsist o n omining ny lultions in n orgnis shion. Th r o glss in this glss strutur n lult using th r o trpziums, tringls n rtngls. Lt s strt: Inorrt lyout Thr stunts writ thir solution to ining th r o this shp on th or. Eh stunt s working is shown in th tl low. 10 m Chris Mtt Moir A = l = 50 + 1 2 πr = 1 2 π 52 = 39.27 + 50 = 89.27 m 2 2 A = 1 2 π 52 = 39.27 + 10 5 = 89.27 m 2 A = l + 1 2 πr = 10 5 + 1 2 π 52 = 89.27 m 2 All thr stunts hv th orrt nswr ut only on stunt rivs ull mrks. Who is it? Explin wht is wrong with th lyout o th othr two solutions. 2 Stg 5.3# 5.3 5.3 5.2 5.2 5.1 4 Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
5D Primtr n r o omposit shps 295 Composit shps r m up o mor thn on si shp. Aition n/or sutrtion n us to in rs n primtrs o omposit shps. Th lyout o th rlvnt mthmtil working ns to mk sns so tht th rr o your work unrstns h stp. Exmpl 9 Fining primtrs n rs o omposit shps Fin th primtr n r o this omposit shp, rouning nswrs to 2 iml pls. SOLUTION P = 2 l + + 1 2 2πr 17 m EXPLANATION 14 m 3 stright sis + smiirl = 2 17 + 14 + 1 2π 7 2 Sustitut l = 17, = 14 n r = 7. = 34 + 14 + π 7 Simpliy. = 69.99 m (to 2.p) Clult n roun to 2 iml pls. A = l + 1 2 πr2 Ar o rtngl + r o smiirl = 17 14 + 1 2 π 72 Sustitut l = 17, = 14 n r = 7. = 238 + 1 π 49 2 Simpliy. = 314.97 m 2 (to 2.p) Clult n roun to 2 iml pls. Ky Exris 5D UNDERSTANDING AND FLUENCY 1, 2, 3 6(½) 2, 3 7(½) 3 7(½) 1 Nm th two irnt shps tht mk up ths omposit shps,.g. squr n smiirl. is Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
296 Chptr 5 Lngth, r, sur r n volum Exmpl 9 2 Copy n omplt th working to in th primtr n r o ths omposit shps. Roun to 1 iml pl. 12 m 2 m 10 m 12 m 6 m 20 m P = 2 + 3 + 1 2 2πr P = 20 + 12 + + + = + 3 + = = m A = h + 1 2 = 5 + 1 2 = + = m 2 = m A = l 1 2 h = 12 1 2 8 = = m 2 3 Fin th primtr n th r o h o ths simpl omposit shps, rouning nswrs to 2 iml pls whr nssry. 1 6 m 2 m 12 m 4 m 4 m 8 m 6 m 4 Fin th r o h o th ollowing omposit shps. 4 m 1. 2 m 3.0 m 1 m 18 m 2. 1 m 10 m 26 mm 4.8 m 3.92 m 1m 12 m 3. 6.2 m 20 mm 7 m 2. 10 m 7 m 0. Unorrt 3r 2.5 smpl m pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 1 Ph 03 8671 1400 8 m
5D Primtr n r o omposit shps 297 5 Fin th r o h o ths omposit shps. Hint: us sutrtion. 4 m 12 m 8 m 9 m 10 m 1 4 m 2 m 20 m 11 m 6 Fin th primtr n th r o h o th ollowing omposit shps orrt to 2 iml pls whr nssry. 2 m 6 m 4 mm 2 m 1 irl 4 1.8 m 6 m 20 m 6 m 6 m 2 m 10 m Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
298 Chptr 5 Lngth, r, sur r n volum 7 Fin th r o th sh rgion o h o th ollowing shps y sutrting th r o th lr shp rom th totl r. Roun to 1 iml pl whr nssry. g 18 m 2.6 m 3. m 7 mm 1. PROBLEM-SOLVING AND REASONING 10 m h 9 m 1.7 m 4 m 0.6 m i 1 3.2 mm 6 m 14 mm 9.01 m 8 An r o lwn is m up o rtngl msuring 10 m y 1 n smiirl o rius. Fin th totl r o lwn orrt to 2 iml pls. 9 Twnty irulr pis o pstry, h o imtr 4 m, r ut rom rtngulr lyr o pstry 20 m long n 16 m wi. Wht is th r, orrt to 2 iml pls, o pstry rmining tr th 20 pis r rmov? 10 Ths shps inlu stors. Fin thir r to 1 iml pl. 18 m 24 m 140 11 A nw r mnuturr is signing logo. It is in th shp o imon insi rtngl. Th imon is to hv horizontl with o n n r qul to on sixth o th r o th rtngl. Fin th rquir hight o th imon. 7 m 11 m 10 m 8 mm 8, 9, 12 9, 10, 12, 13 10, 11, 12(½), 13, 14 12 m 6 m Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
