1 & 2. Similarity & Pythagoras' Theorem · Geometry (Std. X) 1 1 & 2. Similarity & Pythagoras'...

GeometryGeometryGeometryGeometry (Std.(Std.(Std.(Std. X)X)X)X)1111

1111 &&&& 2222.... SimilaritySimilaritySimilaritySimilarity &&&& Pythagoras'Pythagoras'Pythagoras'Pythagoras' TheoremTheoremTheoremTheorem1.1.1.1. AAAA rightrightrightright triangletriangletriangletriangle hashashashas hypotenusehypotenusehypotenusehypotenuse ofofofof lengthlengthlengthlength pppp cmcmcmcm andandandand oneoneoneone sidesidesideside lengthlengthlengthlength qqqq cm.cm.cm.cm. IfIfIfIf pppp qqqq ==== 1,1,1,1, findfindfindfind thethethethe

lengthlengthlengthlength ofofofof thethethethe thirdthirdthirdthird sidesidesideside ofofofof thethethethe triangle.triangle.triangle.triangle.

SolutionSolutionSolutionSolution :::: Let th e th ird s id e b e x cm . Th en , b y Pyth ag oras th eorem , we h ave

p 2 = q 2 + x 2

x 2 = p 2 q 2 = (p q ) (p + q ) = p + q [Q p q = 1 ]

x = qp = 1q2 [ Q p q = 1 p = q + 1 ]

Hen ce , th e len g th of th e th ird s id e is 1q2 cm .

2.2.2.2. ABCABCABCABC isisisis anananan isoscelesisoscelesisoscelesisosceles triangletriangletriangletriangle right-angledright-angledright-angledright-angled atatatat B.B.B.B. SimilarSimilarSimilarSimilar trianglestrianglestrianglestriangles ACDACDACDACD andandandand ABEABEABEABE areareareare constructedconstructedconstructedconstructed

onononon sidessidessidessides ACACACAC andandandand AB.AB.AB.AB. FindFindFindFind thethethethe ratioratioratioratio betweenbetweenbetweenbetween thethethethe areasareasareasareas ofofofof ABEABEABEABE andandandand ACD.ACD.ACD.ACD.

SolutionSolutionSolutionSolution :::: Let AB = BC = x.

It is g iven th a t ABC is rig h t- an g led a t B.

AC2 = AB2 + BC2

AC2 = x 2 + x 2

AC = 2 x

It is g iven th a t

ABE ~ ACD

ACDArea

ABEArea = 2

2

ACAB

ACDArea

ABEArea =

22

x2

x

ACDArea

ABEArea =

21

3.3.3.3. PPPP andandandand QQQQ areareareare pointspointspointspoints onononon thethethethe sidessidessidessides CACACACA andandandand CBCBCBCB respectivelyrespectivelyrespectivelyrespectively ofofofof ABCABCABCABC rightrightrightright angledangledangledangled atatatat C.C.C.C.

ProveProveProveProve thatthatthatthat AQAQAQAQ2222 ++++ BPBPBPBP2222 ==== ABABABAB2222 ++++ PQPQPQPQ2222....

SolutionSolutionSolutionSolution ::::

In rig h t- an g led trian g les ACQ an d PCB, we h ave

AQ2 = AC2 + CQ2 an d PB2 = PC2 + CB2

AQ2 + BP2 = (AC2 + CQ2) + (PC2 + CB2)

AQ2 + BP2 = (AC2 + BC2) + (PC2 + QC2)

D

CB

A

X

E

2 x

X

Q

P

CB

A


AQ2 + BP2 = AB2 + PQ2 [By Pyth ag oras th eorem ., AC2+ BC2= AB2 an d PC2+ QC2= PQ2]

4.4.4.4. InInInIn ABC,ABC,ABC,ABC, IfIfIfIf ADADADAD BCBCBCBC andandandand ADADADAD2222 ==== BDBDBDBD DC,DC,DC,DC, ProveProveProveProve thatthatthatthat BACBACBACBAC ==== 90909090....

SolutionSolutionSolutionSolution :::: We h ave ,

AD2 = BD DC

AD AD = BD DC

D CA D=

ADBD

Th u s , in ABD an d CAD, we h ave

D CA D=

ADBD

an d , BDA ~ ADC [Each eq u al to 9 0 ]

So , b y SAS- crite rio n of s im ila rity, we g et

DBA ~ DAC

DBA an d DAC are eq u ian g u lar

1 = C an d 2 = B

1 + 2 = B + C

A = B + C [Q 1 + 2 = A]

Bu t, A + B + C = 1 8 0

A + A = 1 8 0 [Q B + C = A]

2 A = 1 8 0 A = 9 0

Hen ce , BAC = 9 0 .

5555.... ABCDABCDABCDABCD isisisis aaaa parallelogram,parallelogram,parallelogram,parallelogram, PPPP isisisis aaaa pointpointpointpoint onononon sidesidesideside BCBCBCBC andandandand DPDPDPDP whenwhenwhenwhen productproductproductproduct meetsmeetsmeetsmeets ABABABAB producedproducedproducedproduced atatatat L.L.L.L.

ProveProveProveProve thatthatthatthat

i)i)i)i)BLBLBLBLDCDCDCDC

PLPLPLPLDPDPDPDP

ii)ii)ii)ii)DCDCDCDCALALALAL

DPDPDPDPDLDLDLDL

....

GivenGivenGivenGiven :::: A p ara lle log ram ABCD in wh ich P is a p o in t on s id e BC su ch th a t DP p rod u ced m eets AB p rod u ced

at L.

ToToToTo Prove:Prove:Prove:Prove: i)BLDC

PLDP

ii)DCAL

DPDL

ProofProofProofProof :::: i) In ALD, we h ave

BP | | AD A

D

P

B

C

L

1 2

A

CB D


BALB =

PDLP

ABBL =

DPPL

DCBL =

DPPL [Q AB = DC]

PLDP =

BLDC [Takin g recip ro ca ls of b o th s id es ]

ii) From (i), we h ave

PLDP =

BLDC

DPPL =

DCBL [Takin g recip ro ca ls of b o th s id es ]

DPPL =

ABBL [Q DC = AB]

DPPL + 1 =

ABBL + 1 [Ad d in g 1 on b oth s id es ]

DP

PLDP =AB

ABBL

DPDL =

ABAL

DPDL =

DCAL [Q AB = DC]

6666.... ABCDABCDABCDABCD isisisis aaaa quadrilateralquadrilateralquadrilateralquadrilateral inininin whichwhichwhichwhich ABABABAB ==== AD.AD.AD.AD. TheTheTheThe bisectorbisectorbisectorbisector ofofofof BACBACBACBAC andandandand CADCADCADCAD intersectintersectintersectintersect thethethethe sidessidessidessides

BCBCBCBC andandandand CDCDCDCD atatatat thethethethe pointspointspointspoints EEEE andandandand FFFF respectively.respectively.respectively.respectively. ProveProveProveProve thatthatthatthat EFEFEFEF |||||||| BD.BD.BD.BD.

Given:Given:Given:Given: A q u ad rila te ra l ABCD in wh ich AB = AD an d th e b is ecto rs of BAC an d CAD m eet th e s id es

BC an d CD at E an d F res p ective ly.

ToToToTo Prove:Prove:Prove:Prove: EF | | BD

ConstructionConstructionConstructionConstruction :::: Jo in AC, BD an d EF.

ProofProofProofProof :::: In CAB, AE is th e b is ecto r of BAC.

BECE

ABAC

……………………(i)

In ACD, AF is th e b is ecto r of CAD.

DFCF

ADAC

DFCF

ABAC

……………………(ii) [Q AD = AB]

FD

E

C

BA


From (i) an d (ii),we g et

DFCF

BECE

FDCF

EBCE

Th u s , in CBD, E an d F d ivid e th e s id es CB an d CD res p ective ly in th e sam e ra t io . Th ere fore , b y th e

con verse of Th ale ’s Th eorem , we h ave EF | | BD.

7777.... ABCABCABCABC isisisis aaaa triangletriangletriangletriangle inininin whichwhichwhichwhich ABABABAB ==== ACACACAC andandandand DDDD isisisis aaaa pointpointpointpoint onononon ACACACAC suchsuchsuchsuch thatthatthatthat BCBCBCBC2222 ==== ACACACAC CD.CD.CD.CD. ProveProveProveProve thatthatthatthat

BDBDBDBD ==== BC.BC.BC.BC.

GivenGivenGivenGiven :::: ABC in wh ich AB = AC an d D is a p o in t on th e s id e AC su ch th a t BC2 = AC CD

ToToToTo proveproveproveprove :::: BD = BC

ConstructionConstructionConstructionConstruction :::: Jo in BD

ProofProofProofProof :::: We h ave ,

BC2 = AC CD

CDBC

====BCAC ……(i)

Th u s , in ABC an d BDC, we h ave

BCAC

====CDBC [from (i)]

an d , C = C [Com m on ]

ABC ~ BDC [By SAS crite rion of s im ila rity]

DCBC

BDAB

[Q AB = AC]

CDBC

BDAC

……(ii)

From (i) an d (ii), we g et

CDBC

====CDBD

BD = BC.

8888.... InInInIn trapeziumtrapeziumtrapeziumtrapezium ABCD,ABCD,ABCD,ABCD, ABABABAB |||||||| DCDCDCDC andandandand DCDCDCDC ==== 2AB.2AB.2AB.2AB. EFEFEFEF drawndrawndrawndrawn parallelparallelparallelparallel totototo ABABABAB cutscutscutscuts ADADADAD inininin FFFF andandandand BCBCBCBC inininin EEEE

suchsuchsuchsuch thatthatthatthatECECECEC

BEBEBEBE =44443333.... DiagonalDiagonalDiagonalDiagonal DBDBDBDB intersectsintersectsintersectsintersects EFEFEFEF atatatat G.G.G.G. ProveProveProveProve thatthatthatthat 7777 FEFEFEFE ==== 10101010 AB.AB.AB.AB.

Solution:Solution:Solution:Solution: In DFG an d DAB, we h ave

1 = 2 [Q AB | | DC | | EF 1 an d 2 are corresp on d in g an g les ]

FDG = ADB [Com m on ]

B

D

C

A


So, b y AA- crite rion of s im ila rity, we h ave

DFG ~ DAB

ABFG

DADF

…….(i)

In trap ez iu m ABCD, we h ave

EF | | AB | | DC

ECBE

DFAF

DFAF =

43

Given

43

ECBE

Q

DFAF + 1 =

43 + 1 [Ad d in g 1 on b oth s id es ]

DF

DFAF =47

DFAD =

47

ADDF =

74 …..(ii)


ABFG =

74

FG =74 AB ……(iii)

In BEG an d BCD, we h ave

BEG = BCD [Corres p on d in g An g les ]

B = B [Com m on ]

BEG ~ BCD [By AA- crite rion of s im ila rity]

BCBE =

CDEG

73 =

CDEG

37

BEBC1

341

BEEC

34

BEEC

43

ECBE

Q

EG =73 CD

EG =73

2 AB [Q CD = 2 AB (g iven )]

EG =76 AB ……(iv)

G

CD

E

A

F

B

1

2


Ad d in g (iii) an d (iv), we g et

FG + EG =74 AB +

76 AB

EF =7

1 0 AB

7 EF = 1 0 AB .

9999.... ThroughThroughThroughThrough thethethethe mid-pointmid-pointmid-pointmid-point MMMM ofofofof thethethethe sidesidesideside CDCDCDCD ofofofof aaaa parallelogramparallelogramparallelogramparallelogram ABCD,ABCD,ABCD,ABCD, thethethethe linelinelineline BMBMBMBM isisisis drawndrawndrawndrawn

intersectingintersectingintersectingintersecting ACACACAC inininin LLLL andandandand ADADADAD producedproducedproducedproduced inininin E.E.E.E. ProveProveProveProve thatthatthatthat ELELELEL ==== 2BL.2BL.2BL.2BL.

ProofProofProofProof:::: In BMC an d EMD, we h ave

MC = MD [Q M is th e m id - p o in t o f CD]

CMB = EMD [Vert ica lly op p o s ite an g les ]

an d , MBC = MED [Alte rn a te an g les ]

So , b y AAS- crite rion of con g ru en ce , we h ave

BMC EMD

BC = DE ….(i)

Als o , AD = BC [Q ABCD is a p ara lle log ram ] …(ii)

AD + DE = BC + BC

AE = 2 BC ….(iii)

Now, in AEL an d CBL, we h ave

ALE = CLB [Vert ica lly op p o s ite an g les ]

EAL = BCL [Alte rn a te an g les}

So , b y AA- crite rion of s im ila rity of trian g les , we h ave

AEL ~ CBL

CBAE

BLEL

BCBC2

BLEL

[Us in g eq u atio n s (iii)]

BLEL = 2

EL = 2 BL.

10101010.... ProveProveProveProve thatthatthatthat threethreethreethree timestimestimestimes thethethethe sumsumsumsum ofofofof thethethethe squaressquaressquaressquares ofofofof thethethethe sidessidessidessides ofofofof aaaa triangletriangletriangletriangle isisisis equalequalequalequal totototo fourfourfourfour timestimestimestimes

thethethethe sumsumsumsum ofofofof thethethethe squaressquaressquaressquares ofofofof thethethethe mediansmediansmediansmedians ofofofof thethethethe triangle.triangle.triangle.triangle.

