bmm10234 chapter 5 - solid geometry.pdf
TRANSCRIPT
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
1/31
Chapter 5 | Solid Geometry | BMM10234 | 125 of 204
A solid figure or solid is ny portion of spe ounded y one or more surfe, plne or urve Tese surfe re lled tefes of te solid nd intersetions of djent fes re lled edges
For example:
A polyhedron is solid ounded y plne fes
A parallelopiped is solid ounded y tree pirs of prllel plnes
A parallelopiped, wose fes re retngulr is lled uoid or retngulr solid
A parallelopipedwose fes re ll squre is lled ue
Cuboid and Cube
A
E H
B
CD
F Gb
c
a
A
E
D C
G
H
F
a
B
a
a
Fig.1 - CUBOID Fig.2 - CUBE
Te squre of te digonl of uoid is equl to sum of te squre of tree onurrent edges Let F e te digonl of teretngulr solid in fig () nd fig (), If FG, FD nd FE mesure , nd units of lengt respetively
In fig () F = + + , ie digonl F = + + or F =2 2 2a b c+ +
Te digonls of ue re equl If denotes te edge of ue, ten (digonl)= + + =
Digonl =23a a 3=
Surfe re of uoid = (l + + l) were l, , nd re te lengt, redt nd eigt of te uoid respetively
b
l
h
a
a
a
Preface
Solid Geometry
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
2/31
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
3/31
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
4/31
Chapter 5 | Solid Geometry | BMM10234 | 128 of 204
=3 216 =
digonl = a 3 = 6 3 = (Sine digonl of ue of side = a 3 )
Example 4:
Find te edge of ue wose surfe s te sme re s tt of retngulr solid of te following dimensions : lengt
dm, redt dm, dept dmSolution:
surfe re of te ue = surfe re of te retngulr solid
Terefore surfe re of te retngulr solid = ( + + )= ( + + ) = ()
Surfe re of te retngulr solid = sq dm
Terefore, (edge) = sq dm
(edge) =344 172
6 3=
edge =
172
3dm = dm
Example 5:
Te surfe re of ue is sq m Find its volume
Solution:
(edge) = sq m (edge) =384
6=
edge = 64 = ms
Volume = (edge)= () = u m
Example 6:
Te volume of retngulr solid is u m nd ny one digonl is 244 m If its tikness is ms, find its lengt nd
redt
Solution:
Let , nd mesure lengt, redt, nd tikness respetively of te solid
Terefore volume = l =
Digonl = 2 2 2a b c+ + = 244 + + =
+ = = = (Sine = )
Volume = = ; = =576
6=
We know tt ( + ) = + + = + () = +
( + ) = + = 400 =
( + ) = ()( ) = + = () = =
( ) = 16 =
ie = ()
Adding () nd (), = ; =24
2=
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
5/31
Chapter 5 | Solid Geometry | BMM10234 | 129 of 204
Sustituting te vlue of in eqution ()
+ = , + = , = =
Terefore lengt = ms, redt = ms
Example 7:
Te mesurements of luggge ox re ms, ms nd ms Find te volume of te ox ow mny sq m of lot is
required to mke over for te ox?Solution:
l = , = , = , v = volume = l
Terefore, V = = m
Quntity of lot required to mke te over of te ox = surfe Are (SA) of te ox
SA = (l + + l) = ( + 28 )= ( + + ) = = sq ms of lot is required to over te ox
Example 8:
Te nnul rinfll t ple is m Find te weigt in metri tones of te nnul rinfll tere on etre of lnd, tking
te weigt of wter to e metri tonne equl to u m
Solution:
etre = , sq m
eigt of rinfll = ms or43
m100
Volume of rinfll = Are of te lnd eigt of rinfll = , 43
100= u m
Weigt of wter = metri tonnes = metri tonnes
Example 9:
A ui meter of iron is flttened y mmering, so s to over n re of etres Find te tikness of iron in u m orret
to te first two signifint figures
Solution:
ui meter = ,, ui m
etres = , , sq m
Tikness of Iron =Volume of the iron 10, 00, 000 1
area covered 6 10,000 10,000 600= =
= ms
Example 10:
A retngulr tnk is m long, m rod nd m deep Clulte in litres te mount of wter it will old If litres