5D Primtr n r o omposit shps 299 12 Using xt vlus (.g.10 + 4π ), in th r o th shps givn in Qustion 6. 13 A irl o rius 10 m hs hol ut out o its ntr to orm ring. Fin th rius o th hol i th rmining r is 50% o th originl r. Roun to 1 iml pl. 14 Us Pythgors thorm (illustrt in this igrm) to hlp xplin why ths omposit shps inlu inorrt inormtion. ENRICHMENT Constrution ut-outs 10 m 14 m 12 m 15 Th ront o grnthr lok onsists o timr or with imnsions s shown. A irl o rius 20 m is ut rom th or to orm th lok. Fin th rmining r o th timr or orrt to 1 iml pl. 16 Th numr 10 is ut rom rtngulr pi o ppr. Th imnsions o th sign r shown low. 6 m 9 m 2 Fin th lngth n with o th rtngulr pi o ppr shown. Fin th sum o th rs o th two ut-out igits, 1 n 0, orrt to 1 iml pl. Fin th r o ppr rmining tr th igits hv n rmov (inlu th ntr o th 0 in your nswr) n roun to 1 iml pl. 2 = 2 + 2 8 m 15, 16 0.8 m 60 m 64 m 20 m Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
300 Chptr 5 Lngth, r, sur r n volum Ky is 5E Sur r o prisms n pyrmis Thr-imnsionl ojts or solis hv outsi surs tht togthr orm th totl sur r. Nts r vry hlpul or trmining th numr n shp o th surs o thrimnsionl ojt. For this stion w will l with right prisms n pyrmis. A right prism hs uniorm ross-stion with two intil ns n th rmining sis r rtngls. A right pyrmi hs its px sitting ov th ntr o its s. Right tringulr prism Tringulr ross-stion Lt s strt: Drwing prisms n pyrmis px s Right squr-s pyrmi Prisms r nm y th shp o thir ross-stion n pyrmis r nm y th shp o thir s. Try to rw s mny irnt right prisms n pyrmis s you n. Dsri th irnt kins o shps tht mk up th sur o your solis. Whih solis r th most iiult to rw n why? Th sur r (A) o soli is th sum o th rs o ll th surs. A nt is two-imnsionl illustrtion o ll th surs o soli. A right prism is soli with uniorm ross-stion n rmining sis r rtngls. Thy r nm y th shp o thir ross-stion. Th nts or rtngulr prism (uoi ) n squr-s pyrmi r shown hr. Soli Nt Sur r Rtngulr prism l h l h Squr-s pyrmi A = 2(l + lh + h) Stg 5.3# 5.3 h 1 A = 2 + 4 ( 2 h ) = 2 + 2h 5.3 5.2 5.2 5.1 4 Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
5E Sur r o prisms n pyrmis 301 Exmpl 10 Clulting sur r Fin th sur r o h o th ollowing soli ojts. 2 m SOLUTION A = 2 (5 3) + 2 (5 2) + 2 (2 3) = 30 + 20 + 12 = 62 m 2 A = 2 2 + 4 1 2 2 3 2 m EXPLANATION top si s 2 m 2 m 2 m = 4 + 12 = 16 m 2 2 m Exris 5E UNDERSTANDING AND FLUENCY 2 m 1 Drw suitl nt or ths prisms n pyrmis n nm h soli. 1(½), 2, 3 4(½), 6 2, 3 5(½), 6 3 5(½), 6 Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
302 Chptr 5 Lngth, r, sur r n volum Exmpl 10 Exmpl 10 2 Copy n omplt th working to in th sur r o ths solis. 7 m 8 m 7 m A = 2 8 7 + 2 8 + 2 = + + = m 2 4 m A = 2 1 4 + 5 7 + 4 + 2 = + + + = m 2 3 Fin th sur r o th ollowing rtngulr prisms. Drw nt o th soli to hlp you. 2 m 4 m 8 m 2 m 2.4 m 7 m 10 m 5.9 m 2 m 3.2 m 4 Fin th sur r o h o ths pyrmis. Drw nt o th soli to hlp you. m m m 8 m 8 m 10 m 5 Fin th sur r o h o th ollowing soli ojts. 4 m 1.7 m 3.2 m 5.1 m 1.8 m 1.7 m 4 m 12 m 7 m 6 m 3. 0. 6.2 m 3. 2 m 1 m 2 m 4 m 2 m 2 m 2 m 2 m Unorrt 6 Fin th 3r sur smpl pgs r o Cmrig u o si Univrsity lngth Prss 1 mtr. Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400 4 m 0.6 m