GivenGivenGivenGiven :::: A ABC in wh ich AD, BE an d CF are th ree m ed ian s .

E

C

L M

A

B

D


ToToToTo proveproveproveprove :::: 2 (AB2 + BC2 + CA2) = 4 (AD2 + BE2 + CF2)

ProofProofProofProof :::: Sin ce in an y trian g le , th e su m of th e sq u ares of an y two s id es is eq u al to twice th e sq u are of h a lf o f

th e th ird s id e tog e th e r with twice th e sq u are of th e m ed ian b is ect in g it .

Th ere fore , takin g AD as th e m ed ian b isect in g s id e BC, we h ave

AB2 + AC2 = 2 (AD2 + BD2)

AB2 + AC2 = 2

22

2BCAD

AB2 + AC2 = 2

4

BCAD2

2

2 (AB2 + AC2) = (4 AD2 + BC2) ….(i)

Sim ila rly, b y takin g BE an d CF res p ective ly as th e m ed ian s , we g et

2 (AB2 + BC2) = (4 BE2 + AC2) …..(ii)

an d , 2 (AC2 + BC2) = (4 CF2 + AB2) …..(iii)

Ad d in g (i), (ii) an d (iii), we g et

4 (AB2 + BC2 + AC2) = 4 (AD2 + BE2 + CF2) + (BC2 + AC2 + AB2)

3 (AB2 + BC2 + AC2) = 4 (AD2 + BE2 + CF2)

Hen ce , 3 (AB2 + BC2 + AC2) = 4 (AD2 + BE2 + CF2)

11111111.... PPPP andandandand QQQQ areareareare thethethethe midmidmidmid –––– pointpointpointpoint ofofofof thethethethe sidessidessidessides CACACACA andandandand CBCBCBCB respectivelyrespectivelyrespectivelyrespectively ofofofof aaaa ABC,ABC,ABC,ABC, rightrightrightright angledangledangledangled atatatat C.C.C.C.

ProveProveProveProve that:that:that:that:

i)i)i)i) 4444 AQAQAQAQ2222 ==== 4444 ACACACAC2222 ++++ BCBCBCBC2222

ii)ii)ii)ii) 4444 BPBPBPBP2222 ==== 4444 BCBCBCBC2222 ++++ ACACACAC2222

iii)iii)iii)iii) 4444 (AQ(AQ(AQ(AQ2222 ++++ BPBPBPBP2222)))) ==== 5AB5AB5AB5AB2222

ProofProofProofProof:::: (i) s in ce AQC is a rig h t trian g le rig h t- an g led a t C.

AQ2 = AC2 + QC2

4 AQ2 = 4 AC2 + 4 QC2 [Mu lt ip lyin g b oth s id es b y 4 ]

4 AQ2 = 4 AC2 + (2 QC)2

4 AQ2 = 4 AC2 + BC2 [Q BC = 2 QC]

ii) Sin ce BPC is a rig h t trian g le rig h t- an g led a t C.

BP2 = BC2 + CP2

4 BP2 = 4 BC2 + 4 CP2 [Mu lt ip lyin g b oth s id es b y 4 ]

4 BP2 = 4 BC2 + (2 CP)2

4 BP2 = 4 BC2 + AC2 [Q AC = 2 CP]

iii) From (i) an d (ii), we h ave

E

D

A

F

B C

P

B CQ

A


4 AQ2 = 4 AC2 + BC2 an d , 4 BP2 = 4 BC2 + AC2

4 AQ2 + 4 BP2 = (4 AC2 + BC2) + (4 BC2 + AC2)

4 (AQ2 + BP2) = 5 (AC2 + BC2)

4 (AQ2 + BP2) = 5 AB2 [In ABC, we h ave AB2 = AC2 + BC2 ]

12121212.... AAAA pointpointpointpoint OOOO inininin thethethethe interiorinteriorinteriorinterior ofofofof aaaa rectanglerectanglerectanglerectangle ABCDABCDABCDABCD isisisis joinedjoinedjoinedjoined withwithwithwith eacheacheacheach ofofofof thethethethe verticesverticesverticesvertices A,B,A,B,A,B,A,B, CCCC andandandand D.D.D.D.

ProveProveProveProve thatthatthatthat OBOBOBOB2222 ++++ ODODODOD2222 ==== OCOCOCOC2222 ++++ OAOAOAOA2222

ProofProofProofProof:::: Let ABCD b e th e g iven rectan g le an d le t O b e a p oin t with in it Jo in OA, OB, OC an d OD.

Th rou g h O, d raw EOF | | AB. Th en , ABFE is a rectan g le .

In rig h t trian g les OEA an d OFC, we h ave

OA2 = OE2 + AE2 an d OC2 = OF2 + CF2

OA2 = OC2 = (OE2 + AE2) + (OF2 + CF2)

OA2 = OC2 = OE2 + OF2 + AE2 + CF2 …(i)

Now, in rig h t trian g les OFB an d ODE, we h ave

OB2 = OF2 + FB2 an d OD2 = OE2 + DE2

OB2 + OD2 = (OF2 + FB2) + (OE2 + DE2)

OB2 + OD2 = OE2 + OF2 + DE2 + BF2

OB2 + OD2 = OE2 + OF2 + CF2 + AE2 [Q DE = CF an d AE = BF] …(iii)


OA2 + OC2 = OB2 + OD2 .

13131313.... ABCABCABCABC isisisis aaaa rightrightrightright triangletriangletriangletriangle right-angledright-angledright-angledright-angled atatatat CCCC andandandand ACACACAC ==== 3 BC.BC.BC.BC. ProveProveProveProve thatthatthatthat ABCABCABCABC ==== 60606060....

ProveProveProveProve:::: Let D b e th e m id - p o in t o f AB. Jo in CD. Sin ce ABC is a rig h t trian g le rig h t- an g led a t C.

AB2 = AC2 + BC2

AB2 = 2BC3 + BC2 [Q AC = 3 BC (Given )]

AB2 = 4 BC2

AB = 2 BC

Bu t, BD =21 AB i.e , AB = 2 BD

BD = BC

We kn ow th at th e m id - p o in t o f th e h yp oten u se of a rig h t trian g le is eq u id is tan t from th e vert ice s .

CD = AD = BD

CD = BC [s in ce BD = BC]

D

CB

A

FE

D C

BA

O


Th u s , in BCD, we h ave

BD = CD = BC

BCD is eq u ila te ra l

ABC = 6 0 .

11114444.... InInInIn fig.,fig.,fig.,fig., DDDD andandandand EEEE trisecttrisecttrisecttrisect BC.BC.BC.BC. ProveProveProveProve thatthatthatthat 8888 AEAEAEAE2222 ==== 3AC3AC3AC3AC2222 ++++ 5555 ADADADAD2222....

ProofProofProofProof:::: Sin ce D an d E are th e p o in ts of trisect ion of BC.

Th ere fore ,

BD = DE = CE.

Let BD = DE = CE = x. Th en , BE = 2 x an d BC = 3 x.

In rig h t trian g les ABD, ABE an d ABC, we h ave

AD2 = AB2 + BD2

AD2 = AB2 + x 2 ……..(i)

AE2 = AB2 + BE2

AE2 = AB2 + 4 x 2 ……..(ii)

an d , AC2 = AB2 + BC2

AC2 = AB2 + 9 x 2 ……..(iii)

Now, 8 AE2 3 AC2 5 AD2 = 8 (AB2 + 4 x 2) 3

(AB2 + 9 x 2) 5 (AB2 + x 2)

8 AE2 3 AC2 5 AD2 = 0

8 AE2 = 3 AC2 + 5 AD2 .

11115555.... InInInIn fig.,fig.,fig.,fig., twotwotwotwo chordschordschordschords ABABABAB andandandand CDCDCDCD intersectintersectintersectintersect eacheacheacheach otherotherotherother atatatat thethethethe pointpointpointpoint P.P.P.P. ProveProveProveProve that:that:that:that:

i)i)i)i) APCAPCAPCAPC ~~~~ DPBDPBDPBDPB ii)ii)ii)ii) AP.PBAP.PBAP.PBAP.PB ==== CP.DPCP.DPCP.DPCP.DP

ProofProofProofProof::::

(i) In s PAC an d PDB, we h ave :

APC = DPB [Vert . op p . s ]

CAP = BDP [An g les in th e sam e seg m en t of a circle are eq u al]

By AA crite rion of s im ila rly, we h ave:

APC ~ DPB.

(ii) Sin ce s APC ~ DPB, th erefore

DPAP =

PBCP or PA.PB = CP.DP.

11116666.... InInInIn aaaa rightrightrightright triangletriangletriangletriangle ABCABCABCABC right-angledright-angledright-angledright-angled atatatat C,C,C,C, PPPP andandandand QQQQ areareareare thethethethe pointspointspointspoints onononon thethethethe sidessidessidessides CACACACA andandandand CBCBCBCB

respectively,respectively,respectively,respectively, whichwhichwhichwhich dividedividedividedivide thesethesethesethese sidessidessidessides inininin thethethethe ratioratioratioratio 2:1.2:1.2:1.2:1. ProveProveProveProve thatthatthatthat

D E CB

A

D

B

A

C

P


(i)(i)(i)(i) 9AQ9AQ9AQ9AQ2222 ==== 9AC9AC9AC9AC2222 ++++ 4BC4BC4BC4BC2222 (ii)(ii)(ii)(ii) 9BP9BP9BP9BP2222 ==== 9BC9BC9BC9BC2222 ++++ 4AC4AC4AC4AC2222 (iii)(iii)(iii)(iii) 9(AQ9(AQ9(AQ9(AQ2222 ++++ BPBPBPBP2222)))) ==== 13AB13AB13AB13AB2222....

ProofProofProofProof:::: It is g iven th a t P d ivid es CA in th e ra t io 2 :1 . Th erefore ,

CP =32 AC …..(i)

Also , Q d ivid es CB in th e ra t io 2 :1 .

QC =32 BC …..(ii)

(i) Ap p lyin g Pyth ag oras th eorem in rig h t- an g le ACQ, we h ave

AQ2 = QC2 + AC2

AQ2 =94 BC2 + AC2 [Us in g (ii)]

9999 AQAQAQAQ2222 ==== 4444 BCBCBCBC2222 ++++ 9999 ACACACAC2222 …..(iii)

(ii) Ap p lyin g Pyth ag oras th eorem in rig h t trian g le BCP, we h ave

BP2 = BC2 + CP2

BP2 = BC2 +94 AC2 [Us in g (i)]

9999 BPBPBPBP2222 ==== 9999 BCBCBCBC2222 ++++ 4444 ACACACAC2222 ……(iv)

(iii) Ad d in g (iii) an d (iv), we g et

9 (AQ2 + BP2) = 1 3 (BC2 + AC2)

9999 (AQ(AQ(AQ(AQ2222 ++++ BPBPBPBP2222)))) ==== 13AB13AB13AB13AB2222 [Q BC2 = AC2 + AB2]

17171717.... IfIfIfIf AAAA bebebebe thethethethe areaareaareaarea ofofofof aaaa rightrightrightright triangletriangletriangletriangle andandandand bebebebe oneoneoneone ofofofof thethethethe sidessidessidessides containingcontainingcontainingcontaining thethethethe rightrightrightright angle,angle,angle,angle, proveproveproveprove

thatthatthatthat thethethethe lengthlengthlengthlength ofofofof thethethethe altitudealtitudealtitudealtitude onononon thethethethe hypotenusehypotenusehypotenusehypotenuse isisisis22224444 AAAA4444bbbb

AbAbAbAb2222

....

ProofProofProofProof:::: Let PQR b e a rig h t trian g le rig h t- an g led a t Q su ch th a t QR = b an d A = Area of PQR

Draw QN p erp en d icu lar to PR.

We h ave ,

A = Area of PQR

A =21 (QR PQ)

A =21 (b PQ)

PQ =bA2 ……(i)

Now, in ’s PNQ an d PQR, we h ave

PNQ = PQR [Each eq u al to 9 0 ]

C

P

B

A

Q

N

RQ b

P


an d , QPN = QPR [Com m on ]

So , b y AA- crite rion of s im ila rity, we h ave

PNQ ~ PQR

PRPQ =

QRNQ …..(ii)

Ap p lyin g Pyth ag oras th eorem in PQR, we h ave

PQ2 + QR2 = PR2

2

2

bA4 + b 2 = PR2

PR = 2

42

bbA4 =

bbA4 42

From (i) an d (ii), we h ave

PRbA2

=

bNQ

NQ =PR

A2

NQ =42 bA4

Ab2

bbA4

PR22

Q

18181818.... InInInIn anananan equilateralequilateralequilateralequilateral triangletriangletriangletriangle ABCABCABCABC thethethethe sidesidesideside BCBCBCBC isisisis trisectedtrisectedtrisectedtrisected atatatat D.D.D.D. ProveProveProveProve thatthatthatthat 9999 ADADADAD2222 ==== 7777 ABABABAB2222....

ProofProofProofProof:::: Let ABC b e an eq u ila te ra l trian g le an d le t D b e a p oin t on BC su ch th a t

BD =31 BC.

