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CIRCULAR BASE AND TOP ARE IDENTICAL

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WW HAT IS A CYLINDERHAT IS A CYLINDER ??

A SoLID gENERATED bY THE A SoLID gENERATED bY THE REvoLuTIoN of A RECTANgLE AbouT REvoLuTIoN of A RECTANgLE AbouT oNE of ITS SIDES IS CALLED A oNE of ITS SIDES IS CALLED A CYLINDER.CYLINDER.

CoNSIDER A CYLINDER WITH RADIuS R AND HEIgHT H.

A CYLINDER HAS ATWo CIRCuLAR ENDS WHICH ARE PARALLEL .EACH of THESE IS CALLED bASE of THE CYLINDER.

THE LINE SEgMENT JoININg THE CENTRES of THE TWo bASES IS CALLED THE AXIS of THE CYLINDER.

THE PERPENDICuLAR DISTANCE bETWEEN THE TWo ENDS IS CALLED THE HEIgHT of THE CYLINDER.

AREA of THE LATERAL SuRfACE of THE AREA of THE LATERAL SuRfACE of THE CYLINDER=AREA of THE RECTANguLAR CYLINDER=AREA of THE RECTANguLAR STRIP of PAPERSTRIP of PAPER

=AREA of A RECTANgLE of =AREA of A RECTANgLE of LENgTH 2*22/7*RLENgTH 2*22/7*R

=2*22/7*R*H=2*22/7*R*H =2*22/7*R*H SquARE uNITS=2*22/7*R*H SquARE uNITSTHuS foR A CYLINDER of RADIuS R AND THuS foR A CYLINDER of RADIuS R AND

HEIgHT H WE HAvE,HEIgHT H WE HAvE,LATERAL (CuRvED SuRfACE) LATERAL (CuRvED SuRfACE)

AREA=2*22/7*R*H Sq uNITSAREA=2*22/7*R*H Sq uNITSToTAL SuRfACE AREA ToTAL SuRfACE AREA

=(2*22/7*R*H+2*22/7*R*R)=(2*22/7*R*H+2*22/7*R*R)

LET R and D BE THE EXTERNAL AND INTERNAL RADII OF THE LET R and D BE THE EXTERNAL AND INTERNAL RADII OF THE CYLINDER AND CYLINDER AND h BEh BE ITS HEIGHT .ITS HEIGHT .1-SURFACE AREA OF EACH BASE=22/7(R*R-D*D)1-SURFACE AREA OF EACH BASE=22/7(R*R-D*D)2-curved surface area=(external surface area )+(internal 2-curved surface area=(external surface area )+(internal surfacesurface area)area) =2*22/7*R*H+2*22/7*D*H=2*22/7*R*H+2*22/7*D*H =2*22/7*h(R+D) sq units =2*22/7*h(R+D) sq units 3-total surface area =2*22/7*R*h+2*22/7*D*h+2*22/7(R*R-3-total surface area =2*22/7*R*h+2*22/7*D*h+2*22/7(R*R-D*D)D*D) =2*22/7*h(R+D)+2*22/7(R+D)(R-D)=2*22/7*h(R+D)+2*22/7(R+D)(R-D) =2*22/7(R+D)(h+R-D) sq units=2*22/7(R+D)(h+R-D) sq units

EX 1-A RECTANguLAR SHEET of PAPER 44CM*18CM IS RoLLED ALoNg ITS LENgTH AND A CYLINDER IS foRMED. fIND THE RADIuS of THE CYLINDER.

SoLuTIoN-WHEN THE RECTANguLAR SHEET IS RooLED ALoNg ITS LENgTH ,WE fIND THAT THE LENgTH of THE SHEET foRMS THE CIRCuMfERENCE of ITS bASE AND bREADTH of THE SHEET bECoMES THE HEIgHT of THE CYLINDER.

LET R CM bE THE RADIuS of THE bASE AND HCM bE THE HEIgHT. THEN H=18CM

CIRCuMfERENCE of THE bASE=L of THE SHEET

oR 2*22/7*R=44

oR R= 7CM44cm

18 cm18 cm

EX -2

A CoMPANY PACkAgES ITS MILk PoWDER IN CYLINDRICAL CoNTAINERS WHoSE bASE HAS A DIAMETRE of 16.8CM AND HEIgHT 20.5CM.CoMPANY PLACES A LAbEL ARouND THE CuRvED SuRfACE of THE CoNTAINER. If THE LAbEL IS PLACED 1.5CM fRoM THE ToP AND THE boTToM, WHAT IS THE SuRfACE AREA of THE LAbEL?