of wter is drwn off, find to te nerest mm, to wi te wter level sinks? (l litre = u m)
Solution:
Lengt of te tnk = m = m
redt of tnk = m = m
Dept of te tnk = m = m
Volume of wter =350 180 150
1000
= litres
Volume of wter to e drwn off = litres = u m= ,, u m
Derese in wter level
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
6/31
Chapter 5 | Solid Geometry | BMM10234 | 130 of 204
=vol of water 700000 100
cmsarea of the bottom of the tank 350 180 9
= =
Derese in wter level =1
11 cms9
Example 11:
A rigt pyrmid m ig s squre se, e side of wi is m Find te volume nd surfe re of te pyrmid
AD
CB
P
O
l
h
Q
Solution:
Are of te se = (side) = ()
Are of te se () = sq m
eigt of te pyrmid = OP = ms
Volume of te pyrmid =1 1
BH 100 123 3
= = u m
Surfe re of te pyrmid = Are of ACD + OALet Q e te mid point of AD, ten in te rigt ngled tringle OPQ
OP = ms = , = ms
Slnt eigt = l =
2 22 2a 10h 12 144 25 169 13
2 2
+ = + = + = =
Totl surfe re = +1
2pl ( = Are of te se, p = perimeter of te se)
= +1
4 10 132
(p = , = m) = + = + = sq m
Example 12:
Find te volume of rigt pyrmid on regulr exgonl se, wen e side of te se is m nd te eigt is m
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
7/31
Chapter 5 | Solid Geometry | BMM10234 | 131 of 204
Solution:
volume of pyrmid =1
3se re eigt
Are of exgon =2
3 3.a
2sq units
Are of se () =23 3 10
2 (euse = )
=3 3
100 3 3 50 150 32
= = =
= sq m, eigt = m
Volume =1
3(re of te se eigt) =
1
3 ( )
V = = ui meters
Example 13:
Te se of prism is n equilterl tringle of side ms If its eigt is ms, find its totl surfe re
Solution:
Totl surfe re (TSA) of prism = + p ( is te re of te se, p is te perimeter of te se nd is te eigt of
te prism)
TSA = 23
a4
+ ()
= 23
84
+
= 3
644
+ = 32 3 + = sq m
Terefore totl surfe re of te prism = sq m
Example 14:
Te se of prism is squre wose re is m If te volume of te prism is , find its eigt
Solution:
Volume = ( is te re of te se, is eigt)
Or =volume 2500
B 100= = ms
Example 15:
Te se of prism is rigt ngled tringle Te sides ontining te rigt ngle re ms nd ms If te eigt of te
prism is ms, find te volume of te prism
Solution:
Are of rigt ngle tringle =1
2 (were nd re te sides ontining rigt ngle)
Are of te se of te prism () =ab 3 4
2 2
= = sq m Volume = = =
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
8/31
Chapter 5 | Solid Geometry | BMM10234 | 132 of 204
Example 16:
A tnk s retngulr se of dimension m y m At wt veloity per seond must wter flow into it troug pipe
wose retngulr ross setion mesures dm y dm tt te level my e rised y1
33
dm in ours?
Solution:
Dimensions of te tnk re m, m, 133
dm, ie 103
dm = 10 1 1m m3 10 3
=
Volume of inoming wter in te tnk in ours = l = 1
3m = m
Dimensions of te pipe re l = dm =8
10m, = dm =
5
10m
Te re of ross setion = l =28 5 2 m
10 10 5 =
Lengt of wter olumn in ours
volume of water in tank 15000 5 1500037,500m2area of cross-section of the pipe 2
5
= = = =
Veloity of wter in seond37,500
2 60 60=
m/sec =
375 125 55
72 24 24= = m/sec
eociy o we in second =5
524
m/sec
Example 17:
Te dimensions o ecngu boe e in e io : : nd e dieence beween e cos o coveing i wi see
o ppe e e o Rs nd Rs pe sque is Rs Find e dimensions o e box
Solution: : b : = : : ; = x, b = x, = x wee x is e consn o popoioniy Suce
e o e box = (b + b + )
= (xx + x x + x x)= (x + x + x) = x
Cos o ppe Rs pe m = (x) = x
Cos o ppe Rs pe m = ( x) = x
x x = , x = x =416
26= 16 x =
ence e dimensions o e box e = x = = m, b = x = = m
= x = = m
Example 18:
men ook dip in nk wic is m ong nd m bod W is e ise in e we eve i e vege dispcemen
o we by mn is m?