Draw AE BC. Jo in AD

In AEB, an d AEC, we h ave AB = AC,

AEB = AEC = 9 0

an d , AE = AE

So, b y RHS- crite rion of s im ila rity, we h ave

AEB ~ AEC

BE = EC

Th u s , we h ave

BD =31 BC, DC =

32 BC an d BE = EC =

21 BC …..(i)

Sin ce C = 6 0 . Th ere fore , ADC is an acu te trian g le .

AD2 = AC2 + DC2 2 DC EC

A

EDB C


AD2 = AC2 +2

BC32

232 BC

21 BC [Us in g (i)]

AD2 = AC2 +94 BC2

32 BC2

AD2 = AB2 +94 AB2

32 AB2 [Q AB = BC = AC]

AD2 =9

AB6AB4AB9 222 =97 AB2

9 AD2 = 7 AB2.

19191919.... InInInIn thethethethe fig.,fig.,fig.,fig., DDDD isisisis aaaa pointpointpointpoint onononon sidesidesideside BCBCBCBC ofofofof ABCABCABCABC suchsuchsuchsuch thatthatthatthatCDCDCDCDBDBDBDBD

====

ACACACACABABABAB

.... ProveProveProveProve thatthatthatthat ADADADAD isisisis thethethethe bisectorbisectorbisectorbisector ofofofof BAC.BAC.BAC.BAC.

GivenGivenGivenGiven :::: ABC is a trian g le an d D is p o in t on BC su ch th a t

CDBD

====ACAB

....

ToToToTo proveproveproveprove :::: AD is th e b is ecto r of BAC.

ConstructionConstructionConstructionConstruction :::: Prod u ce BA to E su ch th a t AE = AC. Jo in CE.

ProofProofProofProof :::: In AEC, s in ce AE = AC, h en ce

AEC = ACE ….(i) [Q An g les op p . To eq u al s id es of a are eq u al]

Now,CDBD =

ACAB [Given ]

So ,CDBD =

AEAB [Q AE = AC, b y con s tru ct ion ]

By con verse of Bas ic Prop ort ion a lly Th eorem , we h ave:

DA | | CE

Now, s in ce CA is a tran sve rsa l, we h ave:

BAD = AEC ….(ii) [Corres p on d in g s ]

an d DAC = ACE ….(iii) [ Alte rn a te an g les ]

a lso , AEC = ACE [Fro m (i)]

Hen ce , BAD = DAC [Fro m (ii) an d (iii)]

Th u s , AD b is ects BAC.

20202020.... IfIfIfIf twotwotwotwo sidessidessidessides andandandand aaaa medianmedianmedianmedian bisectingbisectingbisectingbisecting thethethethe thirdthirdthirdthird sidesidesideside ofofofof aaaa triangletriangletriangletriangle areareareare respectivelyrespectivelyrespectivelyrespectively proportionalproportionalproportionalproportional

totototo thethethethe correspondingcorrespondingcorrespondingcorresponding sidessidessidessides andandandand thethethethe medianmedianmedianmedian ofofofof anotheranotheranotheranother triangle,triangle,triangle,triangle, thenthenthenthen proveproveproveprove thatthatthatthat thethethethe twotwotwotwo

trianglestrianglestrianglestriangles areareareare similar.similar.similar.similar.

A

DB C

E


GivenGivenGivenGiven :::: Two trian g les ABC an d DEF, in wh ich AP an d DQ are th e m ed ian s , su ch th a t

DEAB

====DFAC

====DQAP

ToToToTo ProveProveProveProve :::: ABC ~ DEF

ConstructionConstructionConstructionConstruction :::: Prod u ce AP to G, so th a t PG = AP.

Jo in CD. Also , p rod u ce DQ to H,

so th a t QH = DQ. Jo in FH.

ProofProofProofProof :::: In APB an d GPC,

BP = CP [Q AP is th e m ed ian ]

AP = GP [By con s tru ct ion ]

an d APB = CPG [Vert ica lly op p os ite an g les ]

APB GPC [SAS Th eorem of con g ruen ce]

AB = GC …(i) [C.P.C.T.]

Ag ain , in DQE an d HQF,

EQ = FQ [Q DQ is th e m ed ian ]

DQ = HQ [By con s tru ct ion ]

an d DQE = HQF [Vert ica lly op p o s ite an g les ]

DQE HQF [SAS Th eorem of con g ruen ce]

DE = HF ….(ii) [C.P..C.T.]

Now,DEAB =

DFAC =

DQAP [Given ]

HFGC =

DFAC =

DQAP [Fro m (i) an d (ii), AB = GC an d DE = HF]

HFGC =

DFAC =

DQ2AP2

HFGC =

DFAC =

DHAG [Q 2 AP = AG an d 2 DQ = DH].

AGC ~ DHF [By SSS s im ila rity]

1 = 2

Sim ila rly, 3 = 4

Th u s , 1 + 3 = 2 + 4 A = D ….(iii)

Th u s , in ABC an d DEF,

A = D [Fro m (iii)]

3

P

G

C

1

B

A

4

Q

H

F

2

E

D


an dDEAB =

DFAC [Given ]

So , b y SAS s im ila rity, ABCABCABCABC ~~~~ DEFDEFDEFDEF.

21. The areas of two similar triangles are 36 cm2 and 25 cm2 respectively. If the one side of the secondtriangle is 3 cm long, find the length of the corresponding side of the first triangle.

Solution:Let A1, A2 b e th e areas of th e trian g les an d s1, s2 to b e th e ir corresp on d in g s id es .Th en

22222222.... IfIfIfIf thethethethe areasareasareasareas ofofofof twotwotwotwo similarsimilarsimilarsimilar trianglestrianglestrianglestriangles areareareare equal,equal,equal,equal, proveproveproveprove thatthatthatthat theytheytheythey areareareare congruent.congruent.congruent.congruent.GivenGivenGivenGiven:::: Let ABC ~ PQRToToToTo Prove:Prove:Prove:Prove: ABC PQRProof:

ABC ~ PQR (g iven ) A(ABC) = A(PQR) (i) (g iven )

Now

2

2

2

2

2

2

QRBC

PRAC

PQAB

)PQR(A)ABC(A

(ii) (Th m . of area of s im ila r trian g les )

2

2

2

2

2

2

QRBC

PRAC

PQAB1 (from i an d ii)

AB = PQ, AC = PR, BC = QR ABC PQR (SSS tes t )

22223.3.3.3. InInInIn thethethethe givengivengivengiven figure,figure,figure,figure, DEDEDEDE |||| |||| ABABABAB andandandand EFEFEFEF |||| |||| BD,BD,BD,BD, proveproveproveprove thatthatthatthat CDCDCDCD2222 ==== ACACACAC ×××× CF.CF.CF.CF.

Proof:Proof:Proof:Proof: In CBD, EF | | BD

CBCE

CDCF

[By p rovin g CFE an d CDB sim ila r b y AA tes t]

In CAB, DE | | AB

CBCE

CACD

[By p rovin g CDE an d CAB sim ila r b y AA tes t]

CDCF

CACD

CF × AC = CD2

22224.4.4.4. ABCABCABCABC isisisis aaaa triangle.triangle.triangle.triangle. PQPQPQPQ isisisis aaaa linelinelineline segmentsegmentsegmentsegment intersectingintersectingintersectingintersecting ABABABAB inininin PPPP andandandand ACACACAC inininin QQQQ suchsuchsuchsuch thatthatthatthat PQPQPQPQ |||| |||| BCBCBCBC andandandand


dividesdividesdividesdivides ∆∆∆∆ABCABCABCABC intointointointo twotwotwotwo partspartspartsparts equalequalequalequal inininin area.area.area.area. FindFindFindFindABBP ....

ProofProofProofProof::::

22225.5.5.5. ProveProveProveProve thatthatthatthat thethethethe areaareaareaarea ofofofof equilateralequilateralequilateralequilateral triangletriangletriangletriangle describeddescribeddescribeddescribed onononon thethethethe sidesidesideside ofofofof aaaa squaresquaresquaresquare isisisis halfhalfhalfhalf thethethethe areaareaareaarea ofofofofthethethethe equilateralequilateralequilateralequilateral triangletriangletriangletriangle describeddescribeddescribeddescribed onononon itsitsitsits diagonal.diagonal.diagonal.diagonal.

ProofProofProofProof::::Let ABCD is a sq u are .In ABC,By Pyth ag oras th m ,AB2 + BC2 = AC2

2 AB2 = AC2

21

ACAB

2

2

(i)

ABE an d ACF are eq u ila te ra l trian g les . ABE ~ ACF

2

2

ACAB

)ACF(A)ABE(A

21

)ACF(A)ABE(A

(from i) Hen ce p roved .

22226.6.6.6. InInInIn thethethethe givengivengivengiven figure,figure,figure,figure, DEDEDEDE |||| |||| BCBCBCBC andandandand AD/DBAD/DBAD/DBAD/DB ==== 2/32/32/32/3.... Calculate:Calculate:Calculate:Calculate:


(i)

(ii)

An s :(i) ADE ~ ABC [AA Sim ilarity]

32

DBAD

[g iven ]

If AD = 2x, th en DB = 3x AB = 5x

)ABC(A)ADE(A

= 2

2

ABAD

= 2

2

x5x2

= 4 : 2 5

(ii) If a r( ADE) = 4y, th en ar( ABC) = 2 5y a r(trap DBCE) = 2 5y – 4y = 2 1y

y25y21

)ABC(A)DECBtrap(A = 2 1 : 2 5

22227.7.7.7. InInInIn thethethethe givengivengivengiven figure,figure,figure,figure, ABCABCABCABC isisisis aaaa triangletriangletriangletriangle withwithwithwith ∠EDBEDBEDBEDB ==== ∠ACB.ACB.ACB.ACB. ProveProveProveProve thatthatthatthat ∆∆∆∆ABCABCABCABC ~~~~ ∆∆∆∆EBD.EBD.EBD.EBD. IfIfIfIf BEBEBEBE ==== 6666 cm,cm,cm,cm,ECECECEC ==== 4444 cm,cm,cm,cm, BDBDBDBD ==== 5555 cmcmcmcm andandandand areaareaareaarea ofofofof ∆∆∆∆BEDBEDBEDBED ==== 9999 cmcmcmcm2222,,,, calculate:calculate:calculate:calculate:

((((iiii)))) LengthLengthLengthLength ofofofof AB.AB.AB.AB.((((iiiiiiii)))) AreaAreaAreaArea ofofofof ∆∆∆∆ABC.ABC.ABC.ABC.

SolutionSolutionSolutionSolution::::


2 8 . In th e g iven fig u re , DE | | BC an d CD | | EF. Prove th a t AD2 = AB × AF

Proof:Proof:Proof:Proof: In ABC,DE | | BC

ACAE

ABAD

[b y p rovin g ADE an d ABC sim ila r] ...(i)

In ACD, EF | | CD

ACAE

ADAF

[b y p rovin g AFE an d ADC sim ila r] ...(ii)


AD2 = AB × AF Hen ce p roved .

22229.9.9.9. LetLetLetLet XXXX bebebebe anyanyanyany pointpointpointpoint onononon thethethethe sidesidesideside BCBCBCBC ofofofof aaaa ∆∆∆∆ABC.ABC.ABC.ABC. XM,XM,XM,XM, XNXNXNXN areareareare drawndrawndrawndrawn parallelparallelparallelparallel totototo BABABABA andandandand CACACACA meetingmeetingmeetingmeetingCA,CA,CA,CA, BABABABA inininin M,M,M,M, NNNN respectively;respectively;respectively;respectively; MNMNMNMN meetsmeetsmeetsmeets BCBCBCBC producedproducedproducedproduced inininin T.T.T.T. ProveProveProveProve thatthatthatthat TXTXTXTX2222 ==== TBTBTBTB ×××× TCTCTCTC


Proof:Proof:Proof:Proof: In TMX,BN | | XM

TMTN

TXTB

[By p rovin g TBN an d TXM] ...(i)

In TMC,NX | | AC

TMTN

TCTX

[By p rovin g TXN an d TCM] ...(ii)


TX2 = TB × TC Hen ce p roved .

33330.0.0.0. InInInIn thethethethe givengivengivengiven figure,figure,figure,figure, ADADADAD andandandand CECECECE areareareare mediansmediansmediansmedians ofofofof ∆∆∆∆ABC.ABC.ABC.ABC. DFDFDFDF isisisis drawndrawndrawndrawn parallelparallelparallelparallel totototo CE.CE.CE.CE. ProveProveProveProve thatthatthatthat((((iiii)))) EFEFEFEF ==== FBFBFBFB((((iiiiiiii)))) AGAGAGAG :::: GDGDGDGD ==== 2222 :::: 1111

Proof:Proof:Proof:Proof:(i) In BCE, FD | | EC

[Us in g BPT]Bu t BD = DC [ D is th e m id - p o in t o f BC] 1 = BE/ EF EF = BF Hen ce p roved .(ii) E is th e m id - p o in t o f AB AE = EB = x (say)Als o , BF = EF = x/ 2In ADF, DF | | EG

[Us in g BPT]

AG : GD = 2 : 1 Hen ce p roved .

