Volumes of common solids

Transcription

1 Capte 7 Volumes of common solids 7.1 Intoduction Te volume of any solid is a measue of te space occupied by te solid. Volume is measued in cubic units suc as mm,cm and m. Tis capte deals wit finding volumes of common solids; in engineeing it is often impotant to be able to calculate volume o capacity to estimate, say, te amount of liquid, suc as wate, oil o petol, in diffeent saped containes. A pism is a solid wit a constant coss-section and wit two ends paallel. Te sape of te end is used to descibe te pism. Fo example, tee ae ectangula pisms (called cuboids), tiangula pisms and cicula pisms (called cylindes). On completing tis capte you will be able to calculate te volumes and suface aeas of ectangula and ote pisms, cylindes, pyamids, cones and spees, togete wit fusta of pyamids and cones. Volumes of simila sapes ae also consideed. 7. Volumes and suface aeas of common sapes 7..1 Cuboids o ectangula pisms A cuboid is a solid figue bounded by six ectangula faces; all angles ae igt angles and opposite faces ae equal. A typical cuboid is sown in Figue 7.1 wit lengt l, beadt b and eigt. and Volume of cuboid = l b suface aea = b + l + lb = (b + l + lb) Figue 7.1 b A cube is a squae pism. If all te sides of a cube ae x ten Volume = x and suface aea = 6x Poblem 1. A cuboid as dimensions of 1cm by 4cm by cm. Detemine (a) its volume and (b) its total suface aea Te cuboid is simila to tat in Figue 7.1, wit l = 1cm, b = 4cmand = cm. (a) Volume of cuboid= l b = 1 4 = 144 cm (b) Suface aea = (b + l + lb) = ( ) = ( ) = 96 = 19 cm Poblem. An oil tank is te sape of a cube, eac edge being of lengt 1.5 m. Detemine (a) te maximum capacity of te tank in m and lites and (b) its total suface aea ignoing input and output oifices l DOI: /B

2 Volumes of common solids 41 (a) Volume of oil tank=volume of cube = 1.5m 1.5m 1.5m = 1.5 m =.75 m 1m = 100cm 100cm 100cm = 10 6 cm. Hence, volume of tank = cm 1lite = 1000cm, ence oil tank capacity = lites = 75 lites 1000 (b) Suface aea of one side = 1.5m 1.5m =.5m. A cube as six identical sides, ence total suface aea of oil tank = 6.5 = 1.5m Poblem. A wate tank is te sape of a ectangula pism aving lengt m, beadt 75cm and eigt 500mm. Detemine te capacity of te tank in (a) m (b) cm (c) lites Capacity means volume; wen dealing wit liquids, te wod capacity is usually used. Te wate tank is simila in sape to tat in Figue 7.1, wit l = m, b = 75 cm and = 500mm. (a) Capacity of wate tank = l b. To use tis fomula, all dimensions must be in te same units. Tus, l = m, b = 0.75m and = 0.5m (since 1m= 100cm = 1000mm). Hence, (b) (c) capacity of tank = = 0.75 m 1m = 1m 1m 1m = 100cm 100cm 100cm i.e., 1m = =10 6 cm. Hence, capacity = 0.75m = cm = cm 1lite = 1000cm. Hence, cm = 750,000 = 750 lites Cylindes A cylinde is a cicula pism. A cylinde of adius and eigt is sown in Figue 7.. Figue 7. Volume = π Cuved suface aea = π Total suface aea = π+ π Total suface aea means te cuved suface aea plus te aea of te two cicula ends. Poblem 4. A solid cylinde as a base diamete of 1cm and a pependicula eigt of 0cm. Calculate (a) te volume and (b) te total suface aea ( ) 1 (a) Volume = π = π 0 = 70π = 6 cm (b) Total suface aea = π + π = ( π 6 0) + ( π 6 ) = 40π + 7π = 1π = 980 cm Poblem 5. A coppe pipe as te dimensions sown in Figue 7.. Calculate te volume of coppe in te pipe, in cubic metes. 5 cm 1 cm Figue 7..5 m