Solution:
oume o we dispced by mn = m
oume o we dispced by men = m = m
Teeoe voume o we ised = m
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
9/31
Chapter 5 | Solid Geometry | BMM10234 | 133 of 204
Leng o nk = m, bed o nk = m
Teeoe bse e o e nk = b = ( )m = m
Teeoe ise in we eve =volume of water raised 2000 1
marea of the base 4000 2
= =
Volume and surface area of cone, cylinder, sphere and hemi-sphere
Cone
A cone my be deined s imi o pymid Te bse o cone is cice
A
C
B
O
l
h
r
() Le suce Ae o ig cicu cone (LSA) = lwee is e dius o e bse, is e sn eig() To suce Ae (TSA) o ig cicu cone is LSA + e o e bse
= l+ = (l+ ) o ( + l)
() oume o ig cicu cone =21 r h
3 (wee is e dius o e bse, is e veic eig o e cone)
(i) Sn eig o cone, = 2 2h r+
(ii) eic eig o cone, = 2 2l r
(iii) Rdius o e bse, = 2 2l h
Cylinder
A cyinde wose bse cuve is cice is ced cicu cyinde
h
r
A cicu cyinde, in wic e ine joining e cenes o e bses is pependicu o e bses, is ced ig cicu
cyinde
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
10/31
Chapter 5 | Solid Geometry | BMM10234 | 134 of 204
() Le suce Ae (LSA) o cyinde (cicu) = peimee o e bse eig = = () To suce Ae (TSA) o ig cicu cyinde = Le e + Ae o e bse + e o e bse op
= + + = + = ( + ) = ( + )() oume o e ig cicu cyinde = Ae o e bse x eig =
Sphere
A spee is se o poins in spce wic e given disnce om ixed poin Te ixed poin is ced e cene o e
spee
A
B
PO L O
A spee my be seen o be geneed by e evouion o semi cice bou is dimee() Suce e o spee wose dius is =
() oume o spee =4
3
Hemi-sphere
Te pne wic psses oug e cene o e spee divides e spee in wo equ ps Ec p is ced e emi-
spee
O
() Le suce e o emispee =
24 r
2
=
() To suce e o emispee = e o e cicu p + LSA = + =
() oume o emispee =
3
3 3
4r
4 1 23 r r2 3 2 3
= =
Example 19:
Te dius o e bse nd e eig o cone e especivey cm nd cm Find e voume, cuved suce e ndo suce e o e cone
l h
r
O
Solution:
= cms, = cm
L = + = + = + = 9 = 6409= 80.05 cm
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
11/31
Chapter 5 | Solid Geometry | BMM10234 | 135 of 204
oume = 21 1 22
r h 35 35 723 3 7
= = = 9 cm
Cuved suce e = (l + r) =22
7 = = cm
To suce e = ( + ) = 227 ( + ) = = cm
Example 20:
We ows e e o mees pe minue om cyindic pipe mm in dimee ow ong woud i ke o i conic
vesse wose dimee e suce is cm nd dep is cm
Solution:
Rdius o e pipe =d 5 5 5 1
mm cms2 2 2 10 20 4
= = = =
Speed o we = mees pe minue = cm pe minue= cm pe minue
ie voume o we oug pipe pe minue
=
222 1 22 1 1 1375
1000 cu.cm 1000 cubic cm7 4 7 4 4 7
= =
Rdius o e conic vesse =diameter 40
20cm2 2
= =
Cpciy o e conic vesse =1
3
=21 22 1 22 7040020 24 20 20 24
3 7 3 7 7 = = cubic cm
To i
1375
7 cub cm, e ime equied is minue
To i70400
7cu cm e ime equied is:
=
70400
70400 7 256 17 511375 7 1375 5 5
7
= = =
Teeoe, ime equied o i e vesse is1
515
minues
Example 21:
Te eig o ig cicu cone is cm nd e dius o e bse is cm Find e e o coss - secion o e cone
omed by pne pe o e bse o e cone nd e disnce o cm o e veex o e cone
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
12/31
Chapter 5 | Solid Geometry | BMM10234 | 136 of 204
C
BA
V
C
A B
Solution:
A' C' nd VAC re similr
TereforeVC' A 'C ' 10 A'C ' 10 28
A 'C 'VC AC 40 28 40
= = = = ms = r'
Are of te ross setion, r =22
7 = sq m
Example 22:
A onil tent is required to ommodte person If e person requires dm of spe on te floor nd dm of irto rete, find te vertil eigt, slnt eigt nd widt of te tent
Solution:
Floor re of e person = dm
Floor re for person = = dm
Terefore, te se re of te one = r = dm