33331.1.1.1. InInInIn thethethethe givengivengivengiven figure,figure,figure,figure, ∠ABDABDABDABD ==== ∠CDBCDBCDBCDB ==== ∠PQBPQBPQBPQB ==== 90909090°°°°.... IfIfIfIf ABABABAB ==== xxxx units,units,units,units, CDCDCDCD ==== yyyy unitsunitsunitsunits andandandand PQPQPQPQ ==== zzzz units,units,units,units,

proveproveproveprove thatthatthatthat zzzz ====yx

xy

....


33332.2.2.2. ABCDABCDABCDABCD isisisis aaaa trapeziumtrapeziumtrapeziumtrapezium inininin whichwhichwhichwhich ABABABAB |||| |||| CDCDCDCD andandandand ABABABAB ==== 2CD.2CD.2CD.2CD. TheTheTheThe diagonalsdiagonalsdiagonalsdiagonals ACACACAC andandandand BDBDBDBD intersectintersectintersectintersect eacheacheacheachotherotherotherother atatatat O.O.O.O. IfIfIfIf ar(ar(ar(ar(∆∆∆∆AOB)AOB)AOB)AOB) ==== 84848484 cmcmcmcm2222,,,, findfindfindfind ar(ar(ar(ar(∆∆∆∆COD).COD).COD).COD).

SolutionSolutionSolutionSolution::::


3 3 . Eq u ila te ra l trian g les are d rawn on th e s id es of a rig h t an g led trian g le . Sh ow th at th e area of th etrian g le on th e h yp oten u se is eq u al to su m of areas of o th er two trian g les d rawn on th e rem ain in gtwo s id es .


��


CircleCircleCircleCircle

1111.... InInInIn fig.,fig.,fig.,fig., XPXPXPXP andandandand XQXQXQXQ areareareare tangentstangentstangentstangents fromfromfromfrom XXXX totototo thethethethe circlecirclecirclecircle withwithwithwith centrecentrecentrecentre O.O.O.O. RRRR isisisis aaaa pointpointpointpoint onononon thethethethe circle.circle.circle.circle.ProveProveProveProve that,that,that,that, XAXAXAXA ++++ ARARARAR ==== XBXBXBXB ++++ BR.BR.BR.BR.


Sin ce len g th s of tan g en ts from an exte rio rp o in t to a circle are eq u al.

XP = XQ [Fro m X] ….(i)AP = AR [Fro m A] ….(ii)BQ = BR [From B] .…(iii)

Now, XP = XQ XA + AP = XB + BQ XA + AR = XB + BR [Us in g (i) an d (ii)]

2.2.2.2. InInInIn fig.,fig.,fig.,fig., thethethethe incircleincircleincircleincircle ofofofof ABCABCABCABC touchestouchestouchestouches thethethethe sidessidessidessides BC,BC,BC,BC, CACACACA andandandand ABABABAB atatatat D,D,D,D, EEEE andandandand FFFF respectively.respectively.respectively.respectively. ShowShowShowShow

thatthatthatthat AFAFAFAF ++++ BDBDBDBD ++++ CECECECE ==== AEAEAEAE ++++ BFBFBFBF ++++ CDCDCDCD ====2222

1111(Perimeter(Perimeter(Perimeter(Perimeter ofofofof ABC)ABC)ABC)ABC)

Solution:Solution:Solution:Solution:

Sin ce len g th s of th e tan g en ts from an exterio r p o in t to a circle are eq u al.

AF = AE [From A] ….(i)

BD = BF [Fro m B] ….(ii)

an d , CE = CD [From C] ….(iii)

Ad d in g eq u ation s (i), (ii) an d (iii), we g et

AF + BD + CE = AE + BF + CD

Now,

Perim ete r of ABC = AB + BC + AC

Perim ete r of ABC = (AF + FB) + (BD + CD) + (AE + EC)

Perim ete r of ABC = (AF + AE) + (BF + BD) + (CD + CE)

Perim ete r of ABC = 2 AF + 2 BD + 2 CE

Perim ete r ABC = 2 (AF + BD + CE)

CECDan d,BFBD,AFAE

g etwe),iii(an d)ii(),i(From

AF + BD + CE =21 (Perim e ter of ABC)

Hen ce , AF + BD + CE = AE + BF + CD =21 (Perim ete r of ABC).

Q

RX

A

B

P

B

E

D

F

C

A

S

R

O

P

C

Q

D

B

A


3333.... ABCDABCDABCDABCD isisisis aaaa quadrilateralquadrilateralquadrilateralquadrilateral suchsuchsuchsuch thatthatthatthat DDDD ==== 90909090.... AAAA circlecirclecirclecircle C(O,C(O,C(O,C(O, r)r)r)r) touchestouchestouchestouches thethethethe sidessidessidessides AB,AB,AB,AB, BC,BC,BC,BC, CDCDCDCD andandandand

DADADADA atatatat P,P,P,P, Q,Q,Q,Q, RRRR andandandand SSSS respectively.respectively.respectively.respectively. IfIfIfIf BCBCBCBC ==== 38383838 cm,cm,cm,cm, CDCDCDCD ==== 25252525 cmcmcmcm andandandand BPBPBPBP ==== 27cm,27cm,27cm,27cm, findfindfindfind r.r.r.r.


Sin ce tan g en ts to a circle is p erp en d icu lar to

th e rad iu s th rou g h th e p oin t .

ORD = OSD = 9 0

It is g iven th a t D = 9 0 .

Als o , OR = OS. Th ere fore , ORDS is a sq u are .

Sin ce tan g en ts from an exte rio r p o in t to a

circle are eq u al in len g th .

BP = BQ

CQ = CR

an d , DR = DS.

Now,

BP = BQ

BQ = 2 7 [Q BP = 2 7 cm (Given )]

BC CQ = 2 7

3 8 CQ = 2 7 [Q BC = 3 8 cm ]

CQ = 1 1 cm

CR = 1 1 cm [Q CR = CQ]

CD DR = 1 1

2 5 DR = 1 1 [Q CD = 2 5 cm ]

DR = 1 4 cm

Bu t, ORDS is a sq u are . Th ere fore , OR = DR = 1 4 cm .

Hen ce , r = 1 4 cm .

4444.... InInInIn fig.,fig.,fig.,fig., llll andandandand mmmm areareareare twotwotwotwo parallelparallelparallelparallel tangentstangentstangentstangents atatatat AAAA andandandand B.B.B.B. TheTheTheThe tangenttangenttangenttangent atatatat CCCC makesmakesmakesmakes anananan interceptinterceptinterceptintercept DEDEDEDE

betweenbetweenbetweenbetween llll andandandand m.m.m.m. ProveProveProveProve thatthatthatthat DFEDFEDFEDFE ==== 90909090....


Sin ce tan g en ts d rawn from an exte rn a l p o in t

to a circle are eq u al. Th erefore , DA = DC.

Th u s , in trian g les ADF an d DFC, we h ave

DA = DC

DF = DF [Com m on ]

S

R

O

P

C

Q

D

B

A

D

m

l

C

E

F

B

A


AF = CF [Rad ii o f th e sam e circle ]

So , b y SSS- crite rion of con g ru en ce , we h ave

ADF DFC

ADF = CDF

ADC = 2 CDF …..(i)

Sim ila rly, we can p rove th a t

BEF = CEF

CEB = 2 CEF

Now, ADC + CEB = 1 8 0 [su n of th e in te rio r an g les on th e sam e s id e of tran svers a l is 1 8 0 ]

2 CDF + 2 CEB = 1 8 0 [Us in g eq u ation s (i) an d (ii)]

CDF + CEF = 9 0

1 8 0 DFE = 9 0

o

Q

1 8 0DFECEFCDF

trian g leaofan g lesareDFEan dCEF,CDF

DFE = 9 0 .

5555.... TheTheTheThe radiusradiusradiusradius ofofofof thethethethe incircleincircleincircleincircle ofofofof aaaa triangletriangletriangletriangle isisisis 4444 cmcmcmcm andandandand thethethethe segmentssegmentssegmentssegments intointointointo whichwhichwhichwhich oneoneoneone sidesidesideside isisisis

divideddivideddivideddivided bybybyby thethethethe pointpointpointpoint ofofofof contactcontactcontactcontact areareareare 6666 cmcmcmcm andandandand 8888 cm.cm.cm.cm. DetermineDetermineDetermineDetermine thethethethe otherotherotherother twotwotwotwo sidessidessidessides ofofofof thethethethe

triangle.triangle.triangle.triangle.


Let l b e th e in cen tre of ABC su ch th a t in - rad iu s = IL = IM = IN = 4 cm .

Als o , AM = 6 cm ,an d CM = 8 cm .

Let BL = = x cm

We h ave ,

AM = 6 cm an d AM = AN

AN = 6 cm

Sim ilarly, CL = CM = 8 cm

a = BC = BL + CL = (x + 8 )cm

b = AC = AM + CM = (6 + 8 )cm = 1 4 cm

an d , c = AB = AN + = (x + 6 )cm

2 s = a + b + c

2 s = x + 8 + 1 4 + x + 6

s = x + 1 4

Now,

Area of ABC = csbsass

4 cm

C

N

A

M

L

I

B

6 cm

8 cmX cm4 cm

4 cm

X cm


= 6x1 4x1 41 4x8x1 4x)1 4x(

= 8x61 4x

= 1 4xx4 8 …….(i)

Als o ,

Area of ABC = Area of IBC + Area ICA + Area of IAB

=21

BC IL +21

CA IM +21 AB IN

=21 (x + 8 ) 4 +

21 1 4 4 +

21 (x + 6 ) 4

= 2 (x + 8 ) + 2 8 + 2 (x + 6 ) cm 2

= 4 x + 5 6 cm 2


1 4xx4 8 = (4 x + 5 6 )

4 8 x(x + 1 4 ) = (4 x + 5 6 )2

4 8 x(x + 1 4 ) = 1 6 (x + 1 4 )2

3 x(x + 1 4 ) = (x + 1 4 )2

3 x(x + 1 4 ) (x + 1 4 )2 = 0

(x + 1 4 )(3 x x 1 4 ) = 0

2 (x + 1 4 ) (x 7 ) = 0

x 7 = 0

x = 7

BC = (x + 8 )cm = 1 5 cm an d AB = (x + 6 )cm = 1 3 cm .

6666.... PQPQPQPQ isisisis aaaa chordchordchordchord ofofofof lengthlengthlengthlength 8cm8cm8cm8cm ofofofof aaaa circlecirclecirclecircle ofofofof radiusradiusradiusradius 5cm.5cm.5cm.5cm. TheTheTheThe tangentstangentstangentstangents atatatat PPPP andandandand QQQQ intersectintersectintersectintersect atatatat aaaa

pointpointpointpoint T.T.T.T. findfindfindfind thethethethe lengthlengthlengthlength TP.TP.TP.TP.


Let TR = y.

Sin ce OT is p erp en d icu lar b isecto r of PQ.

PR = QR = 4 cm

In rig h t trian g le ORP, we h ave

OP2 = OR2 + PR2

OR2 = OP2 PR2 = 5 2 4 2 = 9

OR = 3 cm .

4 cm

y

5 cm

x

R

P

O

Q

T4 cm

5 cm


In rig h t trian g le PRT an d OPT, we h ave

TP2 = TR2 + PR2

An d , OT2 = TP2 + OP2

OT2 = (TR2 + PR2) + OP2 [Su b s t itu t in g th e va lu e of TP2]

(y + 3 )2 = y2 + 1 6 + 2 5

6 y = 3 2

y =3

1 6

TR =3

1 6

TP2 = TR2 + PR2

TP2 =2

31 6

+ 4 2 =

92 5 6 + 1 6 =

94 0 0

TP =3

2 0 cm .

7777.... InInInIn thethethethe givengivengivengiven figure,figure,figure,figure, OOOO isisisis thethethethe centrecentrecentrecentre ofofofof thethethethe circle.circle.circle.circle. FindFindFindFind thethethethe valuevaluevaluevalue ofofofof xxxx....

SolutionSolutionSolutionSolution::::We have x = ACB [Ang les in the sam e segm en t]Also , AEB = ACB + CBE [ Ex terio r ang le p rop erty]

1 2 0 ° = x + 2 5 ° x = 1 2 0 ° – 2 5 ° = 9 5 °

8888.... InInInIn thethethethe givengivengivengiven figure,figure,figure,figure, ∠ACEACEACEACE ==== 43434343°°°° andandandand ∠CAFCAFCAFCAF ==== 62626262°°°°.... IfIfIfIf ∠AECAECAECAEC ==== xxxx,,,, findfindfindfind thethethethe valuesvaluesvaluesvalues ofofofof aaaa,,,, bbbb andandandand cccc....