5E Sur r o prisms n pyrmis 303 PROBLEM-SOLVING AND REASONING 7, 8, 11 8, 9, 11 9 12 7 A rtngulr ox is to ovr in mtril. How muh is rquir to ovr th ntir ox i it hs th imnsions 1., 1. n 1.9 m? 8 Two woon oxs, oth with imnsions 80 m, 1 m n 2, r pl on th groun, on on top o th othr s shown. Th ntir outsi sur is thn pint. Fin th r o th pint sur. 2 9 Th our wlls n roo o rn (shown) r to pint. Fin th sur r o th rn, not inluing th loor. I 1 litr o pint ovrs 10 m 2, in how mny litrs r rquir to omplt th jo. 3. 1 m 80 m 2 1.9 m 10 An opn top rtngulr ox 20 m wi, 2 long n 10 m high is m rom woo 1 m thik. Fin th sur r: outsi th ox (o not inlu th top g) insi th ox (o not inlu th top g) 2 2 m 2.4 m 1 m thik 11 Drw th stk o 1 m u loks tht givs th minimum outsi sur r n stt this sur r i thr r: 2 loks 4 loks 8 loks 4 m 20 m 10 m Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
304 Chptr 5 Lngth, r, sur r n volum 12 Cus o si lngth on unit r stk s shown. 1 u 2 us 3 us t. Complt this tl. Numr o us (n) 1 2 3 4 5 6 7 8 9 Sur r (A) Cn you in th rul or th sur r (A) or n us stk in this wy? Writ own th rul or A in trms o n. Invstigt othr wys o stking us n look or ruls or sur r in trms o n, th numr o us. ENRICHMENT Pythgors rquir 13 For prisms n pyrmis involving tringls, Pythgors thorm ( 2 = 2 + 2 ) n us. Apply th thorm to hlp in th sur r o ths solis. Roun to 1 iml pl. 10 m 2 m 6 m 9 m 8 m 13 7 m 4 m 7 m 4 m m Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
5F Sur r o ylinrs 5F 305 5F Sur r o ylinrs Th nt o ylinr inlus two irls n on rtngl. Th lngth o th rtngl is qul to th irumrn o th irl. h Lt s strt: Curv r r r h r irumrn r h r Cirumrn = 2πr Roll pi o ppr to orm th urv sur o ylinr. Do not stik th ns togthr so you n llow th ppr to rturn to lt sur. Wht shp is th ppr whn lying lt on tl? Whn urv to orm th ylinr, wht o th sis o th rtngl rprsnt on th ylinr? How os this hlp to in th sur r o ylinr? Ths pips r hollow ylinrs, lik th roll-up ppr shts. r Stg 5.3# 5.3 Sur r o ylinr = 2 irls + 1 rtngl r = 2 πr 2 + 2πr h A = 2πr 2 + 2πrh h 2 irulr ns urv r or A = 2πr(r + h) 5.3 5.2 5.2 5.1 4 Ky is Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
306 Chptr 5 Lngth, r, sur r n volum Exmpl 11 Fining th sur r o ylinr Fin th sur r o this ylinr, rouning to 2 iml pls. SOLUTION A = 2πr 2 + 2πrh = 2 π 5 2 + 2π 5 25 = 50 π + 250 π = 50π + 250π = 300π = 942.48 m 2 (to 2.p.) Exmpl 12 Fining sur rs o ylinril portions Fin th sur r o this hl-ylinr, rouning to 2 iml pls. SOLUTION 1 A = 2 ( 2 πr2 ) + 1 2 (2πr) 8 + 4 8 = 2 1 2 π 22 + 1 2 2 π 2 8 + 32 = 20π + 32 = 94.8 2 (to 2.p.) Exris 5F UNDERSTANDING AND FLUENCY EXPLANATION In ition to th urv sur, inlu th rtngulr s n th smiirulr ns. 4 m 1 2(2 πr) m 4 m 8 m 1 Drw nt suit to ths ylinrs. Ll th sis using th givn msurmnts. C = 26 C = 31.4 4.1 EXPLANATION 2πr 2 32 5 11 2 8 m 1 5 2 6 3(), 4(), 5, 6 Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
5F Sur r o ylinrs 307 Exmpl 11 Exmpl 12 2 Th urv sur o ths ylinrs is llow to lttn out to orm rtngl. Wht woul th lngth n rth o this rtngl in h ylinr? Roun to 2 iml pls whr nssry. C = 22 m 16 m 10 m 2 m 8 m 3 Fin th sur r o ths ylinrs, rouning to 2 iml pls. Us nt to hlp. 1 m 2 m 10 m 4 Fin th sur r o ths ylinrs, rouning to 1 iml pl. Rmmr tht th rius o irl is hl th imtr. 6 m 12 m 7 m 2 m 4 m 10 m 5 Fin th sur r o ylinril plsti ontinr o hight 18 m n with irl o rius t h n, orrt to 2 iml pls. 6 Fin th r o th urv sur only or ths ylinrs, orrt to 2 iml pls. 8 m m 2.4 m 10 m PROBLEM-SOLVING AND REASONING 11 mm 7 Fin th sur r o ths solis, rouning to 2 iml pls. 10 m 2 m 6.2 m 3.8 m 4 m 10 m 20 m 4 m 2 m 3.2 m 7(½), 8, 11 7(½), 8, 9, 11, 12 7(½), 9 12 12 km 6 m Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 6 km Ph 03 8671 1400