r =154 154 154 7
4922 22
7
= =
r = dm
Air spe for e person = dm
Terefore ir spe for person = dm
Volume of tent =
1
3r =
21 22 17 h 176 73 7 3
= =
=176 7 3
22 7
= = dm
l = + r = + = + = l = 625 = dm ie slnt eigt l = dm, widt of te tent is te dimeter ie d =
r = = dm
Example 23:
If te rdii of te ends of uket ms ig re m nd m find te surfe re of te uket Te given uket n
e tougt of s te differene of two rigt irulr ones VA nd VA''
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
13/31
Chapter 5 | Solid Geometry | BMM10234 | 137 of 204
C BA
24 cm
V
A BC
Solution:
Te tringle VAC nd VA'C' re similr
eigt of te uket = CC' = = ms
Rdius of te uket = C = r= ms
Rdius of te ottom of te uket = C'' = r= ms
Let VA = lm VA' = l
= V' ie V' = l
m
Or V = lms, V' = l
, VC =
= VC' + m =
+ m
Terefore tringle VAC nd VA'C' re similr
Terefore,1 1 1 1 2
2 2 2 2 2
l r h l 24 h15
l r h l 5 h
+= = = =
2
2
24 h
h
+ = +
=
=
=
= +
= + = ms
l =
+ r
= + = + = l= 1521 =
l
=
+ r
= + = + = l= 169 l =
Required surfe re of te uket = rlr
l
+ r
= ( + ) = ( + ) = ( )
= () = 22
7= m
Terefore, surfe re of te uket = m
Example 24:
A frustrum of one s dimeter m nd m nd slnt eigt of m Clulte te eigt of te one of wi te
frustrum is prt
C
B
D
A
O
x
3.9
Solution:
In tringle OA nd OCD
OA = ODC = ; O = O {ommon}Terefore tringle OA is similr to tringle OCD
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
14/31
Chapter 5 | Solid Geometry | BMM10234 | 138 of 204
OA AB OB 2.5 x
OC CD OD 4 x 3.9= = =
+
(euse A =5
2,CD =
8
2,OA = x, OC = x + )
2.5 x4 3.9 x= +
x = ( + x) = x x =
x = x =9.75
1.5= OC = + = m
In tringle ODC, OD = OC CD = =
OD = OD = 92.16 = Terefore eigt of te one = ms
Example 25:
Te eigt of rigt irulr ylinder is m nd its se rdius is m Find te totl re of te ylinder
Solution:
Totl re of te ylinder
= r(r + ) = 227
( + ) = = sq m
Example 26:
Te irumferene of te se of rigt irulr ylinder is ms It is m ig Find its volume nd totl surfe re
r
30
Solution:
irumferene of te se of te ylinder
= r = m r =88 88 7
2 2 22
=
= ms
TSA = r (r + ) = 22
7 ( + ) = = sq ms
Volume = r =22
7 =
22
7 = m
Example 27:
Te dimeter of ylinder is m nd is eigt is m Find te
(i) urved surfe re
(ii) te totl surfe re
(iii) te volume of te ylinder
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
15/31
Chapter 5 | Solid Geometry | BMM10234 | 139 of 204
Solution:
r =d 28
2 2= = m, = m
(i) urved surfe re = r = 22
7 = m
(ii) Totl surfe re = r(r + ) = 22
7 ( + ) = m = m
(iii) Volume = r =22
7 = , m
Example 28:
A ylinder rod roller mde of iron is m long Its inner dimeter is ms nd te tikness of te iron seet rolled into te
rod roller is ms Find te weigt of te roller if of iron weigs gm
Solution:
inner rdius r =54
2
= , outer rdius R = r + = + = ms
Volume of iron = (volume of externl ylinder volume of internl ylinder)
= R r = (R r) = 22
7 ( )
= 22
7 m= m
weigt of te roller = = gms = kg
Example 29:
Two ylinder pots ontin te sme mount of milk If teir dimeters re in te rtio : , find te rtio of teir eigts
Solution:
Let r e te rdius of te ylinder ie dimeter = r
Let R e te rdius of te seond ylinder ie dimeter = R
Let e te eigt of te first ylinder
nd e te eigt of te seond ylinder
Terefore,2r 2 r 2
2R 1 R 1= = r = R
Let V
d V
e te volumes of two pots
V= V
ie, r = R
(R) = R R = R = ;h 1
H 4=
ie te rtio of teir eigt = :
Example 30:Wter flows out troug irulr ylindril pipe wose internl dimeter is ms t te rte of meters per seond into
ylindril tnk, te rdius of wose se is m y ow mu will te level of wter rise in lf n our?