Solution:Solution:Solution:Solution:Since CAF = 6 2 °, so c = 6 2 °[Ex terio r ang le of a cyclic q uad rila tera l is eq u al to the in te rior op p os ite ang le]


In ACE, CAE + ACE + AEC = 1 8 0 ° [Sum of ang les of t riang le is 1 8 0 °] 6 2 ° + 4 3 ° + x = 1 8 0 ° 1 0 5 ° + x = 1 8 0 ° x = 7 5 °Also , x = b + c[Ex terio r ang le of a triang le is eq ua l to the sum of two in te rior op p os ite ang les ] 7 5 ° = b + 6 2 ° b = 7 5 ° – 6 2 ° b = 1 3 °Now, a + x = 1 8 0 ° [Op p os ite ang les of a cyclic q uad rila tera l] a + 7 5 ° = 1 8 0 ° a = 1 0 5 °Hen ce , a = 1 0 5 °, b = 1 3 °, c = 6 2 °

9999.... InInInIn thethethethe givengivengivengiven figure,figure,figure,figure, ABCDABCDABCDABCD isisisis aaaa cycliccycliccycliccyclic quadrilateral.quadrilateral.quadrilateral.quadrilateral. IfIfIfIf5z

4y

3x

findfindfindfind thethethethe valuevaluevaluevalue ofofofof xxxx,,,, yyyy andandandand zzzz....


Given5z

4y

3x

= k (s ay)

If x = 3k, then y = 4k and z = 5kNow, BCP = DCQ = x [Vert ica lly op p os ite ang les ]Also , ABC = x + y [Ex terio r ang le p rop e rty]and ADC = x + z [Ex te rior ang le p rop e rty]Bu t ABC + ADC = 1 8 0 ° [Op p os ite ang les of a cyclic q uad rila tera l] x + y + x + z = 1 8 0 ° 2x + y + z = 1 8 0 ° 6k + 4k + 5k = 1 8 0 ° 1 5k = 1 8 0 ° k = 1 2 ° x = 3 × 1 2 ° = 3 6 °y = 4 × 1 2 ° = 4 8 °z = 5 × 1 2 ° = 6 0 °

10101010.... InInInIn aaaa cycliccycliccycliccyclic quadrilateral,quadrilateral,quadrilateral,quadrilateral, ifififif oneoneoneone pairpairpairpair ofofofof oppositeoppositeoppositeopposite sidessidessidessides isisisis equal,equal,equal,equal, thethethethe otherotherotherother pairpairpairpair isisisis parallel.parallel.parallel.parallel.ProveProveProveProve it.it.it.it.

Solution:Solution:Solution:Solution:Jo in O to A, B, C, D and jo in A to C

AD = BC [Given ]


AOD = BOC [Eq ua l chord s of a circle sub tend s eq u al ang les a t the cen tre of the circle] 2 ACD = 2 BAC[Ang le sub tend ed b y an arc a t the cen tre is d ou b le the ang le sub ten d ed b y it a t any othe r p oin t on

the rem ain ing p art of the circle]or ACD = BACBut they are a lte rna te in terio r ang les AB | | DC Hen ce p roved .

1 1 . ABCD is a cyclic q uad rila tera l in which DAC = 2 7 °, DBA = 5 0 ° and ADB = 3 3 °. Calcu la te :(i) DBC (ii) DCB (iii) CAB

Solution:Solution:Solution:Solution:(i) DBC = DAC = 2 7 ° [Ang les in the sam e seg m en t](ii) DCA = DBA = 5 0 ° [An g les in the sam e segm en t]

ACB = ADB = 3 3 ° [Ang les in the sam e segm en t] DCB = DCA + ACB = 5 0 ° + 3 3 ° = 8 3 °

(iii) In ADB, ADB + DAB + ABD = 1 8 0 ° [Sum of ang les of t riang le is 1 8 0 °] 3 3 ° + 2 7 ° + CAB + 5 0 ° = 1 8 0 ° [ DAB = DAC + CAB] CAB + 1 1 0 ° = 1 8 0 ° CAB = 7 0 °

12121212.... InInInIn thethethethe givengivengivengiven figure,figure,figure,figure, PQPQPQPQ andandandand RSRSRSRS areareareare twotwotwotwo straightstraightstraightstraight lineslineslineslines throughthroughthroughthrough thethethethe centrecentrecentrecentre OOOO ofofofof aaaa circle.circle.circle.circle. IfIfIfIf∠PORPORPORPOR ==== 80808080°°°° andandandand ∠RSKRSKRSKRSK ==== 40404040°°°°,,,, findfindfindfind thethethethe numbernumbernumbernumber ofofofof degreesdegreesdegreesdegrees inininin((((iiii)))) ∠SRKSRKSRKSRK ((((iiiiiiii)))) ∠PQRPQRPQRPQR

Solution:Solution:Solution:Solution:(i) Jo in PR. SKR = 9 0 ° [An g le in a sem icircle]

In SRK, SRK + SKR + RSK = 1 8 0 ° [Sum of ang les of t riang le is 1 8 0 °]


SRK + 9 0 ° + 4 0 ° = 1 8 0 ° SRK + 1 3 0 ° = 1 8 0 ° SRK = 5 0 °(ii) Also , OP = OR [Rad ii of sam e circle] ORP = OPR = x (s ay) [An g les op p os ite to eq u al s id es of a triang le are eq u al]In POR, ORP + OPR + POR = 1 8 0 ° [Sum of ang les of t riang le is 1 8 0 °] x + x + 8 0 ° = 1 8 0 ° 2x = 1 0 0 ° x = 5 0 °In PRQ, P + PRK + PQR = 1 8 0 ° [Sum of ang les of t riang le is 1 8 0 °] 5 0 ° + ORP + SRK + PQR = 1 8 0 ° 5 0 ° + 5 0 ° + 5 0 ° + PQR = 1 8 0 ° 1 5 0 ° + PQR = 1 8 0 ° PQR = 1 8 0 ° – 1 5 0 ° PQR = 3 0 °

13131313.... InInInIn thethethethe givengivengivengiven figure,figure,figure,figure, OOOO isisisis thethethethe centrecentrecentrecentre ofofofof thethethethe circle,circle,circle,circle, ∠BADBADBADBAD ==== 75757575°°°° andandandand chordchordchordchord BCBCBCBC ==== chordchordchordchord CD.CD.CD.CD.Find:Find:Find:Find: ((((iiii)))) ∠BCDBCDBCDBCD ((((iiiiiiii)))) ∠OBDOBDOBDOBD ((((iiiiiiiiiiii)))) ∠BOCBOCBOCBOC

Solution:Solution:Solution:Solution:Jo in O to C and B to D.

(i) We have BCD + BAD = 1 8 0 ° [Op p os ite ang les of a cyclic q uad rila tera l] BCD + 7 5 ° = 1 8 0 ° BCD = 1 0 5 °(ii) Now, BOD = 2 BAD = 2 × 7 5 ° = 1 5 0 °[Ang le sub tend ed b y an arc a t the cen tre is d ou b le the ang le sub ten d ed b y it a t any othe r p oin t onthe rem ain ing p art of the circle]In BOD, BO = OD [Rad ii of sam e circle] 2 = 1 = x(say) [An g les op p os ite to eq u al s id es of t riang le are eq u al]


Now, 1 + 2 + BOD = 1 8 0 ° [Sum of ang les of t riang le is 1 8 0 °] x + x + 1 5 0 ° = 1 8 0 ° 2x = 3 0 ° x = 1 5 ° OBD = x = 1 5 °(iii) Eq ua l chord s sub tend eq ua l ang les a t the cen tre BOC = CODBut BOC + COD = BOD BOC + BOC = 1 5 0 ° 2 BOC = 1 5 0 ° BOC = 7 5 °

14141414.... InInInIn thethethethe givengivengivengiven figure,figure,figure,figure, ABABABAB ==== ACACACAC ==== CD.CD.CD.CD. IfIfIfIf ∠ADCADCADCADC ==== 38383838°°°°,,,, calculate:calculate:calculate:calculate:((((iiii)))) ∠ABCABCABCABC ((((iiiiiiii)))) ∠BCEBCEBCEBCE

Solution:Solution:Solution:Solution:(i) AC = CD [Given ] CAD = CDA [Ang les op p os ite to eq u al s id es of t riang le are eq ua l] CAD = 3 8 °

Now, ACB = CAD + CDA [Exte rior ang le p rop e rty]= 3 8 ° + 3 8 ° ACB = 7 6 °Also , AB = AC [Given] ABC = ACB [Ang les op p os ite to eq ua l s id es of t riang le are eq ua l] ABC = 7 6 °(ii) Now, CED = ABC [Ex terio r ang le of a cyclic q uad rila tera l is eq ua l to in terio r op p os ite

ang le ) CED = 7 6 °Now, BCE = CED + EDC [Ex te rior ang le p rop e rty] BCF = 7 6 ° + 3 8 ° BCE = 1 1 4 °

15151515.... InInInIn triangletriangletriangletriangle PQR,PQR,PQR,PQR, PQPQPQPQ ==== 24242424 cm,cm,cm,cm, QRQRQRQR ==== 7777 cmcmcmcm andandandand ∠PQRPQRPQRPQR ==== 90909090°°°°....FindFindFindFind thethethethe radiusradiusradiusradius ofofofof thethethethe inscribedinscribedinscribedinscribed circle.circle.circle.circle.


Solution:Solution:Solution:Solution:In righ t- ang led PQR, PR2 = PQ2 + QR2 = (2 4 )2 + (7 )2 = 5 7 6 + 4 9 = 6 2 5PR = 2 5 cmOA = OB (Rad ii of sam e circle)and QA = QB (Tangen ts from exte rnal p oin t)Also , OB PQ, OA QR (Rad ius is to the tangen t)and PQR = 9 0 ° Fourth ang le AOB = 9 0 °

Each ang le = 9 0 ° and ad jacen t s id es are eq u al. OAQB is a sq uare .OA = QA = BQ = OB = x cm (say)CR = AR = QR – QA = (7 – x) cm [ Tang en ts from ex terna l p oin t are eq ua l]CP = PB = PQ – BQ = (2 4 – x) cmNow, CR + CP = PR 7 – x + 2 4 – x = 2 5 3 1 – 2x = 2 5 3 1 – 2 5 = 2x 6 = 2x x = 3 cm

11116666.... InInInIn thethethethe givengivengivengiven figure,figure,figure,figure, quadrilateralquadrilateralquadrilateralquadrilateral ABCDABCDABCDABCD isisisis circumscribedcircumscribedcircumscribedcircumscribed andandandand ADADADAD ⊥ DC.DC.DC.DC. FindFindFindFind xxxx,,,, ifififif radiusradiusradiusradiusofofofof thethethethe incircleincircleincircleincircle isisisis 10101010 cm.cm.cm.cm.

Solution:Solution:Solution:Solution:Jo in OR. Rad ius is p erp end icu lar to the tangen t OSD = ORD = 9 0 °,


OS = OR and each ang le = 9 0 ° OSDR is a sq ua re . DR = 1 0 cmBQ = BP = 2 7 cm CQ = 3 8 – 2 7 = 1 1 cmCR = CQ = 1 1 cmx = DR + RC = 1 0 cm + 1 1 cm = 2 1 cm

11117777.... InInInIn thethethethe givengivengivengiven figure,figure,figure,figure, twotwotwotwo circlescirclescirclescircles touchtouchtouchtouch eacheacheacheach otherotherotherother externallyexternallyexternallyexternally atatatat aaaa pointpointpointpoint P.P.P.P. ABABABAB isisisis thethethethe directdirectdirectdirectcommoncommoncommoncommon tangenttangenttangenttangent ofofofof thesethesethesethese circles.circles.circles.circles. ProveProveProveProve that:that:that:that:((((iiii)))) ∠APBAPBAPBAPB ==== 90909090°°°° ((((iiiiiiii)))) tangenttangenttangenttangent atatatat pointpointpointpoint PPPP bisectsbisectsbisectsbisects ABABABAB

Solution:Solution:Solution:Solution:(i) AM = MP [Tan gen ts from an ex terna l p oin t are eq u al] MAP = MPASim ilarly, MP = MB MBP = MPBNow, in ABP, PAB + ABP + APB = 1 8 0 ° [Sum of ang les of t riang le is 1 8 0 °] MPA + MPB + MPA + MPB = 1 8 0 °2 ( MPA + MPB) = 1 8 0 ° APB = 9 0 °(ii) AM = MP and MP = MB AM = MB Tang en t a t P b isects AB. Hence p roved .

11118888.... InInInIn thethethethe givengivengivengiven figure,figure,figure,figure, proveproveproveprove that:that:that:that: APAPAPAP ++++ BQBQBQBQ ++++ CRCRCRCR ==== BPBPBPBP ++++ CQCQCQCQ ++++ ARARARAR

Solution:Solution:Solution:Solution:Tang en ts from ex ternal p oin t have eq u al leng th .

AP = AR ...(i)BQ = BP ...(ii)

and CR = CQ ...(iii)Ad d ing (i), (ii) and (iii), we ge t


AP + BQ + CR = BP + CQ + AR Hence p roved .

19.19.19.19. ThreeThreeThreeThree circlescirclescirclescircles touchtouchtouchtouch eacheacheacheach otherotherotherother externally.externally.externally.externally. AAAA triangletriangletriangletriangle isisisis formedformedformedformed bybybyby joiningjoiningjoiningjoining thethethethe centrescentrescentrescentres ofofofofthesethesethesethese threethreethreethree circles.circles.circles.circles. FindFindFindFind thethethethe radiiradiiradiiradii ofofofof thethethethe circles,circles,circles,circles, ifififif thethethethe sidessidessidessides ofofofof thethethethe triangletriangletriangletriangle formedformedformedformed areareareare 6666cm,cm,cm,cm, 8888 cmcmcmcm andandandand 9999 cm.cm.cm.cm.