3 4 Basic Engineeing Matematics Oute diamete, D = 5cm = 0.5m and inne diamete, d = 1cm = 0.1m. Aea of coss-section of coppe = π D πd 4 4 = π(0.5) π(0.1) 4 4 = = 0.078m Hence, volume of coppe 1 mm 16 mm 40 mm = (coss-sectional aea) lengt of pipe = = m 7.. Moe pisms A igt-angled tiangula pism is sown in Figue 7.4 wit dimensions b, and l. Figue 7.5 Te solid sown in Figue 7.5 is a tiangula pism. Te volume V of any pism is given by V = A, wee A is te coss-sectional aea and is te pependicula eigt. Hence, volume = = 840mm =.840 cm (since 1cm = 1000mm ) Poblem 7. Calculate te volume of te igt-angled tiangula pism sown in Figue 7.6. Also, detemine its total suface aea I b Figue cm and Volume = 1 bl suface aea = aea of eac end + aea of tee sides Notice tat te volume is given by te aea of te end (i.e. aea of tiangle = 1 b) multiplied by te lengt l. In fact, te volume of any saped pism is given by te aea of an end multiplied by te lengt. Poblem 6. Detemine te volume (in cm )ofte sape sown in Figue 7.5 A 6cm B C 8cm Figue 7.6 Volume of igt-angled tiangula pism = 1 bl = i.e. volume = 960 cm

4 Volumes of common solids 4 Total suface aea = aea of eac end + aea of tee sides. In tiangle ABC, AC = AB + BC fom wic, AC = AB + BC = = 10cm Hence, total suface aea ( ) 1 = b + (AC 40) + (BC 40) + (AB 40) = (8 6) + (10 40) + (8 40) + (6 40) = i.e. total suface aea = 1008 cm Poblem 8. Calculate te volume and total suface aea of te solid pism sown in Figue 7.7 Figue 7.7 4cm 5cm 11 cm 5cm 5cm 15 cm Te solid sown in Figue 7.7 is a tapezoidal pism. Volume of pism = coss-sectional aea eigt = 1 (11 + 5)4 15 = 15 = 480 cm Suface aea of pism = sum of two tapeziums + 4 ectangles = ( ) + (5 15) + (11 15) + (5 15) = = 454 cm Now ty te following Pactice Execise Pactice Execise 105 Volumes and suface aeas of common sapes (answes on page 51) 1. Cange a volume of cm to cubic metes.. Cange a volume of 5000mm to cubic centimetes.. A metal cube as a suface aea of 4 cm. Detemine its volume. 4. A ectangula block of wood as dimensions of 40mm by 1mm by 8mm. Detemine (a) its volume, in cubic millimetes (b) its total suface aea in squae millimetes. 5. Detemine te capacity, in lites, of a fis tank measuing90cmby60cmby1.8m, given 1lite = 1000cm. 6. A ectangula block of metal as dimensions of 40mm by 5mm by 15mm. Detemine its volume in cm. Find also its mass if te metal as a density of 9g/cm. 7. Detemine te maximum capacity, in lites, of a fis tank measuing 50cm by 40cm by.5m(1lite = 1000cm ). 8. Detemine ow many cubic metes of concete ae equied fo a 10m long pat, 150mm wide and 80mm deep. 9. A cylinde as a diamete 0mm and eigt 50mm. Calculate (a) its volume in cubic centimetes, coect to 1 decimal place (b) te total suface aea in squae centimetes, coect to 1 decimal place. 10. Find (a) te volume and (b) te total suface aea of a igt-angled tiangula pism of lengt 80cm and wose tiangula end as a base of 1cm and pependicula eigt 5cm. 11. A steel ingot wose volume is m is olled out into a plate wic is 0mm tick and 1.80m wide. Calculate te lengt of te plate in metes.