308 Chptr 5 Lngth, r, sur r n volum g h i 9 m 16 m 6 m 7 m 18 m 6 m 4 mm m 2 mm m 8 A wtr trough is in th shp o hl-ylinr. Its smiirulr ns hv imtr 40 m n th trough lngth is 1 m. Fin th outsi sur r in m 2 o th urv sur plus th two smiirulr ns, orrt to 2 iml pls. 9 A log with imtr 60 m is in lngth. Its hollow ntr is 20 m in imtr. Fin th sur r o th log in m 2, inluing th ns 20 m n th insi, orrt to 1 iml pl. 60 m 10 A ylinril rollr is us to prss rush rok in prprtion or tnnis ourt. Th rtngulr tnnis ourt r is 30 m long n 1 wi. Th rollr hs rth o 1 m n imtr 60 m. Fin th sur r o th urv prt o th rollr in m 2 orrt to 60 m 3 iml pls. Fin th r, in m 2 to 2 iml pls, o rush rok tht n prss tr: i 1 rvolution ii 20 rvolutions Fin th minimum numr o omplt rvolutions rquir to prss th ntir tnnis ourt r. 11 It is mor pris to giv xt vlus or lultions involving π,.g. 24π. Giv th xt nswrs to th sur r o th ylinrs in Qustion 3. 12 A ylinr ut in hl givs hl th volum ut not hl th sur r. Explin why. ENRICHMENT Soli stors 13 Th stor r rul A = 360 πr2 n ppli to in th sur rs o solis tht hv ns tht r stors. Fin th xt sur r o ths solis. 225 80 13 2 m 145 1 m 6 m 2 m 1 m Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
5G Volum o prisms 309 5G Volum o prisms Volum is th numr o ui units ontin within thr-imnsionl ojt. To in th volum w n ount 2 m 24 ui ntimtrs (24 m 3 ) or 4 m multiply 3 4 2 = 24 m 3. 4 m W n s tht th r o th s (3 4 = 12 m 2 ) lso givs th volum o th s lyr 1 m high. Th numr o lyrs quls th hight, hn, multiplying th r o th s y th hight will giv th volum. This i n ppli to ll right prisms provi uniorm ross-stion n intii. In suh solis, th hight lngth us to lult th volum is th lngth o th g running prpniulr to th s or ross-stion. Lt s strt: Cui units Consir this 1 m u ivi into ui millimtrs. How mny ui mm sit on on g o th 1 m u? How mny ui mm sit on on lyr o th 1 m u? How mny lyrs r thr? How mny ui mm r thr in totl in th 1 m u? Complt this sttmnt 1 m 3 = mm 3 Explin how you n in how mny: m 3 in 1 m 3 m 3 in 1 km 3 1 m = 10 mm 1 m = 10 mm 1 m = 10 mm Common mtri units or volum inlu ui kilomtrs (km 3 ), ui mtrs (m 3 ), ui ntimtrs (m 3 ) n ui millimtrs (mm 3 ). 1000 3 100 3 10 3 1000 3 = 1000 000 000 km m 3 100 3 = 1000 000 1000 3 100 3 10 3 10 3 = 1000 For pity ommon units inlu: mglitrs (ML) 1 ML = 1000 kl kilolitrs (kl) 1 kl = 1000 L litrs (L) 1 L = 1000 ml millilitrs (ml) Also 1 m 3 = 1 ml so 1 L = 1000 m 3 n 1 m 3 = 1000 L Stg 5.3# 5.3 5.3 5.2 5.2 5.1 4 Ky is Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
310 Chptr 5 Lngth, r, sur r n volum Ky is Volum o solis with uniorm ross-stion: V = r o ross-stion (A) prpniulr hight (h). V = A h Th hight is th lngth o th g tht runs prpniulr to th ross-stion. Som ommon ormuls or volum inlu: Rtngulr prism (uoi) l V = A h = l h = lh h (hight) Exmpl 13 Convrting units o volum h 1 Tringulr prism V = A h 1 = ( 2 h 1 ) h 2 1 = ( 2 h 1 ) h 2 Convrt th ollowing volum msurmnts into th units givn in th rkts. 2. 3 (m 3 ) 458 mm 3 (m 3 ) SOLUTION 2. 3 = 2.5 100 3 = 2500000 m 3 458 mm 3 = 458 10 3 = 0.458 m 3 Cross-stion h Cross-stion EXPLANATION 100 3 = 1 000 000 h 2 (hight) m 3 2.500000 m m 3 10 3 = 1000 458. h Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