Solution:Solution:Solution:Solution:Let rad ius of the circle with cen tre C b e r cm . Also , AB = 6 cm , BC = 8 cm and AC = 9 cm

BQ = (8 – r) cm BP = (8 – r) cm ,

AR = (9 – r) cm AP = (9 – r) cm

Now, AB = 6 cm

AP + BP = 6 cm

9 – r + 8 – r = 6 cm

1 7 – 6 = 2r

1 1 = 2 r

r = 5 .5 cm

BQ = 8 cm – 5 .5 cm = 2 .5 cm

AP = 9 – r = 9 cm – 5 .5 cm = 3 .5 cm

Hence , rad ii of the circles are 2 .5 cm , 3 .5 cm and 5 .5 cm .

��

1.

Ans :

2.

Ans :

3.

Ans :

4.

What point on the x-axis is equidistant from (7, 6) and (– 3, 4)?

Let A(7, 6), B(– 3, 4) be the given points and P(x, 0) be the required point.Since P is equidistant from A and B, therefore,

AP = BP AP2 = BP2

(x – 7)2 + (0 – 6)2 = (x + 3)2 + (0 – 4)2 x2 + 49 – 14x + 36 = x2 + 9 + 6x + 16

– 14x – 6x = 25 – 85 – 20x = – 60

x = = 3.

Required point on x-axis is (3, 0).

The centre of a circle is (2 – 1, 7) and it passes through the point (– 3, –1). If the diameter of

the circle is 20 units, then find the value of .

OA = 10 units

OA =

10 =

Squaring 100 = 4 2 + 8 + 68

4 2 + 8 – 32 = 0 2 + 2 – 8 = 02 + 4 – 2 – 8 = 0 ( + 4) – 2( + 4) = 0 ( +

4) ( – 2) = 0 = – 4, = 2

The point R divides the line segment AB where A(–4, 0), B(0, 6) are such that AR = AB. Find

the coordinates of R.

Let coordinates of R be (x, y).

AR = AB [Given]

But AR + RB = AB AB + RB = AB

RB = AB – =

Thus, R divides AB in the ratio 3 : 1.

x =

Thus, coordinates of R are .

If A(4, – 8), B(3, 6) and C(5, –4) are the vertices of ∆ABC, D is the mid point of BC and P is a

point on AD joined such that = 2, find the coordinates of P.

Co-ordinate Geometry

Ans :

5.

Ans :

6.

Ans :

7.

Ans :

A(4, – 8), B(3, 6) and C(5, – 4) are vertices of ∆ABC and D is the mid-pointof BC [Given]

Coordinates of D are

= 2 [Given]

AP : PD = 2 : 1

Coordinates of P are

If C is a point lying on the line segment AB joining A(1, 1) and B(2, – 3) such that 3AC = CB,

then find the coordinates of C.

= [Given]

Coordinates of C (x, y) = ,

x =

(x, y) =

If two vertices of a parallelogram are (3, 2), (– 1, 0) and the diagonals cut at (2, – 5), find theother vertices of the parallelogram.

Let coordinates of C be (x1, y1) and D be (x2, y2).

So, = 2 ... (i) and = – 5 ... (ii) [Mid-point

theorem]

Also, = 2 ... (iii) and = – 5 ... (iv)

[Mid-point theorem]

From equation (i), we get

= 2 x1 + 3 = 4 x1 = 4 – 3 = 1

Solving equation (ii), we get

y1 + 2 = –10

y1 = – 10 – 2 y1 = – 12

Solving equation (iii), we get

x2 – 1 = 4

x2 = 4 + 1 = 5

Solving equation (iv), we get

y2 + 0 = –10 y2 = –10

Coordinates of C are (1, –12) and D are (5, –10).

In what ratio does the line x – y – 2 = 0 divide the line segment joining (3, –1) and (8, 9)?

Let the line x – y – 2 = 0, divides the line segment joining (3, – 1) and (8, 9) in the ratio k

: 1 and let the coordinates of the required point be (x1, y1).

Then x1 =

and y1 =

This point (x1, y1) lies on the line whose equation is x – y – 2 = 0.

It must satisfy the equation of the given line

− 2 = 0

8k + 3 – (9k – 1) – 2(k + 1) = 0

8k + 3 – 9k + 1 – 2k – 2 = 0

– 3k + 2 = 0 k =

Therefore, the required ratio is k : 1 = : 1 or 2 : 3.

8.

Ans :

9.

Ans :

10.

Ans :

11.

Ans :

The points (p, q); (m, n) and (p – m, q – n) are collinear, show that pn = qm.

Given points are collinear

x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0

p(n – (q – n)) + m(q – n – q) + (p – m)(q – n) = 0

p(n – q + n) + m(– n) + (p – m)(q – n) = 0

2 pn – pq – mn + pq – pn – mq + mn = 0

pn – mq = 0 pn = mq

If the point (x, y) be equidistant from the points (a + b, b – a) and (a – b, a + b), prove that bx =

ay.

Let P(x, y), A(a + b, b – a) and B(a – b, a + b) be the given points.

Since AP = BP,

AP2 = BP2

(x – a – b)2 + (y – b + a)2 = (x – a + b)2 + (y – a – b)2

(x – a – b)2 – (x – a + b)2 = (y – a – b)2 – (y – b + a)2

(x – a – b + x – a + b) (x – a – b – x + a – b) = (y – a – b + y – b + a) (y – a – b – y + b

– a)

(2x – 2a)(– 2b) = (2y – 2b)(– 2a) – 4bx + 4ab = – 4ay + 4ab

– 4bx = – 4ay bx = ay. Hence proved.

Find the lengths of the medians of ABC having vertices at A(5, 1), B(1, 5) and C(– 3, – 1).

Let P, Q and R be the mid-points of the sides BC, AB and ACrespectively.

So, P =

and R =

P (–1, 2), Q (3, 3) and R (1, 0). AP, BR and CQ are the medians.

AP =

BR =

CQ =

If R(x, y) is a point on the line segment joining the points P(a, b) and Q(b, a), then prove that x

+ y = a + b.

R(x, y) lies on the line segment joining the points P(a, b) and Q(b, a). Then P, Q and Rlie on a line.

x(b – a) + a(a – y) + b(y – b) = 0 bx – ax + a2 – ay + by – b2 = 0

b(x + y) – a(x + y) + (a2 – b2) = 0 (b – a) (x + y) – (b – a) (b + a) = 0

(b – a) {(x + y) – (b + a)} = 0 (x + y) = (b + a) [Assuming a ≠ b]

Trigonom etry

1111.... IfIfIfIf tantantantan (A(A(A(A ++++ B)B)B)B) ==== 3 andandandand tantantantan (A(A(A(A B)B)B)B) ====3

1;;;; 0000 <<<< AAAA ++++ BBBB 90909090 ;;;; AAAA >>>> B,B,B,B, findfindfindfind AAAA andandandand B.B.B.B.

SolutionSolutionSolutionSolution :::: tan (A + B) = 3 ==== tan 6 0 A + B = 6 0 ……(i)

tan (A B) =3

1==== tan 3 0 A B = 3 0 ……(ii)

ad d in g (i) an d (ii)2 A = 9 0 A = 4 5 From (i) 4 5 + B = 6 0 B = 6 0 4 5 = 1 5 ; Hen ce A = 4 5 , B = 1 5 .

2.2.2.2. IfIfIfIf coscoscoscos (40(40(40(40 ++++ x)x)x)x) ==== sinsinsinsin 30303030 ,,,, FindFindFindFind x.x.x.x.SolutionSolutionSolutionSolution ::::

cos (4 0 + x) = s in 3 0

cos (4 0 + x) =21 . [Q s in 3 0 =

21 ]

cos (4 0 + x) = cos 6 0 [Q cos 6 0 =21 ]

4 0 + x = 6 0 x = 2 0 .

3333.... IfIfIfIf tantantantan 2A2A2A2A ==== cot(Acot(Acot(Acot(A 18181818),),),), wherewherewherewhere 2A2A2A2A isisisis anananan acuteacuteacuteacute angle,angle,angle,angle, findfindfindfind thethethethe valuevaluevaluevalue ofofofof A.A.A.A.SolutionSolutionSolutionSolution ::::

tan 2 A = cot (A 1 8 )tan 2 A = tan {9 0 (A 1 8 )} [Q tan (9 0 A) = cot A]tan 2 A = tan (1 0 8 A)

2 A = 1 0 8 A 3 A = 1 0 8 A = 3 6 .

4444.... IfIfIfIf tantantantan AAAA ==== cotcotcotcot B,B,B,B, ProveProveProveProve thatthatthatthat AAAA ++++ BBBB ==== 90909090....SolutionSolutionSolutionSolution ::::

tan A = cot Btan A = tan (9 0 B) [Q tan (9 0 ) = co t ]

A = 9 0 B A + B = 9 0 .

5555.... IfIfIfIf ADADADAD ==== 10101010 cm,cm,cm,cm, findfindfindfind BCBCBCBCSolution:Solution:Solution:Solution:

In ABD,

s in 3 0 =ADAB

21 =

10AB

AB = 5 cm .In ABC,

4 5 DB

3 0

A

C

tan 4 5 =BCAB

1 =BC5

BC = 5 cm .

6666.... IfIfIfIf A,A,A,A, B,B,B,B, CCCC areareareare internalinternalinternalinternal anglesanglesanglesangles ofofofof aaaa triangletriangletriangletriangle ABC,ABC,ABC,ABC, thenthenthenthen showshowshowshow thatthatthatthat sinsinsinsin

2

CB==== coscoscoscos

2A....


A + B + C = 1 8 0 [An g le su m p rop erty of a trian g le] B + C = 1 8 0 A

2

CB = 9 0 2A

s in

2

CB = s in

2A9 0

s in

2

CB====

2A [s in ce s in (9 0 A) = cos A]

7777.... ProveProveProveProve :::: 0Bs inAs inBcosAcos

BcosAcosBs inAs in


L.H.S. = 0Bs inAs inBcosAcos

BcosAcosBs inAs in

L.H.S. = Bs inAs inBcosAcos

BcosAcosBcosAcosBs inAs inBs inAs in

L.H.S. = Bs inAs inBcosAcosBcosAcosBs inAs in 2222

L.H.S. = Bs inAs inBcosAcos

BcosBs inAcosAs in 2222

L.H.S. = Bs inAs inBcosAcos11

= 0 = R.H.S.

8888.... IfIfIfIf sinsinsinsin ++++ coscoscoscos ==== pppp andandandand secsecsecsec ++++ coseccoseccoseccosec ==== q,q,q,q, showshowshowshow thatthatthatthat q(pq(pq(pq(p2222 –––– 1)1)1)1) ==== 2p.2p.2p.2p.SolutionSolutionSolutionSolution ::::

L.H.S. = q (p 2 – 1 ) L.H.S. = (sec + cosec ) {(s in + cos )2 1 }

L.H.S. =

s in

1cos

1 {s in 2 + cos 2 + 2 s in cos - 1 }

L.H.S. =

s incoscoss in (1 + 2 s in cos - 1 )

L.H.S. =

s incoscoss in (2 s in cos )

L.H.S. = 2 (s in + cos ) = 2 p = R.H.S.

9999.... ProveProveProveProve thatthatthatthat :::: 9999 secsecsecsec2222 5555 tantantantan2222 ==== 5555 ++++ 4444 secsecsecsec2222 SolutionSolutionSolutionSolution ::::

L.H.S. = 9 sec 2 5 tan 2 = 9 (1 + tan 2 ) 5 tan 2 [Q sec2 = 1 + tan 2 ]= 9 + 9 tan 2 5 tan 2 = 9 + 4 tan 2 = 5 + (4 + 4 tan 2 )= 5 + 4 (1 + 1 tan 2 )= 5 + 4 sec2 [Q sec2 = 1 + tan 2 ]

= R.H.S.

11110000.... ProveProveProveProve that:that:that:that: (sin(sin(sin(sin ++++ coseccoseccoseccosec ))))2222 ++++ (cos(cos(cos(cos ++++ secsecsecsec ))))2222 ==== 7777 ++++ tantantantan2222 ++++ cotcotcotcot2222 ....SolutionSolutionSolutionSolution ::::

L.H.S. = (s in + cosec )2 + ( cos + sec )2

= s in 2 + cosec 2 + 2 s in cosec + cos 2 + sec2 + 2 cos sec

= (s in 2 + cos 2 )+ 1 + cot 2 + 2 s in s in

1 + 1 + tan 2 + 2 cos cos

1

[s in ce sec2 = 1 + tan 2 ; cosec 2 = 1 + cot 2 )= 1 + 1 cot 2 + 2 + 1 tan 2 + 2 [s in ce s in 2 + cos 2 = 1 ]= tan 2 + cot 2 + 7= R.H.S.