5 44 Basic Engineeing Matematics 1. Te volume of a cylinde is 75cm. If its eigt is 9.0cm, find its adius. 1. Calculate te volume of a metal tube wose outside diamete is 8cm and wose inside diamete is 6cm, if te lengt of te tube is 4m. 14. Te volume of a cylinde is 400cm. If its adius is 5.0cm, find its eigt. Also detemine its cuved suface aea. 15. A cylinde is cast fom a ectangula piece of alloy 5cm by 7cm by 1cm. If te lengt of te cylinde is to be 60cm, find its diamete. 16. Find te volume and te total suface aea of a egula exagonal ba of metal of lengt m if eac side of te exagon is 6 cm. 17. A block of lead 1.5m by 90cm by 750mm is ammeed out to make a squae seet 15mm tick. Detemine te dimensions of te squae seet, coect to te neaest centimete. 18. How long will it take a tap dipping at a ate of 800mm /s to fill a -lite can? 19. A cylinde is cast fom a ectangula piece of alloy 5.0cm by 6.50cm by 19.cm. If te eigt of te cylinde is to be 5.0cm, detemine its diamete, coect to te neaest centimete. 0. How muc concete is equied fo te constuction of te pat sown in Figue 7.8, if te pat is 1 cm tick? 7..4 Pyamids Volume of any pyamid = 1 aea of base pependicula eigt A squae-based pyamid is sown in Figue 7.9 wit base dimension x by x and te pependicula eigt of te pyamid. Fo te squae-base pyamid sown, Figue 7.9 volume = 1 x x Poblem 9. A squae pyamid as a pependicula eigt of 16cm. If a side of te base is 6cm, detemine te volume of a pyamid Volume of pyamid = 1 aea of base pependicula eigt x m 8.5 m = 1 (6 6) 16 = 19 cm 1. m.1 m Poblem 10. Detemine te volume and te total suface aea of te squae pyamid sown in Figue 7.10 if its pependicula eigt is 1cm. Volume of pyamid.4 m = 1 (aea of base) pependicula eigt Figue 7.8 = 1 (5 5) 1 = 100 cm

6 Volumes of common solids 45 A fom wic, = = 7cm i.e. pependicula eigt of pyamid = 7cm 7..5 Cones A cone is a cicula-based pyamid. A cone of base adius and pependicula eigt is sown in Figue cm D B C 5cm E Volume = 1 aea of base pependicula eigt Figue 7.10 Te total suface aea consists of a squae base and 4 equal tiangles. Aea of tiangle ADE = 1 base pependicula eigt l = 1 5 AC Te lengt AC may be calculated using Pytagoas teoem on tiangle ABC, wee AB = 1 cm and BC = 1 5 =.5cm. AC = Hence, AB + BC = = 1.6cm aea of tiangle ADE = = 0.65cm Total suface aea of pyamid = (5 5) + 4(0.65) = cm Poblem 11. A ectangula pism of metal aving dimensions of 5cm by 6cm by 18cm is melted down and ecast into a pyamid aving a ectangula base measuing 6cm by 10cm. Calculate te pependicula eigt of te pyamid, assuming no waste of metal Volume of ectangula pism= = 540cm Volume of pyamid Hence, = 1 aea of base pependicula eigt 540 = 1 (6 10) Figue 7.11 i.e. Volume = 1 π Cuved suface aea = πl Total suface aea = πl + π Poblem 1. Calculate te volume, in cubic centimetes, of a cone of adius 0mm and pependicula eigt 80mm Volume of cone = 1 π = 1 π 0 80 = mm 1cm= 10mm and 1cm = 10mm 10mm 10mm = 10 mm,o 1mm = 10 cm Hence, mm = cm i.e., volume = cm

7 46 Basic Engineeing Matematics Altenatively, fom te question, = 0mm = cmand = 80mm = 8 cm. Hence, volume = 1 π = 1 π 8 = cm Poblem 1. Detemine te volume and total suface aea of a cone of adius 5cm and pependicula eigt 1cm Te cone is sown in Figue 7.1. Poblem 14. Find te volume and suface aea of a spee of diamete 10cm Since diamete= 10cm, adius, = 5cm. Volume of spee = 4 π = 4 π 5 = 5.6 cm Suface aea of spee = 4π = 4 π 5 = 14. cm 1 cm l Poblem 15. Te suface aea of a spee is 01.1cm. Find te diamete of te spee and ence its volume Figue 7.1 5cm Suface aea of spee= 4π. Hence, 01.1cm = 4 π, Volume of cone = 1 π = 1 π 5 1 = 14. cm Total suface aea=cuved suface aea+aea of base = πl + π Fom Figue 7.1, slant eigt l may be calculated using Pytagoas teoem: l = = 1cm Hence, total suface aea = (π 5 1) + (π 5 ) = 8.7 cm. fom wic = π = 16.0 and adius, = 16.0 = 4.0cm fom wic, diamete = = 4.0 = 8.0 cm Volume of spee = 4 π = 4 π (4.0) = 68.1 cm Now ty te following Pactice Execise 7..6 Spees Fo te spee sown in Figue 7.1: Volume = 4 π and suface aea = 4π Pactice Execise 106 Volumes and suface aeas of common sapes (answes on page 51) 1. If a cone as a diamete of 80mm and a pependicula eigt of 10mm, calculate its volume in cm and its cuved suface aea. Figue 7.1. A squae pyamid as a pependicula eigt of 4cm. If a side of te base is.4cm long, find te volume and total suface aea of te pyamid.

