BMM10234 Chapter 5 - Solid Geometry.pdf

7/28/2019 BMM10234 Chapter 5 - Solid Geometry.pdf

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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


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3/31


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=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=


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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


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=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


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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 = = =


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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


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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


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() 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


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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


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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''


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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


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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


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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


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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


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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


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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 = :


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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

= = = =


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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


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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


22/31


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


23/31


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


24/31


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 - )


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 - )

() % () % () % () % () %


25/31


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 - )


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


26/31


SCORE SEET

Use HB pencil only. Abide by the time-limit


27/31


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



28/31


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


29/31


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


30/31


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


31/31


() () () () () ()

() () () () () ()

() () () () () ()

() 9 () () 9 () () 9 ()

() () () () () ()

() () () () ()

() () () () ()

() () 9 () () ()


Answers


Documents

BMM10234 Chapter 5 - Solid Geometry.pdf