11111111.... ProveProveProveProve that:that:that:that: IfIfIfIf coscoscoscos ++++ sinsinsinsin ==== 2 coscoscoscos ,,,, proveproveproveprove thatthatthatthat coscoscoscos sinsinsinsin ==== 2 sinsinsinsin ....


cos + s in = 2 cos ……(i)

(cos + s in )2 = 2 cos 2

cos 2 + s in 2 + 2 cos s in = 2 cos 2

2 cos s in = 2 cos 2 cos 2 - s in 2

cos 2 s in 2 = 2 cos s in …….(ii)

Now d ivid in g (ii) b y (i), we g et

s incoss incos 22

=

cos2s incos2

s incoss incoss incos = 2 s in

Hen ce cos s in = 2 sinsinsinsin ....

12121212.... InInInIn anananan acuteacuteacuteacute angledangledangledangled triangletriangletriangletriangle ABC,ABC,ABC,ABC, ifififif tantantantan (A(A(A(A ++++ BBBB C)C)C)C) ==== 1111 and,and,and,and, secsecsecsec (B(B(B(B ++++ CCCC A)A)A)A) ==== 2,2,2,2, findfindfindfind thethethethe

valuevaluevaluevalue ofofofof A,A,A,A, B,B,B,B, andandandand C.C.C.C.


tan (A + B C) = 1 an d sec (B + C A) = 2

tan (A + B C) = tan 4 5 an d sec (B + C A) = sec 6 0

A + B C = 4 5 an d B + C A = 6 0

(A + B C) + (B + C A) = 4 5 + 6 0

2 B = 1 0 5

B = 5 22

1

Pu tt in g B = 5 22

1 in B + C A = 6 0 , we g et

5 22

1 + C A = 6 0

C A = 72

1 ……….(i)

Also , in ABC, we h ave

A + B + C = 1 8 0

A + 5 22

1 + C = 1 8 0

215 2BQ

C + A = 1 2 72

1 ……….(ii)

Ad d in g an d su b tract in g (i) an d (ii), we g et

2 C = 1 3 5 an d 2 A = 1 2 0

C = 6 72

1 an d A = 6 0

Hen ce , A = 6 0 , B = 5 22

1 an d C = 6 72

1 .

13131313.... ProveProveProveProve that:that:that:that:1sectan1sectan

=

coss in1


L.H.S. =1sectan1sectan

L.H.S. = 1sectan

1sectan

L.H.S. = 1sectan

tansectansec 22

[Q sec 2 tan 2 = 1 ]

L.H.S. = 1sectan

tansectansectansec

L.H.S. = 1sectan

tansec1tansec

L.H.S. = 1sectan

tansec1tansec

L.H.S. = 1sectan

1sectantansec

L.H.S. = sec + tan

L.H.S. =cos

1 +

coss in =

coss in1 = R.H.S.

14141414.... IfIfIfIf

coscos

==== mmmm andandandand

s incos

==== nnnn showshowshowshow thatthatthatthat (m(m(m(m2222 ++++ nnnn2222)))) coscoscoscos2222 ==== nnnn2222


LHS = (m 2 + n 2) cos 2

LHS =

2

2

2

2

s incos

coscos cos 2

s incosnan d

coscosmQ

LHS =

22

2222

s incoscoscoss incos cos 2

LHS = cos 2

22

22

s incoscoss in cos 2

LHS = cos 2

22 s incos1 cos 2

LHS =

2

2

s incos

LHS =2

s incos

= n 2

LHS = RHS

15151515.... IfIfIfIf xxxx ==== rrrr sinsinsinsin AAAA coscoscoscos C,C,C,C, yyyy ==== rrrr sinsinsinsin AAAA sinsinsinsin CCCC andandandand zzzz ==== rrrr coscoscoscos A,A,A,A, proveproveproveprove thatthatthatthat rrrr2222 ==== xxxx2222 ++++ yyyy2222 ++++ zzzz2222 ....


X2 + y2 + z 2 = r2 s in 2 A cos 2 C + r2 s in 2 A sin 2 C + r 2 cos 2 A

x 2 + y2 + z 2 = r2 s in 2 A (cos 2 C + sin 2 C) + r2 cos 2 A

x 2 + y2 + z 2 = r2 s in 2 A + r2 cos 2 A [Q cos 2 C + sin 2 C = 1 ]

x 2 + y2 + z 2 = r2 (s in 2 A + cos 2 A)

x 2 + y2 + z 2 = r2

Hen ce , r2 = x 2 + y2 + z 2

16161616.... IfIfIfIf tantantantan AAAA ==== nnnn tantantantan BBBB andandandand sinsinsinsin AAAA ==== mmmm sinsinsinsin B,B,B,B, proveproveproveprove thatthatthatthat coscoscoscos2222 AAAA ====1n1m

2

2

SolutionSolutionSolutionSolution :::: We h ave to fin d cos 2 A in term s of m an d n . Th is m ean s th a t th e an g le B is to b e

e lim in a ted from th e g iven re la t io n s .

Now,

tan A = n tan B tan B =n1 tan A co t B =

Atann

an d , s in A = m sin B s in B =m1 s in A cosec B =

Asinm

Su b s titu t in g th e va lu es of co t B an d cosec B in cosec2 B co t 2 B = 1 , we g et

Asin

m2

2

Atann

2

2= 1

Asin

m2

2

AsinAcosn

2

22= 1

Asin

Acosnm2

222 = 1

m 2 n 2 cos 2 A = sin 2 A

m 2 n 2 cos 2 A = 1 cos 2 A [Q s in 2 A + cos 2 A = 1 ]

m 2 1 = n 2 cos 2 A cos 2 A

m 2 1 = (n 2 1 ) cos 2 A

1n1m

2

2

= cos 2 A

17171717.... ProveProveProveProve that:that:that:that: 2(sin2(sin2(sin2(sin6666 ++++ coscoscoscos6666 )))) 3(sin3(sin3(sin3(sin4444 ++++ coscoscoscos4444 )))) ++++ 1111 ==== 0000


LHS = 2 (s in 6 + cos 6 ) 3 (s in 4 + cos 4 ) + 1

LHS = 2

3232 coss in 3

2222 coss in + 1

LHS = 2

22222222 coss incoss incoss in

3

22222222 coss in2coss in2coss in + 1

LHS = 2

22222222 coss in3coss in2coss in

3

22222 coss in2coss in + 1

LHS = 2 [(s in 2 + cos 2)2 3 s in 2 cos 2 ] 3 [1 2 s in 2 cos 2 ] + 1

LHS = 2 (1 3 s in 2 cos 2 ) 3 (1 2 s in 2 cos 2) + 1

LHS = 2 6 s in 2 cos 2 3 + 6 s in 2 cos 2 + 1

LHS = 0 = RHS.

11118888.... ProveProveProveProve that:that:that:that: sinsinsinsin6666 ++++ coscoscoscos6666 ++++ 3sin3sin3sin3sin2222 coscoscoscos2222 ==== 1.1.1.1.


LHS = s in 6 + cos 6 + 3 s in 2 cos 2

LHS = [(s in 2 )3 + (cos 2 )3] + 3 s in 2 cos 2

LHS = (s in 2 + cos 2) {(s in 2)2 + (cos 2)2 s in 2 cos 2 } + 3 s in 2 cos 2

LHS = (s in 2 + cos 2) {(s in 2)2 + (cos 2)2 + 2 s in 2 cos 2 2 s in 2 cos 2 s in 2 cos 2}

+ 3 s in 2

cos 2

LHS = (s in 2 + cos 2 ) {(s in 2 + cos 2)2 3 s in 2 cos 2} + 3 s in 2 cos 2

LHS = (1 3 s in 2 cos 2 ) + 3 s in 2 cos 2

LHS = 1 = RHS.

11119999.... ProveProveProveProve that:that:that:that: secsecsecsec2222 (1(1(1(1 sinsinsinsin4444)))) 2222 tantantantan2222 ==== 1.1.1.1.


LHS = sec 4 (1 s in 4) 2 tan 2

= sec 4 [1 (s in 2)2] 2 tan 2

= sec 4 (1 s in 2 )(1 + s in 2) 2 tan 2

= sec 4 cos 2 (1 + s in 2 ) 2 tan 2

= sec 2 (sec2. cos 2 )(1 + s in 2 )

2

2

coss in2

=

2

2

coss in1

2

2

coss in2

1cos.s ecan d

cos1s ec 22

22Q

=

2

22

coss in2s in1 =

2

2

coss in1

=

2

2

coscos = 1

LHS = RHS.

20202020.... IfIfIfIf secsecsecsec ==== xxxx ++++x4

1,,,, proveproveproveprove that:that:that:that: secsecsecsec ++++ tantantantan ==== 2x2x2x2x or,or,or,or,

x21....


sec = x +x4

1

tan 2 = sec2 1

tan 2 =2

x41x

1

tan 2 = x 2 +2x1 6

1++++

21

1

tan 2 = x 2 +2x1 6

1

21

tan 2 =2

x41x

tan =

x4

1x

tan =

x4

1x or, tan =

x4

1x

Wh en tan =

x4

1x , we h ave

sec + tan = x +x4

1++++ x

x41 = 2 x

Wh en tan =

x4

1x , we h ave

sec + tan =

x4

1x

x4

1x =x4

2 =x2

1

Hen ce , sec + tan = 2 x or,x2

1

2 1 . Show that : =

Proof: LHS =

= RHS

2 2 . Prove the fo llowin g id en t ity : = sec . cos ec + cot

Proo f:

= sec . cos ec + cot = RHS.2 3 . Prove the fo llowin g id en t ity : cos 4 A – cos 2 A = sin 4 A – sin 2 A

Proof:LHS = cos 4 A – cos 2 A = cos 2 A (cos 2 A – 1 ) = (1 – s in 2 A) (– s in 2 A)

= s in 4 A – sin 2 A = RHS.

2 4 . Prove the fo llowin g id en t ity : If x = a sec , y = b tan , p rove tha t 1yb

xa

2

2

2

2

(Question(Question(Question(Question modified)modified)modified)modified)

Proof:x = a sec

axsec ....(i)

y = b tan

bytan ....(ii)

We know,1tansec 22

1by

ax

2

2

2

2

22225.5.5.5. IfIfIfIf aaaa coscoscoscos θθθθ –––– bbbb sinsinsinsin θθθθ ==== xxxx andandandand aaaa sinsinsinsin θθθθ ++++ bbbb coscoscoscos θθθθ ==== yyyy.... ProveProveProveProve thatthatthatthat aaaa2222 ++++ bbbb2222 ==== xxxx2222 ++++ yyyy2222....

Proof:a cos – b s in = x

and a s in + b cos = y

RHS = x2 + y2

= (a cos – b s in )2 + (a s in + b cos )2

= a2 cos 2 + b2 s in 2 – 2ab cos s in + a2 s in 2 + b2 cos 2 + 2ab cos s in = a2 (cos 2 + s in 2 ) + b2 (cos 2 + s in 2 )= a2 (1 ) + b2(1 )

= a2 + b2 = LHS

2 6 .Prove tha t : = 2 sec A

Proof:

2 7 . Prove

Proof:

2 8 . Prove tha t : =

Proof:

LHS =

ins

1cotcosec

1

=

coseccotcoseccotcosec

cotcosec

=

cosec

cotcoseccotcosec

22

=

cosec

1cotcosec

= coseccotcosec= cot

RHS =

cotcosec

1ins1

=

cotcoseccotcosec

cotcoseccosec

=

22 cotcosec

cotcoseccosec

=

1cotcoseccosec

= cotcoseccosec= cot

22229.9.9.9. ProveProveProveProve:::: ==== 2222 coseccoseccoseccosec AAAA .... cotcotcotcot AAAA

Proof:

= 2 cot A . cos ec A = RHS33330.0.0.0. IfIfIfIf tantantantan θθθθ ++++ sinsinsinsin θθθθ ==== mmmm andandandand tantantantan θθθθ –––– sinsinsinsin θθθθ ==== nnnn,,,, showshowshowshow thatthatthatthat mmmm2222 –––– nnnn2222 ==== ....

Solu t ion : LHS = m2 – n2

= (tan + s in )2 – (tan – s in )2 (Given)= tan 2 + s in 2 + 2 tan s in – tan 2 – s in 2 + 2 tan s in = 4 tan s in

RHS = mn4

= sintansintan4

= 22 sintan4

= 2

2

2

sincossin4

=

1cos1sin4 2

2

= 1secsin4 22

= 22 tan.sin4= 4 tan s in LHS = RHS

3 2 . Prove tha t = 2 sec

Proof:

LHS =

= 2 sec = RHS

33333333.... IfIfIfIf secsecsecsec ++++ tantantantan ==== pppp,,,, findfindfindfind thethethethe valuevaluevaluevalue ofofofof coseccoseccoseccosec ....