5G Volum o prisms 311 Exmpl 14 Fining volums o prisms n othr solis Fin th volum o h o ths thr-imnsionl ojts. Ar = 10 m 2 1 m 1 m SOLUTION Exris 5G UNDERSTANDING AND FLUENCY 4 m 1 Drw th ross-stionl shp or ths prisms n stt th hight (prpniulr to th ross-stion). V= l h = 1 1 3 = 3 V = r o ross-stion prp. hight = 10 5 = 50 m 3 V = r o ross-stion prp. hight 1 = ( 2 h 1 ) h 2 = 1 (4 3) 6 2 = 36 m 2 6 m m 8 mm 10 m 10 m EXPLANATION 8 m 14 m 6 m Th soli is rtngulr prism. Lngth = 1 m, rth = 1 m n hight = Sustitut ross-stionl r = 10 n hight = 5. Th ross-stion is tringl. h 1 m m h 2 m 1, 2, 3(½), 4, 5(½), 6, 7 2, 3(½), 4, 5(½), 6 8 3(½), 5(½), 6, 8 2 m 10 m Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
312 Chptr 5 Lngth, r, sur r n volum Exmpl 13 Exmpl 14 Exmpl 14 2 Wht is th nm givn to th shp o th sh ross-stion o h o th ollowing solis? 3 Convrt th ollowing volum msurmnts into th units givn in th rkts. 3 (mm 3 ) 2000 mm 3 (m 3 ) 8.7 m 3 (m 3 ) 5900 m 3 (m 3 ) 0.00001 km 3 (m 3 ) 21700 m 3 (km 3 ) g 3 L (ml) h 0.2 kl (L) i 3500 ml (L) j 0.021 L (ml) k 37000 L (kl) l 42900 kl (ML) 4 Fin th volums o ths thr-imnsionl rtngulr prisms. 2 m 2 m 2 m 6 m 2 m 7 m m 7 mm 5 Fin th volum o h o ths thr-imnsionl ojts. Th ross-stionl r hs n givn. Ar = 2 m 2 2 g 2 m 5.8 m 2 18 m 2 6 m h 3. 4.8 m 3.1 m 2 4.2 m 2 i 7.1 m 2.6 m 2 4.6 m 0.372 m 2 3.9 2.07 m 2 1.027 m 1 mm Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
5G Volum o prisms 313 Exmpl 14 6 Fin th volums o ths prisms. 10 m 4 m 7 Fin th volum o h o ths rtngulr prisms (uois). 2 m 2 m 4 m 3. 3. 3. 8 Fin th volum o ths solis onvrting your nswr to litrs. 10 m 20 m 40 m PROBLEM-SOLVING AND REASONING 8 m 18 m 2. 1 m 12 m 1 m 6 m 1 m 10 m 9 A rik is 10 m wi, 20 m long n 8 m high. How muh sp woul iv o ths riks oupy? 10 How muh ir sp is ontin insi rtngulr ror ox tht hs th imnsions 8 y 62 m y 36 m. Answr using ui mtrs (m 3 ) orrt to 2 iml pls. 11 25 L o wtr is pour into rtngulr ish tnk whih is 50 m long, 20 m wi n 20 m high. Will it ovrlow? 12 Fin th volum o h o th ollowing solis, rouning to 1 iml pl whr nssry. 4 m 2.1 km 4 m 11.2 m 2 m 2 m 4 m 0.72 m 2 m 1.1 9, 10, 15 10 12, 15 11 16 7 m 1.07 m 21 mm 2.5 km 48 mm 1.7 km 29 mm Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
314 Chptr 5 Lngth, r, sur r n volum 13 Us units or pity to in th volum o ths solis in litrs. 2 1 40 m 14 Th givn igrm is skth o nw 2 swimming pool to instll in shool sports omplx. Fin th r o on si o th pool (sh). Fin th volum o th pool in litrs. 15 Wht singl numr o you multiply y to onvrt rom: i L to m 3? ii L to m 3? iii ml to mm 3? Wht singl numr o you ivi y to onvrt rom: i mm 3 to L? ii m 3 to ML? iii m 3 to kl? 16 Writ ruls or th volum o ths solis using th givn pronumrls. A rtngulr prism with lngth = rth = x n hight h. A u with si lngth s. A rtngulr prism with squr s (with si lngth t ) n hight 6 tims th si lngth o th s. ENRICHMENT Volum o pyrmi 1 1. 1 17 Erlir w look t ining th sur r o right pyrmi lik th on shown hr. Imgin th pyrmi sitting insi prism with th sm s. Mk n ut guss s to wht rtion o th prism s volum is th pyrmi s volum. Us th intrnt to in th tul nswr to prt. Drw som pyrmis n in thir volum using th rsults rom prt. 2 m 2. 2 10 m 17 2. Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