8 Volumes of common solids 47. A spee as a diamete of 6cm. Detemine its volume and suface aea. Cylinde 4. If te volume of a spee is 566cm, find its adius. Volume = π Total suface aea = π + π 5. A pyamid aving a squae base as a pependicula eigt of 5cm and a volume of 75cm. Detemine, in centimetes, te lengt of eac side of te base. Tiangula pism 6. A cone as a base diamete of 16mm and a pependicula eigt of 40mm. Find its volume coect to te neaest cubic millimete. 7. Detemine (a) te volume and (b) te suface aea of a spee of adius 40mm. 8. Te volume of a spee is 5cm. Detemine its diamete. b Pyamid I Volume = 1 bl Suface aea = aea of eac end + aea of tee sides 9. Given te adius of te eat is 680km, calculate, in engineeing notation (a) its suface aea in km. (b) its volume in km. A Volume = 1 A Total suface aea = sum of aeas of tiangles foming sides + aea of base 10. An ingot wose volume is 1.5m is to be made into ball beaings wose adii ae 8.0cm. How many beaings will be poduced fom te ingot, assuming 5% wastage? Cone l Volume = 1 π Cuved suface aea = πl Total suface aea = πl + π 7. Summay of volumes and suface aeas of common solids Spee A summay of volumes and suface aeas of egula solids is sown in Table 7.1. Table 7.1 Rectangula pism (o cuboid) Volumes and suface aeas of egula solids Volume = 4 π Suface aea = 4π 7.4 Moe complex volumes and suface aeas Hee ae some woked poblems involving moe complex and composite solids. b l Volume = l b Suface aea = (b + l + lb) Poblem 16. A wooden section is sown in Figue Find (a) its volume in m and (b) its total suface aea

9 48 Basic Engineeing Matematics F 5 8cm Figue 7.14 m 1 cm 15.0 cm 15.0 cm 15.0 cm 15.0 cm (a) (b) Te section of wood is a pism wose end compises a ectangle and a semicicle. Since te adius of te semicicle is 8 cm, te diamete is 16 cm. Hence, te ectangle as dimensions 1 cm by 16cm. Aea of end = (1 16) + 1 π8 = 9.5cm Volume of wooden section = aea of end pependicula eigt = = cm = m, since 1 m = 10 6 cm = m Te total suface aea compises te two ends (eac of aea 9.5cm ), tee ectangles and a cuved suface (wic is alf a cylinde). Hence, total suface aea = ( 9.5) + (1 00) Figue cm D G A E H 5.40 cm Using Pytagoas teoem on tiangle BEF gives BF = EB + EF (BF fom wic EF = EB ) = = 14.64cm Volume of pyamid = 1 (aea of base)(pependicula eigt) C B + (16 00) + 1 (π 8 00) = π = 0 15 cm o.015 m Poblem 17. A pyamid as a ectangula base.60cm by 5.40cm. Detemine te volume and total suface aea of te pyamid if eac of its sloping edges is 15.0cm Te pyamid is sown in Figue To calculate te volume of te pyamid, te pependicula eigt EF is equied. Diagonal BD is calculated using Pytagoas teoem, [.60 i.e. BD = ] = 6.490cm Hence, EB = BD = =.45cm = 1 ( )(14.64) = cm Aea of tiangle ADF (wic equals tiangle BCF) = 1 (AD)(FG), wee G is te midpoint of AD. Using Pytagoas teoem on tiangle FGA gives FG = [ ] = 14.89cm Hence, aea of tiangleadf = 1 (.60)(14.89) = 6.80cm Similaly, if H is te mid-point of AB, FH = = 14.75cm Hence, aea of tiangle ABF (wic equals tiangle CDF) = 1 (5.40)(14.75) = 9.8cm