Solution:
Lengt of wter olumn in seond = m
Lengt of wter olumn in minute = ( ) m
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
16/31
Chapter 5 | Solid Geometry | BMM10234 | 140 of 204
Lengt of wter olumn in1
2our = ( ) m = m
Are of ross setion of pipe = r
=22
7 m (d = m, r =
d 2
2 2= = m)
=222 1 22 m
7 10000 70000 =
Volume of wter in1
2our = re of ross setion eigt =
22
70000 =
396
100m
Are of te se of te tnk = r =22 40 40
7 100 100
(r = ms =
40
100m)
=
2222 2 22 4 88 m
7 5 7 25 175
= =
eigt rised =
396volume 396 175 63100
88Area 100 88 8
175
= = = m = m
Example 31:
A well wit m inside dimeter is m deep Ert tken out of it s een spred evenly round it to widt of m Find
te eigt of te emnkment so formed
A 5 m5 m
14m
Solution:
Rdius of te well =
d 10
2 2= = m
Dept of te well = eigt of te ylinder = = m
Volume of te ert dug out = r = = = Rdius of te se of te given ylinder r = m
Rdius of te se of te ylinder wit emnkment
= R = + = m
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
17/31
Chapter 5 | Solid Geometry | BMM10234 | 141 of 204
Are of te emnkment = Rr = (R r) = ( )= ( ) = Volume of te ert dug out = volume of emnkment
= 350 14 2
475 3 3
= =
m
Example 32:
Find te volume nd surfe re of spere of rdius ms
Solution:
Volume =3 34 4 22r 4.2
3 3 7 = ;
24 22v 4.2 4.2 4.2 310 cm3 7
= =
Surfe re = r = 22
7 =
22
7 = m
Example 33:
Te outer dimeter of speril sell is m nd te inner dimeter is m Find te volume of te metl ontined in te
sell
Solution:
Outer rdius = R =1d 10
2 2= = ms
Inner rdius = r =2d 9
2 2= = ms
Volume of te metl ontined in te sell = outer volume inner volume
3 34 4R r3 3
=
3 3 3 34 4 4 22(R r ) (5 4.5 ) (125 91.125)3 3 3 7
= = =
34 22 88 33.87533.875 88 1.613 142 cm3 7 21
= = = =
Example 34:
A ylindril tu of rdius m ontins wter to dept of m A speril iron ll is dropped into te tu nd tus te
level of wter is rised y m Wt is te rdius of te ll?
Solution:
Let te rdius of te iron ll = r ms
V= volume of te iron ll =
4
3r
Let R e te rdius of te ylindril tu = m
Let e te eigt of te wter efore dropping of iron ll = m
e te inrese in eigt of te wter fter dropping of iron ll = m
= new eigt of wter level in te tu
= +
= + = m
V= volume of wter displed y te iron ll
V= R R
= R(
) = ( )
V= m
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
18/31
Chapter 5 | Solid Geometry | BMM10234 | 142 of 204
Volume of te iron ll = volume of wter displed y te ll
4
3r =
r =3 12 12 6.75
4
= =
= =
r = 3 39 =
Rdius of te spere of iron ll = m
Example 35:
Te internl nd externl dimeters of ollow emi-speril vessel re m nd m respetively Te ost to pint per
m of te surfe is Rs Find te totl ost to pint te vessel ll over
Solution:
Internl dimeter = m, internl rdius r =24
2= m
Externl dimeter = m, externl rdius R =25
2= m
Let A
e te inner surfe re = r = 22
7 =
6336
7m
nd Ae te externl surfe re = R =
22
7 =
6875
7m
Totl re to e pinted is A
+ A
=26336 6875 cm
7 7
+
26336 6875 13211 cm7 7
+= =
For m, ost of pinting is Rs
For13211
7m, ost of pinting is
132110.14 Rs.264.22 .
7
=
Example 36:
A one nd emispere ve equl ses nd equl volume Find te rtio of teir eigts
Solution:
Let V
e te volume of te one, r e te rdius of te one, e te eigt of te one
V
=1
3r, sine one nd emispere ve equl ses
Rdius of emispere = r
Let Ve te volume of emispere =
2
3r
Volume of one = volume of emispere (given)
1
3r =
2
3r, = r
h 2
r 1= , : r = :
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
19/31
Chapter 5 | Solid Geometry | BMM10234 | 143 of 204
Example 37:
A vessel is in te form of emisperil owl, mounted y ollow ylinder Te dimeter of te spere is m nd te
totl eigt of te vessel is m Find te pity of te vessel
7
6
Solution:Dimeter of te emisperil owl is m, r =d 14
2 2= = ms
V= volume of te emisperil owl =
2
3r
V
=
23 22 22 44 7 44 49 21567 718.66 cm
3 7 3 3 3
= = = =
Rdius of te ylinder = rdius of te emispere
r = rdius of te ylinder = mseigt of te ylinder =
Totl eigt of te vessel = + r = ms
= r = = ms
Let Ve te volume of te ylinder, V
= r =
22
7 = = m
Totl pity = V+ V
= + = m
Example 38:
Te lrgest possile spere is rved out of ue of side m Find te volume of te spere
Solution:
Te lrgest spere will e rved out of ue only wen te dimeter of te spere = edge of te ue
D = r = m, r =7
2ms V = volume of spere
=
33 34 4 22 7 4 22 7 7 7 539r 179.6 cm
3 3 7 2 3 7 2 2 2 3
= = = =
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
20/31
Chapter 5 | Solid Geometry | BMM10234 | 144 of 204
Example 39:
A ylinder is irumsried out emispere nd one is insried in te ylinder so s to ve its vertex t te enter
on one end nd te oter end s its se Find te rtio of te volume of te ylinder, emispere nd te one
r
rh
Solution:
Let r e te rdius of te emispere, e te eigt of te ylinder s well s of te rigt irulr one
From te figure = reuse is lso te rdius of emispere; te ylinder eing irumsried out it nd te one eing
insried in te ylinder
Te rtios of te Volume of ylinder : Volume of emispere : Volume of one
r :2
3r :
1
3r (euse = r)
r :2
3r :
1
3r
:2
3:
1
3= : :
Example 40:
Wts wrong wit tis figure?