5H Volum o ylinrs 315 5H Volum o ylinrs Thnilly, ylinr is not right prism us its sis r not rtngls. It os, howvr, hv uniorm ross-stion ( irl) n so ylinr s volum n lult in similr wy to tht o right prism. Cylinril ojts r ommonly us to stor gss n liquis n so working out th volum o ylinr is n importnt msurmnt lultion. Lt s strt: Writing th rul Prviously w us th ormul V = A h to in th volum o solis with uniorm ross-stion. Disuss ny similritis twn th two givn solis. How n th rul V = A h vlop urthr to in th rul or th volum o ylinr? Exmpl 15 Fining th volum o ylinr Fin th volum o ths ylinrs orrt to 2 iml pls. SOLUTION Th volum o ylinr is givn y: V = πr 2 h or V = πr 2 h, whr r is th rius o th irulr ns h is th lngth or istn twn th irulr ns. 10 m V = πr 2 h = π (3) 2 10 = 90π = 282.74 m 3 (to 2.p.) V = πr 2 h = π (0.9) 2 0.3 = 0.76 m 3 (to 2.p.) Th volum o liqui in this tnkr n lult using th volum ormul or ylinr. 0. EXPLANATION 1.8 m Sustitut r = 3 n h = 10 into th rul. 90π m 3 woul th xt nswr. Inlu volum units. Th imtr is 1.8 m so r = 0.9. A h Cylinr r h Stg 5.3# 5.3 5.3 5.2 5.2 5.1 4 Ky is Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
316 Chptr 5 Lngth, r, sur r n volum Exmpl 16 Fining th pity o ylinr Exmpl 15 Fin th pity in litrs o ylinr with rius 30 m n hight 90 m. Roun to th nrst litr. SOLUTION V = πr 2 h = π (30) 2 90 = 254469 m 3 = 254 L (to th nrst litr) Exris 5H UNDERSTANDING AND FLUENCY 1 Stt th rius n th hight o ths ylinrs. 4 m 2.6 m 2 10 m 18 m EXPLANATION Sustitut r = 30 n h = 90. Thr r 1000 m 3 in 1 L so ivi y 1000 to onvrt to litrs, thn roun to th nrst litr. 11.6 m 15.1 m 11.1 m 2.9 m 2 Fin th r o ths irls orrt to 2 iml pls. 10 m 2 m 1.6 m 1(½), 2, 3, 4 5(½) 12.8 m 10.4 m 21. 3 Convrt th ollowing into th units givn in th rkts. Rmmr, 1 L = 1000 m 3 n 1 m 3 = 1000 L. 2000 m 3 (L) 4. 3 (ml) 3.7 L (m 3 ) 1 m 3 (L) 38000 L (m 3 ) 0.0002 m 3 (ml) 4 Fin th volum o ths ylinrs orrt to 2 iml pls. 1 m 3, 4 5(½) 2. 6 m 8 m 8 m 3.8 m 6 m 2 m 1.8 km 4 5(½) 7 m 1.2 m Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
5H Volum o ylinrs 317 Exmpl 16 5 Fin th pity in litrs o ths ylinrs. Roun to th nrst litr. Rmmr, 1 L = 1000 m 3 n 1 m 3 = 1000 L. 1 50 m 1. 2 m 2 PROBLEM-SOLVING AND REASONING 40 m 100 m 30 m 6 A ylinril wtr tnk hs rius o 2 m n hight o 2 m. Fin its pity in m 3 roun to 3 iml pls. Fin its pity in L roun to th nrst litr. 10 m 1. 80 m 7 How mny litrs o gs n tnkr rry i its tnk is ylinril with 2 m imtr n is 12 m in lngth? Roun to th nrst litr. 8 Whih hs iggr volum n wht is th irn in volum to 2 iml pls: u with si lngth 1 m or ylinr with rius 1 m n hight 0.? 9 Fin th volum o ths ylinril portions orrt to 2 iml pls. 20 m 1.2 m 8 m 30 m 50 m 1.8 m 2 m 6, 7, 10 7, 8, 9(½), 10, 11(½) 8 10, 11(½), 12 12 m 1 6 m 12 m 10 Th rul or th volum o ylinr is V = πr 2 h. Show how you oul us this rul to in, orrt to 3 iml pls: h i V = 20 n r = 3 r i V = 100 n h = 5 Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
318 Chptr 5 Lngth, r, sur r n volum 11 Using xt vlus (.g. 20π ) in th volum o ths solis. 8 m 4 m 10 m 1 m 20 m 20 m 10 m 6 km 8 mm 12 Drw ylinr with its irumrn qul to its hight. Try to rw it to sl. ENRICHMENT Soli stors 13 You will rll tht th r o stor is givn y A = 360 πr 2. Us this t to hlp in th volum o ths solis orrt to 2 iml pls. 45 2.2 m 10 m 6 m 1.8 m 60 8 m 4 m 20 m 120 2 6 m 30 4 m 20 m 20 m 8 m 7 m 12 km m 9 mm 13 θ r Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