10 Volumes of common solids 49 Total suface aea of pyamid = (6.80) + (9.8) + (.60)(5.40) = = 15.7 cm Poblem 18. Calculate te volume and total suface aea of a emispee of diamete 5.0 cm Volume of emispee = 1 (volume of spee) = π = ( ) 5.0 π Total suface aea =.7 cm = cuved suface aea + aea of cicle = 1 (suface aea of spee) + π = 1 (4π ) + π ( 5.0 = π + π = π = π = 58.9cm Poblem 19. A ectangula piece of metal aving dimensions 4cm by cm by 1cm is melted down and ecast into a pyamid aving a ectangula base measuing.5cm by 5cm. Calculate te pependicula eigt of te pyamid ) Volume of ectangula pism of metal = 4 1 = 144cm Poblem 0. A ivet consists of a cylindical ead, of diamete 1cm and dept mm, and a saft of diamete mm and lengt 1.5cm. Detemine te volume of metal in 000 suc ivets Radius of cylindical ead= 1 cm = 0.5cm and eigt of cylindical ead= mm= 0.cm. Hence, volume of cylindical ead = π = π(0.5) (0.) = cm Volume of cylindical saft ( ) 0. = π = π (1.5) = cm Total volume of 1 ivet= = 0.04cm Volume of metal in 000 suc ivets = = cm Poblem 1. A solid metal cylinde of adius 6 cm and eigt 15 cm is melted down and ecast into a sape compising a emispee sumounted by a cone. Assuming tat 8% of te metal is wasted in te pocess, detemine te eigt of te conical potion if its diamete is to be 1cm Volume of cylinde = π = π 6 15 = 540π cm If 8% of metal is lost ten 9% of 540π gives te volume of te new sape, sown in Figue Volume of pyamid = 1 (aea of base)(pependicula eigt) Assuming no waste of metal, 144 = 1 (.5 5)(eigt) i.e. pependicula eigt of pyamid = = 4.56 cm Figue cm

11 50 Basic Engineeing Matematics Hence, te volume of (emispee + cone) = π cm ( ) i.e. 1 4 π + 1 π = π Dividing tougout by π gives + 1 = Since te diamete of te new sape is to be 1cm, adius = 6cm, ence (6) + 1 (6) = = i.e. eigt of conical potion, = = 9.4 cm (c) Aea of cicle = π o πd 4 ence, 0.11= πd 4 (4 ) 0.11 fom wic, d = =0.780cm π i.e. diamete of coss-section is.780 mm. Poblem. A boile consists of a cylindical section of lengt 8 m and diamete 6 m, on one end of wic is sumounted a emispeical section of diamete 6m and on te ote end a conical section of eigt 4m and base diamete 6m. Calculate te volume of te boile and te total suface aea Te boile is sown in Figue P Poblem. A block of coppe aving a mass of 50kg is dawn out to make 500m of wie of unifom coss-section. Given tat te density of coppe is 8.91g/cm, calculate (a) te volume of coppe, (b) te coss-sectional aea of te wie and (c) te diamete of te coss-section of te wie (a) A density of 8.91g/cm means tat 8.91g of coppe as a volume of 1cm, o 1g of coppe as a volume of (1 8.91)cm. Density = mass volume fom wic volume = mass density 8m 4m Figue 7.17 Volume of emispee, 6m Q A m B R I C P = π = π = 18π m (b) Hence, 50kg, i.e g, as a volume = mass density = cm = 561 cm Volume of wie = aea of cicula coss-section lengt of wie. Hence, 561cm = aea ( cm) fom wic, aea = cm = 0.11 cm Volume of cylinde, Q = π = π 8 = 7π m Volume of cone, R = 1 π = 1 π 4 = 1π m Total volume of boile = 18π + 7π + 1π = 10π = 0.4 m Suface aea of emispee, P = 1 (4π ) = π = 18πm