SoLuTIoN—CLEARLY SuRfACE AREA of THE LAbEL IS EquAL To THE CuRvED SuRfACE AREA of A CYLINDER of bASE RADIuS

R=16.8/2CM=8.4CMAND HEIgHT H=(20.5-1.5-1.5)CM=17.5CM

THEREfoRE SuRfACE AREA of THE LAbEL=2*22/7*8.4*17.5CM²

=2*22/7*1.2*17.5CM²=924CM²

POWDERED MILK20.5C20.5CMM

1.5CM1.5CM

16.816.8CMCM

TAkE CIRCuLAR SHEETS of TAkE CIRCuLAR SHEETS of RADIuS R AND STuCk THEM uP RADIuS R AND STuCk THEM uP vERTICALLY To foRM A CYLINDER vERTICALLY To foRM A CYLINDER of HEIgHT H.of HEIgHT H.

voLuME of THE CYLINDER=voLuME of THE CYLINDER=MEASuRE of THE SPACE oCCuPIED MEASuRE of THE SPACE oCCuPIED

bY THE CYLINDER=bY THE CYLINDER= THE AREA of EACH CIRCuLAR THE AREA of EACH CIRCuLAR SHEETSHEET =22/7*R²H=22/7*R²H

r

hh

Volume of a hollow cylinder=Volume of a hollow cylinder=Exterior volume-interior Exterior volume-interior

volumevolume=22/7R²h-22/7r²h=22/7R²h-22/7r²h=22/7*h(R²-r²)=22/7*h(R²-r²)

r

R

h

A WELL WITH 10M INSIDE DIAMETER IS Dug ouT 14M DEEP. EARTH TAkEN ouT of IT IS SPREAD ALL ARouND To A WIDTH of 5MTo foRM AN EMbANkMENT. fIND ITS HEIgHT of EMbANkMENT.

voLuME of THE EARTH Dug ouT=22/7*R²*H*M³

=22/7*5*14M³=1100M³

AREA of THE EMbANkMENT=22/7(R²-R²)

=22/(10²-5²)M²=22/7*75M²

THEREfoRE THE HEIgHT of THE EMbANkMENT=voLuME of THE EARTH Dug ouT/AREA of THE EMbANkMENT

=1100/22/7*75=7*1100/22*75=4.66M

15m15m

10m10m

EX-2:THE RAIN WATER THAT fALLS oN A Roof of AREA 6160M² IS CoLLECTED IN A CYLINDRICAL TANk of DIAMETER 14M AND HEIgHT 10M AND THuS THE TANk IS CoMPLETELY fILLED. fIND THE HEIgHT of RAIN WATER oN THE Roof.

SoLuTIoN-LET THE HEIgHT of RAIN WATER oN THE Roof bE H METRE. CLEARLY RAIN WATER ACCuMuLATED oN THE Roof foRMS A CuboID of bASE AREA 6160M² AND HEIgHT H METRE.THEREfoRE THE voLuME of WATER ACCuMuLATED oN THE Roof= 6160*H M³

THIS WATER CoMPLETELY fILLS A CYLINDER TANk of RADIuS 7M AND HEIgHT 10M.

THEREfoRE WATER ACCuMuLATED oN THE Roof=voLuME of THE CYLINDER TANk=

6160H=22/7*7²*10 oR

H=1540/6160M=0.25M=25CM

HENCE, THE HEIgHT of THE RAIN WATER oN THE Roof =25CM.

Ex 3-how Many Cubic Metres Of Earth Must Be Dug Out To Sink A Well 22.5m Deep And Of Diametre 7m?Also,find The Cost Of Plastering The Inner Curved Surface At Rs3/M².

Solution-volume Of Earth To Be Dug Out =Volume Of The Well =22/7*7/2*7/2*22.5 M³=866.25 M³Therefore,area Of Inner Curved

Surface=2*22/7*r*h=2*22/7*7/2*22.5 M²=495m²

Cost Of Platering The Inner Curved Surface =Rs (495*3)=rs 1485m²

WRITTEN BY:ARISHA KHAN AND ANJALI

JOSEPH

ANIMATIONS BY :DRISHTI NIGAM AND DHEERAJ

DINESH

BACKGROUND MUSIC BY:

DAN OWEN JONATHAN

SPEAKER:BHARGAVI THAKUR

THANK YOU