t Invstigtion 319 Cpity n pth Fining pity Fin th pity in litrs o ths ontinrs (i.. in th totl volum o lui thy n hol). 30 m 20 m 10 m Dsigning ontinrs 1 m 30 m Dsign ontinr with th givn shp tht hs 10 -litr pity. You will n to stt th imnsions lngth, rth, hight, rius t. n lult its pity. rtngulr prism ylinr Fining pth Th pth o wtr in prism n oun i th s (ross-stionl) r n volum o wtr r givn. Consir uoi, s shown, with 2.4 litrs o wtr. To in th pth o wtr: Convrt th volum to m 3 : 2.4 L = 2.4 1000 Fin th pth: Volum = r o s = 2400 m 3 2400 = 20 20 2400 = 400 = 6 pth is 6 m Us th ov mtho to in th pth o wtr in ths prisms. 40 m 60 m 80 m Volum = 96 L 20 m 40 m Volum = 8 L Volums o o-shp ojts 30 m Rmmr: 1 millilitr (ml) o lui oupis 1 m 3 o sp; thror, 1 litr (L) oupis 1000 m 3 s thr r 1000 ml in 1 litr. pth () 60 m 20 m 150 m Volum = 500 L 10 m 20 m Som solis my pulir in shp n thir volum my iiult to msur. A rr pi o rok is pl into ylinril jug o wtr n th wtr pth riss rom 10 m to 11 m. Th rius o th jug is. i Fin th r o th irulr s o th ylinr. ii Fin th volum o wtr in th jug or th rok is pl in th jug. iii Fin th volum o wtr in th jug inluing th rok. iv Hn in th volum o th rok. Us th prour outlin in prt i iv ov to in th volum o n ojt o your hoi. Explin n show your working n ompr your rsults with thos o othr stunts in your lss i thy r msuring th volum o th sm ojt. Invstigtion Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
320 Chptr 5 Lngth, r, sur r n volum Puzzls n hllngs 1 A 100 m 2 tory lt roo s ll th wtr ollt to rinwtr tnk. I thr is 1 mm o rinll, how mny litrs o wtr go into th tnk? 2 Wht is th rltionship twn th sh n non-sh rgions in this irulr igrm? 3 A got is tthr to th ntr o on si o sh with 10 m lngth o rop. Wht r o grss n th got grz? 16 m 8 m Sh 4 A rin gug is in th shp o tringulr prism with imnsions s shown. Wht is th pth o wtr whn it is hl ull? 20 m 10 m 5 A rtngulr ish tnk hs s r 0. 2 n hight 30 m n is ill with 80 L o wtr. Tn lrg ish with volum 50 m 3 h r pl into th tnk. By how muh os th wtr ris? 8 m 6 I it tks 4 popl 8 ys to knit 2 rugs, how mny ys will it tk or 1 prson to knit 1 rug? 7 Fin th rul or th volum o ylinr in trms o r only i th hight is qul to its irumrn. 8 Th sur r o ylinr is 2π squr units. Fin rul or h in trms o r. 9 Giv th imnsions o ylinr tht n hol on litr o liqui. Whn signing ontinrs or ylinrs, wht tors n to onsir? 8 m Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
Chptr summry 321 Units 1000 3 100 3 10 3 km 3 3 mm 1000 3 100 3 10 3 1 L = 1000 ml = 1000 m 3 Volum Prism or ylinrs V = Ah A = r o ross-stion h = hight V = πr 2 h V = lh = π(2) 2 7 = 11 6 5 = 28π = 330 m 3 = 87.96 m 3 2 m 7 m 6 m 11 m Squr-s prism or pyrmi Us nt to hlp in th sum o ll sur rs. A = 2 + 4 ( 1 h) 2 = 3 2 + 4 1 3 5 = 39 m 2 2 Cylinr 4 m A = 2πr 2 + 2πrh = 2π(3) 2 + 2π(3)(4) = 18π + 24π = 42π = 131.9 2 Cirl irumrn C = 2πr or π Lngth, r, sur r n volum Sur r Cirl A = πr 2 stor A = θ πr 2 360 = 120 360 π 52 120 = 1 π 25 3 = 26.18 m 2 Ar Lngth n primtr Stor P = 2r + θ 2πr 360 = 2 3 + 90 360 2π 3 = 6 + 1.5π (xt) = 10.71 m (2.p.) Units Tringl A = 1 h 2 Quriltrls Squr A = l 2 Rtngl A = l Prlllogrm A = h Trpzium A = 1 h ( + ) 2 Rhomus or kit A = 1 xy 2 Composit shps 8 m 10 m A = l 1 πr 2 2 = 10 8 1 π 4 2 2 = 80 8π = 54.87 m 2 1 P = 2 10 + 8 + 2π(4) 2 = 28 + 4π = 40.57 m 1000 2 100 2 10 2 km 2 m 2 m 2 2 mm 1000 2 100 2 10 2 Chptr summry Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