12 Volumes of common solids 51 Cuved suface aea of cylinde, Q = π = π 8 = 48π m Te slant eigt of te cone, l, is obtained by Pytagoas teoem on tiangle ABC, i.e. l = (4 + ) = 5 Cuved suface aea of cone, R = πl = π 5 = 15π m Total suface aea of boile= 18π + 48π + 15π = 81π = 54.5 m Now ty te following Pactice Execise Pactice Execise 107 Moe complex volumes and suface aeas (answes on page 51) 1. Find te total suface aea of a emispee of diamete 50 mm.. Find (a) te volume and (b) te total suface aea of a emispee of diamete 6 cm.. Detemine te mass of a emispeical coppe containe wose extenal and intenal adii ae 1 cm and 10 cm, assuming tat 1cm of coppe weigs 8.9g. 4. A metal plumb bob compises a emispee sumounted by a cone. If te diamete of te emispee and cone ae eac 4 cm and te total lengt is 5cm, find its total volume. 5. A maquee is in te fom of a cylinde sumounted by a cone. Te total eigt is 6mand te cylindical potion as a eigt of.5m wit a diamete of 15m. Calculate te suface aea of mateial needed to make te maquee assuming 1% of te mateial is wasted in te pocess. 6. Detemine (a) te volume and (b) te total suface aea of te following solids. (i) a cone of adius 8.0cm and pependicula eigt 10 cm. (ii) a spee of diamete 7.0cm. (iii) a emispee of adius.0cm. (iv) a.5cm by.5cm squae pyamid of pependicula eigt 5.0cm. (v) a 4.0cm by 6.0cm ectangula pyamid of pependicula eigt 1.0cm. (vi) a 4.cm by 4.cm squae pyamid wose sloping edges ae eac 15.0cm (vii) a pyamid aving an octagonal base of side 5.0cm and pependicula eigt 0cm. 7. A metal spee weiging 4kg is melted down and ecast into a solid cone of base adius 8.0cm. If te density of te metal is 8000kg/m detemine (a) te diamete of te metal spee. (b) te pependicula eigt of te cone, assuming tat 15% of te metal is lost in te pocess. 8. Find te volume of a egula exagonal pyamid if te pependicula eigt is 16.0cmand te side of te base is.0cm. 9. A buoy consists of a emispee sumounted by a cone. Te diamete of te cone and emispee is.5m and te slant eigt of te cone is 4.0m. Detemine te volume and suface aea of te buoy. 10. A petol containe is in te fom of a cental cylindical potion 5.0m long wit a emispeical section sumounted on eac end. If te diametes of te emispee and cylinde ae bot 1. m, detemine te capacity of te tank in lites (1lite = 1000cm ). 11. Figue 7.18 sows a metal od section. Detemine its volume and total suface aea cm adius.50 cm Figue m

13 5 Basic Engineeing Matematics 1. Find te volume (in cm ) of te die-casting sown in Figue Te dimensions ae in millimetes Figue ad 1. Te coss-section of pat of a cicula ventilation saft is sown in Figue 7.0, ends AB and CD being open. Calculate (a) te volume of te ai, coect to te neaest lite, contained in te pat of te system sown, neglecting te seet metal tickness (given 1lite = 1000cm ). (b) te coss-sectional aea of te seet metal used to make te system, in squae metes. (c) te cost of te seet metal if te mateial costs pe squae mete, assuming tat 5% exta metal is equied due to wastage. 7.5 Volumes and suface aeas of fusta of pyamids and cones Te fustum of a pyamid o cone is te potionemaining wen a pat containing te vetex is cut off by a plane paallel to te base. Te volume of a fustum of a pyamid o coneis given by te volume of te wole pyamid o cone minus te volume of te small pyamid o cone cut off. Te suface aea of te sides of a fustum of a pyamid o cone is given by te suface aea of te wole pyamid o cone minus te suface aea of te small pyamid o cone cut off. Tis gives te lateal suface aea of te fustum. If te total suface aea of te fustum is equied ten te suface aea of te two paallel ends ae added to te lateal suface aea. Tee is an altenative metod fo finding te volume and suface aea of a fustum of a cone. Wit efeence to Figue 7.1, Figue 7.1 I R Volume = 1 π(r + R + ) Cuved suface aea = πl(r + ) Total suface aea = πl(r + )+ π + πr A 500 mm B m Poblem 4. Detemine te volume of a fustum of a cone if te diamete of te ends ae 6.0cm and 4.0cm and its pependicula eigt is.6cm Figue 7.0 C 800 mm D 1.5 m 1.5 m (i) Metod 1 A section toug te vetex of a complete cone is sown in Figue 7.. AP Using simila tiangles, DP = DR BR Hence, AP.0 = fom wic AP = (.0)(.6) 1.0 = 7.cm