Sol:
Altoug te impossile tringle ertinly looks possile t e orner, you will egin to notie prdox wen you view
te tringle s wole Te ems of te tringle simultneously pper to reede nd ome towrd you Yet, someow,
tey meet in n impossile onfigurtion! It is diffiult to oneive ow te vrious prts n fit togeter s rel tree-
dimensionl ojet
It is not te drwing itself tt is impossile, ut only your tree-dimensionl interprettion of it, wi is onstrined y
ow you interpret pitoril representtion into tree-dimensionl mentl model Given te ne to interpret drwing
or imge s tree-dimensionl, your visul system will do so It does not generlly tke perspetive drwing nd
reinterpret it s flt, euse tere is sptil prdox
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
21/31
Chapter 5 | Solid Geometry | BMM10234 | 145 of 204
Sl.No Concept Formula
Lterl surfe re of uoid (l + )
Totl surfe re of ue
Lterl surfe re of ue
Volume of uoid l or ( re of te se eigt)
Volume of ue
Volume of prllelopiped re of te se eigt
Lterl surfe re of prism Perimeter of te se eigt
Totl surfe re of prism (Are of te se) + lterl
surfe re
Lterl surfe re of pyrmid1
2 perimeter of te se slnt
eigt
Totl surfe re of pyrmid Are of te se + Lterl surfe
Are
Volume of prism Are of te se eigt
Volume of pyrmid1
3 re of te se eigt
Slnt eigt of squre
sed pyrmid
Summary Chart
Table of formulae
2
2 ah
2
+
= vertil eigt,
= side of te se
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
22/31
Chapter 5 | Solid Geometry | BMM10234 | 146 of 204
Lterl surfe Are of rigt
irulr one rl (l = slnt eigt)
Totl surfe Are of rigt
irulr one r(l + r)
Volume of one
1
3 r
( = vertil eigt)
Slnt eigt of one l = 2 2h r+
Vertil eigt of one =2 2l r
Rdius of te se r =2 2l h
Lterl surfe Are of ylinder r
Totl surfe re of ylinder r ( + r)
Volume of ylinder r
Surfe re of spere r
Volume of spere4
3r
Lterl surfe Are of emispere r
Totl surfe Are of emispere r
Volume of emispere2
3r
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
23/31
Chapter 5 | Solid Geometry | BMM10234 | 147 of 204
A re trk is in te form of ring wose inner nd outer irumferene re metre nd meter respetively Find te
widt of te trk (Aess Code - )
() metres () metres () 14 metres () 7 metres ()5
Find te re of qudrnt of irle wose irumferene is m (Aess Code - )
() m () m () m () 219.25 cm () None of tese
A pit metre long, metre wide nd metre deep is dug in field Find te volume of soil removed in ui metres
(Aess Code - )
() m () m () m () m () m
Te volume of ue is m Prt of tis ue is ten melted to form ylinder of lengt m Find te volume of te
ylinder (Aess Code - )
() m () m () m () Dt indequte () None of tese
Find urved nd totl surfe re of onil flsk of rdius m nd eigt m (Aess Code - )
() 60 , 96 () 20 , 96 ()60 , 48 ()30 , 48 () None of tese
Te dimeters of two ones re equl If teir slnt eigt e in te rtio : , Find te rtio of teir urved surfe res
(Aess Code - )
() : () : () : () : () :
Te outer nd inner dimeters of speril sell re m nd m respetively Find te volume of te metl ontined
in te sell (Use 22 / 7) = (Aess Code - )
() m () m () m () m () m
Find te numer of riks, e mesuring m m m, required to onstrut wll m long, m ig nd
m tik, wile te snd nd ement mixture oupies % of te totl volume of wll (Aess Code - )() () () () ()
In sower, m of rin flls Wt will e te volume of te volume of wter tt flls on etre re of ground? (Aess
Code - )
() m () m () m () m () m
ow mny metres of lot m wide will e required to mke onil tent, te rdius of wose se is m nd eigt is
m? (Aess Code - )
() m () m () m () m () m
A onil tent is to ommodte persons E person must ve m spe to sit nd m of ir to ret Wt will
e te eigt of te one? (Aess Code - )
() m () m () m () m () None of tese
Tree ues e of volume of m re joined end to end Find te surfe re of te resulting figure (Aess Code -
)
() m () m () m () m () m
Practice Exercise - 1
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
24/31
Chapter 5 | Solid Geometry | BMM10234 | 148 of 204