322 Chptr 5 Lngth, r, sur r n volum Multipl-hoi qustions Chptr rviw 1 I th r o squr il is 25 km 2, its primtr is: A 10 km B 20 km C 5 km D 50 km E 25 km 2 2.7 m 2 is th sm s: A 270 m 2 B 0.0027 km 2 C 27000 m 2 D 2700 mm 2 E 27 m 2 3 Th primtr o this shp is: A 21 m B 14 m C 24 m D 20 m E 22 m 4 A prlllogrm hs r 10 m 2 n s. Its prpniulr hight is: A 50 m B 2 m C 50 m 2 D 2 m 2 E 0. 5 This omposit shp oul onsir s rtngl with n r in th shp o trpzium rmov. Th shp s r is: A 16 km 2 4 km B 12 km 2 1 km C 20 km 2 3 km 5 km D 24 km 2 E 6 km 2 6 A smiirulr gol r hs imtr 20 m. Its primtr orrt to th nrst mtr is: A 41 m B 36 m C 8 D 51 m E 52 m 7 Th sur r o th pyrmi shown is: A 18 2 B 10 2 C 6 2 D 100 m 2 E 12 2 8 Th r o th urv sur only o hl-ylinr with rius m n hight 12 mm is losst to: A 942.m 2 B 94.2 mm 2 C 377 mm 2 D 471.2 mm 2 E 188.m 2 8 m 9 A prism s ross-stionl r is 100 m 2. I its volum is 6500 m 3, th prism s totl hight woul : A 0.6 B 650000 m C 6 D 6. E 650 m 10 Th xt volum o ylinr with rius n hight 10 m is: A 60π m 2 B 80π m C 45π m 3 D 30π m 3 E 90π m 3 Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
Chptr rviw 323 Short-nswr qustions 1 Convrt th ollowing msurmnts into th units givn in rkts. 3.8 m (m) 1.27 km (m) 27m 2 (m 2 ) 5.2 m 2 (m 2 ) 0.01 m 3 (m 3 ) 53 100 mm 3 (m 3 ) g 3100 ml (L) h 0.043 L (ml) i 2.83 kl (L) 2 Fin th primtr o h o th ollowing shps. 18 mm 6 m 1m 3 Fin th r o h o th ollowing pln igurs. 2 m 3.7 mm 8.2 mm 8 m 2 m 51 mm 3.6 m 22 mm 2 m 1.7 m 4. 4 Insi rtngulr lwn r o lngth 10. n with 3.8 m, nw grn is to onstrut. Th grn is to th shp o tringl with s 2 m n hight 2.. With th i o igrm, in th r o th: grn lwn rmining roun th grn 5 Fin th r n irumrn or primtr o h o th ollowing shps orrt to 2 iml pls. 1 m 3.7 m Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
324 Chptr 5 Lngth, r, sur r n volum 6 Fin th primtr n r o h o ths stors. Roun to 2 iml pls. 4 km 60 2.1 m 120 7 Fin th primtr n r o h o th ollowing omposit shps orrt to 2 iml pls. 2. 1. 8 Fin th sur r o h o th ollowing soli ojts. 1 m 4 m 10 m 14.14 m 8. 9 Fin th sur r o h o th ollowing soli ojts orrt to 2 iml pls. m 16 mm Hl-ylinr 10 Fin th volum o h o ths soli ojts, rouning to 2 iml pls whr nssry. Ar = 2 10 mm 6 m 6 m 2 mm Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400
Chptr rviw 325 Extn-rspons qustions 1 An oi rivs iv nw sks with nh shp m up o rtngl n qurtr-irl s shown. Th g o th nh is lin with rur strip t ost o $2.50 pr mtr. 1 m 80 m Fin th lngth o th rur ging strip in ntimtrs or on sk orrt to 2 iml pls. By onvrting your nswr in prt to mtrs, in th totl ost o th rur strip or th iv sks. Roun to th nrst ollr. Th mnuturr lims tht th sk top r sp is mor thn 1. 2. Fin th r o th sk top in m 2 orrt to 2 iml pls. Convrt your nswr to m 2 n trmin whthr or not th mnuturr s lim is orrt. 2 Cirulr stl riling o imtr 6 m is to us to n th si o rig. Th riling is hollow n th rius o th hollow irulr sp is 2 m. thiknss By ing th givn inormtion to this igrm o th ross-stion o th riling, trmin th thiknss o th stl. Dtrmin, orrt to 2 iml pls, th r o stl in th ross-stion. Eight lngths o riling t 10 m h r rquir or th rig. Using your rsult rom prt, in th volum o stl rquir or th rig in m 3. Convrt your nswr in prt to m 3. Th urv outsi sur o th stl rilings is to pint to hlp prott th stl rom th wthr. Fin th outr irumrn o th ross-stion o th riling orrt to 2 iml pls. Fin th sur r in m 2 o th ight lngths o riling tht r to pint. Roun to th nrst m 2. Th ost o th riling pint is $80 pr m 2. g Using your nswr rom prt, in th ost o pinting th rig rils to th nrst ollr. Unorrt 3r smpl pgs Cmrig Univrsity Prss Plmr, t l. 2018 978-1-108-46816-9 Ph 03 8671 1400