14 Volumes of common solids 5 A Poblem 5. Find te total suface aea of te fustum of te cone in Poblem 4. (i) Metod 1 Cuved suface aea of fustum = cuved suface aea of lage cone cuved suface aea of small cone cut off. Fom Figue 7., using Pytagoas teoem, AB = AQ + BQ D 4.0 cm.0 cm P E fom wic AB = [ ] = 11.1cm and AD = AP + DP B 1.0 cm Figue 7. R.0 cm Q 6.0 cm Te eigt of te lage cone= = 10.8cm Volume of fustum of cone = volume of lage cone volume of small cone cut off C.6 cm = 1 π(.0) (10.8) 1 π(.0) (7.) = = 71.6 cm (ii) Metod Fom above, volume of te fustum of a cone = 1 π(r + R + ) wee R =.0cm, =.0cm and =.6cm Hence, volume of fustum = 1 [ ] π(.6) (.0) + (.0)(.0) + (.0) = 1 π(.6)(19.0) = 71.6 cm fom wic AD = [ ] = 7.47cm Cuved suface aea of lage cone = πl = π(bq)(ab) = π(.0)(11.1) = cm and cuved suface aea of small cone = π(dp)(ad) = π(.0)(7.47) = 46.94cm Hence, cuved suface aea of fustum = = 58.71cm Total suface aea of fustum = cuved suface aea + aea of two cicula ends = π(.0) + π(.0) = = 99.6 cm (ii) Metod Fom page 5, total suface aea of fustum = πl(r + ) + π + π R wee l = BD = =.74cm, R =.0cmand =.0 cm. Hence, total suface aea of fustum = π(.74)(.0 +.0) + π(.0) + π(.0) = 99.6 cm

15 54 Basic Engineeing Matematics Poblem 6. A stoage oppe is in te sape of a fustum of a pyamid. Detemine its volume if te ends of te fustum ae squaes of sides 8.0mand 4.6 m, espectively, and te pependicula eigt between its ends is.6m Te fustum is sown saded in Figue 7.(a) as pat of a complete pyamid. A section pependicula to te base toug te vetex is sown in Figue 7.(b). 4.6 m 8.0 m P U Q 4.6 m R O T 8.0 m S Figue cm C OT = 1.7m (same as AH in Figue 7.(b) and OQ =.6m. By Pytagoas teoem, 8.0 m Figue cm 8.0 m (a) B. m A H 1.7 m. m (b) G D. m.6 m F E 4.0 m QT = (OQ + OT ) = [ ] =.98m Aea of tapezium PRSU = 1 ( )(.98) = 5.07m Lateal suface aea of oppe = 4(5.07) = 100. m By simila tiangles CG BG = BH AH Fom wic, eigt ( ) BH CG = BG = (.)(.6) = 4.87m AH 1.7 Heigt of complete pyamid = = 8.47m Volume of lage pyamid = 1 (8.0) (8.47) = m Volume of small pyamid cut off = 1 (4.6) (4.87) = 4.5m Poblem 8. A lampsade is in te sape of a fustum of a cone. Te vetical eigt of te sade is 5.0cm and te diametes of te ends ae 0.0cm and 10.0 cm, espectively. Detemine te aea of te mateial needed to fom te lampsade, coect to significant figues Te cuved suface aea of a fustum of a cone = πl(r + ) fom page 5. Since te diametes of te ends of te fustum ae 0.0cm and 10.0cm, fom Figue 7.5, cm Hence, volume of stoage oppe = = 146. m Poblem 7. Detemine te lateal suface aea of te stoage oppe in Poblem cm I Te lateal suface aea of te stoage oppe consists of fou equal tapeziums. Fom Figue 7.4, R cm 5.0 cm Aea of tapezium PRSU = 1 (PR + SU )(QT ) Figue 7.5