Te lrgest one is formed t te se of ue of side mesuring m Find te rtio of volume of one to ue (Aess
Code - )
() : () : () : () : () None of tese
Find te rdius of te irle insriped in tringle wose sides re m, m nd m (Aess Code - )
() m () m () m () 2 2 cm () m
A ue wose edge is m long s irle on e of its fes pinted lk Wt is te totl re of te unpinted surfe
of te ue if te irle re of te lrgest re possile? (Aess Code - )
() m () m () m () m () None of tese
Te digrm represents te re swept y te wiper of r Wit te dimensions given in te figure, lulte te sded
re swept y te wiper (Aess Code - )
() m () m () m () Cnnt e determined () None of tese
A retngulr wter reservoir is m y m t te se Wter flows into it troug pipe wose ross-setion is m y
m t te rte of m per seond Find te eigt to wi te wter will rise in te reservoir in minutes: (Aess Code
- )
() m () m () m () Cnnt e determined () None of tese
Wi of te following pirs is not orretly mted: (Aess Code - )
Geometrical object Number of vertices
(A) Tetredron
() Pyrmid wit retngulr se
(C) Cue
(D) Tringel
() A () () C () D () None of tese
From irulr seet of pper of rdius m, setor re % is removed If t e remining prt is used to mke onil
surfe, ten te rtio of te redius nd eigt of te one is: (Aess Code - )
() : () : () : () : () :
A spere of rdius m is dropped into ylindril vessel prtly filled wit wter Te rdius of te vessel is m If tespere is sumerged ompletely, ten te surfe of te wter rises y: (Aess Code - )
() m () m () m () () m
If te se of rigt retngulr prism remins onstnt nd te mesures of te lterl edges re lved, ten its volume will
e redued y: (Aess Code - )
() % () % () % () % () %
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
25/31
Chapter 5 | Solid Geometry | BMM10234 | 149 of 204
If e te eigt of pyrmid stnding on se wi is n equilterl tringle of side units, ten slnt eigt is:
(Aess Code - )
() 2 2h a / 4+ () 2 2h a /8+ () 2 2h a / 3+ () 2 2h a+ () None of tese
In te djoining figure A = CD = C = P = CQ In te middle, irle wit rdius m is drwn In te rest figure ll re
te semiirulr rs Wt is te perimeter of te wole figure? (Aess Code - )
() 4 () 8 () 10 ()5 () None of tese
If from irulr seet of pper of rdius m, setor of o is removed nd te remining is used to mke onil
surfe, ten te ngle t te vertex will e: (Aess Code - )
()1 3sin
10
() 16
sin5
() 13
2sin5
()1 42sin
5
() None of tese
A emispere of led rdius dimeter m is st into rigt irulr one of eigt m Te redius of te se of te one
is: (Aess Code - )
() m () m () m () m () None of tese
A speril steel ll ws silver polised ten it ws ut into similr piees: Wt is rtio of te polised re to te non
polised re: (Aess Code - )
() : () : () : () Cnt e determined () None of tese
If , s, V e te eigt, urved surfe re nd volume of one respetively, ten 3 2 2 2(3 Vh 9V s h ) + is equl to:
(Aess Code - )
() ()
()
V
sh ()
36
V () None of tese
Direction for (Que. 28 - 30): E edge of ue is eqully into n prts, tus tere re totl n smller ues Let,
N Numer of smller ues wit no exposed surfes
N Numer of smller ues wit one exposed surfes
N Numer of smller ues wit two exposed surfes
N Numer of smller ues wit tree exposed surfes
Wt is te numer of unexposed smller ues (N)? (Aess Code - )
() 3(n 2) () 3n () n! () () None of tese
Wt is te vlue of (N)? (Aess Code - )() 28 (n 2) () 6 (n 2) () 12 (n 2) () 23( n 3) () None of tese
Wt is te vlue of (N)? (Aess Code - )
() (n 1)! () 2(n 2) () n(n 1)
2
+ () () None of tese
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
26/31
Chapter 5 | Solid Geometry | BMM10234 | 150 of 204
SCORE SEET
Use HB pencil only. Abide by the time-limit
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
27/31
Chapter 5 | Solid Geometry | BMM10234 | 151 of 204
Tere re two onentri exgons E of te side of ot te exgons re prllel E side of n internl regulr
exgon is m Wt is te re of te sded region, if te distne etween orresponding prllel sides is 2 3 cm:
(Aess Code - )
() 2120 3 cm () 2148 3 cm
()
2
126 cm () Cnt e determined () None of te ove
Iniiy e dimee o boon is cm I cn expode wen e dimee becomes / imes o e inii dimee Ai is
bown cc/s I is known e spe o boon wys emins speic In ow mny seconds e boon wi
expode? (Access Code - )
() s () s () s () 9 () None o ese
Te dius o cone is 2 imes e eig o e cone A cube o mximum possibe voume is cu om e sme cone W
is e io o e voume o e one cone o e voume o e cube? (Access Code - )
()3.18 () 2.25 () 2.35 () Cn be deemined () None o ese
A bcksmi s ecngu ion see ong e s o cu ou cicu discs om is see W is e minimum
possibe wid o e ion see i e dius o ec disc is ? (Access Code - )
() 2 3 ft () (2 3 ) ft+ () (3 2 ) ft+ () (2 2 3 ) ft+ () None o ese
Tee e six cicu ings o ion, kep cose o ec oe A sing binds em s igy s possibe I e dius o ec
cicu ion ing is cm W is e minimum possibe eng o sing equied o bind em? (Access Code - )
() 2(6 3 3 ) cm+ + ()6 (2 3) cm+
() 2 (6 ) cm+ () 3 (3 2) cm+ () None o e bove
Practice Exercise - 2
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
28/31
Chapter 5 | Solid Geometry | BMM10234 | 152 of 204
A cyindic cocob s is dius uni nd eig uni I we wis o incese e voume by sme uni eie by
incesing is dius one o is eig one, en ow mny uni we ve o incese e dius o eig? (Access Code
- )
()
2r 2r
h
+()
2r 2rh
h
()
2
2
2r rh
h
()
2r
2h
() None o ese
Rju s sm cubes o cm e wns o nge o em in cuboid spe, suc e suce e wi be
minimum W is e digon o is ge cuboid? (Access Code - )
()8 2 cm () 273 cm () 4 3 cm () 129 cm () None o ese
A cube o side cm is pined on is ces wi ed coou I is en boken up ino sme idenic cubes W
is e io o N: N
: N
(Access Code - )
Wee, N Numbe o sme cubes wi no cooued suce
N Numbe o sme cubes wi ed ce
N Numbe o sme cubes wi ed ce
() : : () : : () : : () Cn be deemined () None o ese
Direction for (Que. 9 - 10): Five spees e kep in cone in suc wy ec spee ouc ec oe nd so ouc ee suce o e cone, is is due o incesing dius o e spees sing om e veex o e cone Te dius o
e smes spee is cm
9 I e dius o e i (ie, ges) spee be cm, en ind e dius o e id (ie, middemos) spee: (Access Code
- 9)
() cm () 25 3 cm () cm () D insuicien () None o ese
W is e es disnce beween e smes spee nd e veex o e cone? (Access Code - )
() cm () cm () cm () D insuicien () None o ese
A cubic cke is cu ino seve sme cubes by dividing ec edge in equ ps Te cke is cu om e op ong e
wo digons oming ou pisms Some o em ge cu nd es emined in e cubic spe A compee cubic (sme)
cke ws given o dus nd e cu o p o sme cke is given o cid (wic is no n du) I e ckes wee
given equy ec piece o peson, o ow mny peope coud ge e cke? (Access Code - )
() () () () () None o ese
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
29/31
Chapter 5 | Solid Geometry | BMM10234 | 153 of 204
Find e sum o e es o e sded seco given ABCDEF is ny exgon nd e cices e o sme dius wi
dieen veices o e exgon s ei cenes s sown in e igue (Access Code - )
() 2r () 22 r () 25 r / 4 () 23 r / 2 () None o ese
Cices e dwn wi ou veices s e cene nd dius equ o e side o sque I e sque is omed by joining
e e mid-poins o noe sque o side 2 6, ind e e common o e ou cices (Access Code - )
() ( )3 1 / 2 6 () 4 3 3 () 1/ 2( 3 3) () 4 12( 3 1) () None o ese
ABCD is cice nd cices e dwn wi AO, CO, DO nd OB s dimees Aes E nd F e sded E/F is equ o
(Access Code - )
() ()/ () / () / 4 () None o ese
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
30/31
Chapter 5 | Solid Geometry | BMM10234 | 154 of 204
Te bse o pymid is ecnge o sides m m nd is sn eig o e soe side o bse is m Find isvoume (Access Code - )
()156 407 ()78 407 ()312 407 () D insuicien () None o ese
SCORE SEET
Use HB pencil only. Abide by the time-limit
9
-
7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf
31/31
Chapter 5 | Solid Geometry | BMM10234 | 155 of 204
() () () () () ()
() () () () () ()
() () () () () ()
() 9 () () 9 () () 9 ()
() () () () () ()
() () () () ()
() () () () ()
() () 9 () () ()
Practice Exercise - 1
Answers
Practice Exercise - 2