16 Volumes of common solids 55 = 5.0cm, R = 10.0cm and l = [ ] = 5.50cm fom Pytagoas teoem. Hence, cuved suface aea = π(5.50)( ) = 101.7cm i.e., te aea of mateial needed to fom te lampsade is 100 cm, coect to significant figues. Poblem 9. A cooling towe is in te fom of a cylinde sumounted by a fustum of a cone, as sown in Figue 7.6. Detemine te volume of ai space in te towe if 40% of te space is used fo pipes and ote stuctues Figue m 5.0 m 1.0 m 0.0 m Volume of cylindical potion = π ( ) 5.0 = π (1.0) = 5890m Volume of fustum of cone = 1 π(r + R + ) wee and Hence, volume of fustum of cone = = 18.0m, R = 5.0 = 1.5m = 1.0 = 6.0m. = 1 π(18.0) [ (1.5) + (1.5)(6.0) + (6.0) ] = 508m Total volume of cooling towe = = 10 98m If 40% of space is occupied ten volume of ai space = = 6557 m Now ty te following Pactice Execise Pactice Execise 108 Volumes and suface aeas of fusta of pyamids and cones (answes on page 5) 1. Te adii of te faces of a fustum of a cone ae.0cm and 4.0cm and teticknessofte fustum is 5.0cm. Detemine its volume and total suface aea.. A fustum of a pyamid as squae ends, te squaes aving sides 9.0cm and 5.0cm, espectively. Calculate te volume and total suface aea of te fustum if te pependicula distance between its ends is 8.0cm.. A cooling towe is in te fom of a fustum of a cone. Te base as a diamete of.0m,te top as a diamete of 14.0 m and te vetical eigt is 4.0m. Calculate te volume of te towe and te cuved suface aea. 4. A loudspeake diapagm is in te fom of a fustum of a cone. If te end diametes ae 8.0 cm and 6.00 cm and te vetical distance between te ends is 0.0 cm, find te aea of mateial needed to cove te cuved suface of te speake. 5. A ectangula pism of metal aving dimensions 4.cm by 7.cm by 1.4cm is melted down and ecast into a fustum of a squae pyamid, 10% of te metal being lost in te pocess. If te ends of te fustum ae squaes of side cm and 8cm espectively, find te tickness of te fustum. 6. Detemine te volume and total suface aea of a bucket consisting of an inveted fustum of a cone, of slant eigt 6.0cm and end diametes 55.0cm and 5.0cm. 7. A cylindical tank of diamete.0 m and pependicula eigt.0 m is to be eplaced by a tank of te same capacity but in te fom of a fustum of a cone. If te diametes of te ends of te fustum ae 1.0m and.0m, espectively, detemine te vetical eigt equied.

17 56 Basic Engineeing Matematics 7.6 Volumes of simila sapes Figue 7.7 sows two cubes, one of wic as sides tee times as long as tose of te ote. x x (a) Figue 7.7 x x x (b) x Poblem 0. A ca as a mass of 1000kg. A model of te ca is made to a scale of 1 to 50. Detemine te mass of te model if te ca and its model ae made of te same mateial Volume of model Volume of ca ( ) 1 = 50 since te volume of simila bodies ae popotional to te cube of coesponding dimensions. Mass= density volume and, since bot ca and model ae made of te same mateial, ( ) Mass of model 1 = Mass of ca 50 Hence, mass of model = (mass of ca) ( 1 50 ) = 1000 = kg o 8g 50 Now ty te following Pactice Execise Volume of Figue 7.7(a) = (x)(x)(x) = x Volume of Figue 7.7(b) = (x)(x)(x) = 7x Hence, Figue 7.7(b) as a volume (),i.e.7,times te volume of Figue 7.7(a). Summaizing, te volumes of simila bodies ae popotional to te cubes of coesponding linea dimensions. Pactice Execise 109 Volumes of simila sapes (answes on page 5) 1. Te diamete of two speical beaings ae in te atio :5. Wat is te atio of tei volumes?. An engineeing component as a mass of 400 g. If eac of its dimensions ae educed by 0%, detemine its new mass.