Solu t ion :sec + tan = p ...(i)Als o sec2 – tan 2 = 1(s ec + tan ) (sec - tan ) = 1p(s ec - tan ) = 1 [us ing eq uat ion (i)]

sec - tan =p1

...(ii)

(ii) + (i) we ge tsec + tan = p

+ sec - tan =p1

2 sec = p +p1

2 sec =p1p2

sec =p21p2

(i) - (ii) we ge tsec + tan = p

- sec - tan =p1

2 tan = p -p1

2 tan =p1p2

tan =p21p2

cossin

=p21p2

sec.sin =p21p2

p21p.sin

2 =

p21p2

sin =1p1p

2

2

cosec =1p1p

2

2

34343434.... AAAA verticalverticalverticalvertical towertowertowertower standsstandsstandsstands onononon aaaa horizontalhorizontalhorizontalhorizontal planeplaneplaneplane andandandand isisisis surmountedsurmountedsurmountedsurmounted bybybyby aaaa verticalverticalverticalvertical flagflagflagflag ––––

staffstaffstaffstaff ofofofof heightheightheightheight h.h.h.h. AtAtAtAt aaaa pointpointpointpoint onononon thethethethe plane,plane,plane,plane, thethethethe anglesanglesanglesangles ofofofof elevationelevationelevationelevation ofofofof thethethethe bottombottombottombottom andandandand thethethethe

toptoptoptop ofofofof thethethethe flagflagflagflag –––– staffstaffstaffstaff areareareare andandandand respectively.respectively.respectively.respectively. ProveProveProveProve thatthatthatthat thethethethe heightheightheightheight ofofofof thethethethe towertowertowertower isisisis

tantantantantantantantan

tantantantanhhhh....


Let AB b e th e tower an d BC b e th e flag – s ta ff. Le t O b e a p oin t on th e p lan e con ta in in g th e

foo t of th e tower su ch th a t th e an g les of e leva tion of th e b o ttom B an d top C of th e flag –

s ta ff a t O are an d res p ective ly. Let OA = x m etres , AB = y m etres an d BC = h m etres .

In OAB, we h ave

tan =OAAB

tan =xy

x =tan

y ……..(i)

x = y co t

In OAC, we h ave

tan =x

hy

x =

tan

hy

x = (y + h ) co t

Oh eq u atin g th e valu es of x g iven in eq u atio n s (i) an d (ii), we g et

y co t = (y + h ) co t

hhhh

xxxx

OOOO

CCCC

BBBB

AAAA

yyyy

(y co t y cot ) = h co t

y (co t co t ) = h co t

y =

co tco t

co th

y =

tan1

tan1

tanh

=

tantan

tanh

Hen ce , th e h e ig h t of th e tower is

tantan

tanh .

33335555.... TheTheTheThe anglesanglesanglesangles ofofofof elevationelevationelevationelevation ofofofof thethethethe toptoptoptop ofofofof aaaa towertowertowertower fromfromfromfrom twotwotwotwo pointspointspointspoints atatatat distancesdistancesdistancesdistances aaaa andandandand bbbb

metresmetresmetresmetres fromfromfromfrom thethethethe basebasebasebase andandandand inininin thethethethe samesamesamesame straightstraightstraightstraight linelinelineline withwithwithwith itititit areareareare complementary.complementary.complementary.complementary. ProveProveProveProve

thatthatthatthat thethethethe heightheightheightheight ofofofof thethethethe towertowertowertower isisisis ab meters.meters.meters.meters.


Let AB b e th e tower. Let C an d D b e two p oin ts a t d is tan ces a an d b res p ective ly from th e

b ase of th e tower. Th en , AC = a an d AD = b . Let ACB = an d ADB = 9 0 .

Le t h b e th e h e ig h t of th e tower AB.

In CAB,

tan =ACAB

tan =ah ……..(i)

In DAB,

tan (9 0 ) =ADAB

co t =bh …….(ii)

From (i) an d (ii), we h ave

tan co t =abh 2

1 =abh 2

h 2 = ab h = ab m eters .

Hen ce , th e h e ig h t of th e tower is ab m ate rs .

36363636.... TwoTwoTwoTwo stationsstationsstationsstations duedueduedue southsouthsouthsouth ofofofof aaaa leaningleaningleaningleaning towertowertowertower whichwhichwhichwhich leansleansleansleans towardstowardstowardstowards thethethethe northnorthnorthnorth areareareare atatatat

distancesdistancesdistancesdistances aaaa andandandand bbbb fromfromfromfrom itsitsitsits foot.foot.foot.foot. IfIfIfIf ,,,, bebebebe thethethethe elevationselevationselevationselevations ofofofof thethethethe toptoptoptop ofofofof thethethethe towertowertowertower fromfromfromfrom

thesethesethesethese stations,stations,stations,stations, proveproveproveprove thatthatthatthat itsitsitsits inclinationinclinationinclinationinclination totototo thethethethe horizontalhorizontalhorizontalhorizontal isisisis givengivengivengiven bybybyby cotcotcotcot ====

abco taco tb


Let AB b e th e lean in g tower an d le t C an d D b e two s ta t ion s a t d is tan ces a an d b resp ective ly

from th e foo t A of th e tower.

Let AE = x an d BE = h

In AEB, we h ave

hhhh

DDDDCCCCAAAA

(90(90(90(90---- ))))

BBBB

aaaa

bbbb

B

E

xa

h

D C A

b

tan =AEBE

tan =xh

x = h cot ……(i)

In CEB, we h ave

tan =CEBE

tan =xa

h

`

a + x = h cot

x = h cot a ……..(ii)

In DEB, we h ave

tan =DEBE

tan =xb

h

b + x = h cot

x = h cot b ……(iii)

On eq u atin g th e valu es of x ob ta in ed from eq u ation s (i), an d (ii), we h ave

h co t = h cot a

h (co t co t ) = a

h = co tco t

a …….(iv)

On eq u atin g th e valu es of x ob ta in ed from eq u ation s (i) an d (ii), we g et

h co t = h cot b

h (co t co t ) = b

h = co tco t

b

Eq u atin g th e valu es of h from eq u ation s (iv) an d (v), we g et

co tco ta =

co tco tb

a(co t co t ) = b (co t co t )

(b a) co t = b cot a co t

co t =ab

cotaco tb

37373737.... TheTheTheThe angleangleangleangle ofofofof elevationelevationelevationelevation ofofofof aaaa jetjetjetjet planeplaneplaneplane fromfromfromfrom aaaa pointpointpointpoint AAAA onononon thethethethe groundgroundgroundground isisisis 60606060.... AfterAfterAfterAfter aaaa flightflightflightflight

ofofofof 15151515 seconds,seconds,seconds,seconds, thethethethe angleangleangleangle ofofofof elevationelevationelevationelevation changeschangeschangeschanges totototo 30303030.... IfIfIfIf thethethethe jetjetjetjet planeplaneplaneplane isisisis flyingflyingflyingflying atatatat aaaa

constantconstantconstantconstant heightheightheightheight ofofofof 1500150015001500 3 m . fin d th e sp eed of th e je t p lan e .


Let P an d Q b e th e two p os it ion s of th e p lan e an d le t A b e th e p oin t o f ob se rva tio n . Let ABC

b e th e h oriz on ta l lin e th rou g h A. It is g iven th a t an g les of e leva tion of th e p lan e in two

p os it ion s P an d Q from a p oin t A are 6 0 an d 3 0 res p ective ly.

PAB = 6 0 , QAB = 3 0 .

It is a lso g iven th a t PB = 1 5 0 0 3 m eters

In ABP,

tan 6 0 =ABBP

3 =AB

31 5 0 0

AB = 1 5 0 0 m

In ACQ,

tan 3 0 =ACCQ

3

1 =AC

31 5 0 0

AC = 1 5 0 0 3 = 4 5 0 0 m

PQ = BC = AC AB = 4 5 0 0 1 5 0 0 = 3 0 0 0 m

Th u s , th e p lan e trave ls 3 0 0 0 m in 1 5 secon d s .

Hen ce , Sp eed of p lan e =1 5

3 0 0 0 = 2 0 0 m / sec =1 0 0 02 0 0

6 0 6 0 = 7 2 0 km / h r.

38383838.... AAAA roundroundroundround balloonballoonballoonballoon ofofofof radiusradiusradiusradius rrrr subtendssubtendssubtendssubtends anananan angleangleangleangle atatatat thethethethe eyeeyeeyeeye ofofofof thethethethe observerobserverobserverobserver whilewhilewhilewhile thethethethe

angleangleangleangle ofofofof elevationelevationelevationelevation ofofofof itititit’’’’ssss centrecentrecentrecentre isisisis .... ProveProveProveProve thatthatthatthat thethethethe heightheightheightheight ofofofof thethethethe centrecentrecentrecentre ofofofof thethethethe balloonballoonballoonballoon isisisis

rrrr sinsinsinsin coseccoseccoseccosec /2./2./2./2.


Let O b e th e cen tre of th e b a lloon of rad iu s r an d P th e eye of th e ob s erve r. Le t PA, PB b e

tan g en ts from P to th e b alloon . Th en , APB = .

APO = BPO =2

Let OL b e p erp en d icu lar from O on th e h oriz on ta l PX.

We are g iven th a t th e an g le of th e e leva tion of th e

cen tre of th e b a lloon is i.e ., OPL = .

In OAP,

s in2 =

OPOA

s in2 =

OPr

OP = r cosec2 …….(i)

P

CBA

Q

3 0

1 5 0 0 3 m

6 0

A

B

XL

2

O

r

P

2

In OPL,

s in =OPOL

OL = OP s in = r cosec2 s in [Us in g eq u ation (i)]

Hen ce , th e h e ig h t of th e cen tre of th e b a lloon is r s in cosec2 .

39393939.... TheTheTheThe angleangleangleangle ofofofof elevationelevationelevationelevation ofofofof aaaa cliffcliffcliffcliff fromfromfromfrom aaaa fixedfixedfixedfixed pointpointpointpoint isisisis .... AfterAfterAfterAfter goinggoinggoinggoing upupupup aaaa distancedistancedistancedistance ofofofof kkkk

metersmetersmetersmeters towardstowardstowardstowards thethethethe toptoptoptop thethethethe cliffcliffcliffcliff atatatat anananan angleangleangleangle ofofofof ,,,, itititit isisisis foundfoundfoundfound thatthatthatthat thethethethe angleangleangleangle ofofofof elevationelevationelevationelevation

isisisis .... ShowShowShowShow thatthatthatthat thethethethe heightheightheightheight ofofofof thethethethe cliffcliffcliffcliff isisisis

cotcotcotcotcotcotcotcotcotcotcotcotsinsinsinsincoscoscoscoskkkk

metres.metres.metres.metres.


Let AB b e th e cliff an d O an d b e th e fixed p oin t su ch th a t th e an g le of e leva tio n of th e cliff

from O is i.e ., AOB = . Le t AOC = an d OC = k m etres . From C d raw CD an d CE

p erp en d icu lars on AB an d OA resp ective ly. Th en , DCB = . Le t h b e th e h e ig h t of th e cliff

AB.

In OCE,

s in =OCCE

s in =k

CE

CE = k s in

AD = k s in ……(i) [Q CE = AD]

An d , cos =OCOE

cos =k

OE

OE = k cos ……(ii)

In OAB,

tan =OAAB

tan =OAh

OA = h cot

CD = EA = OA OE = h cot k cos ……(vi) [Us in g (ii) an d (iii)]

an d BD = AB AD = AB CE = h k s in ……(v) [Us in g (i)]

kC D

E

B

O A

h

In BCD,

tan =CDBD

tan =

coskco th

s inkh [Us in g (vi) an d (v)]

co t

1 =

coskco th

s inkh

h cot k s in co t = h cot k cos

h (co t co t ) = k (cos s in co t )

h =

co tco t

co ts incosk

Roll No. :

Date :

Time -MM - 31

3

Ans :

42.

Ans :

43.

The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving adistance of 20 metres towards the foot of the tower to a point B, the angle of elevation increases to

60°. Find the height of the tower and distance of the tower from the point A. ( = 1.732)

Let ‘h’ m be height of the tower PQ.

AB = 20 m. Let BQ = x mIn rt. ∆PQB,

...(i)

In rt. ∆PQA,

...(ii)

From (i) and (ii)

= 20 + x x = 10

Distance of tower from A is 20 m + 10 m = 30 m

Put x = 10 in (i), we get

h = × 10 = 1.732 × 10 = 17.32 m

Height of tower is 17.32 m.

The angle of elevation of a jet fighter from a point A on the ground is 60°. After a flight of 15seconds, the angle of elevation changes to 30°. If the jet is flying at a speed of 720 km/hr, find the

constant height. ( = 1.732).

Speed of jet fighter = 720 km/h = 200 m/s Distance covered in 15 seconds = 200 × 15 = 3000 m PB = 3000 m

PB = QC = 3000 m

In right ∆PQA, = tan 60°

x = y ...(i)

In right ∆BCA, = tan 30°

x = ...(ii)

From (i) and (ii),

3y = y + 3000 2y = 3000 y = 1500 m

x = 1500 × m

x = 1500 = 1500 × 1.732 m = 2598 m

From the top of a building 60 m high, the angles of depression of the top and bottom of avertical lamp post are observed to be 30° and 60° respectively. Find(i) The horizontal distance between the building and the lamp post.

CREATIVE

Rectangle

CREATIVE

Rectangle

CREATIVE

Rectangle

CREATIVE

Rectangle

CREATIVE

Typewriter

40

CREATIVE

Rectangle

CREATIVE

Typewriter

41

CREATIVE

Rectangle

CREATIVE

Typewriter

42

Ans :

44.

Ans :

45.

(ii) The height of the lamp post, = 1.732

Let AB = 60 m is height of building and CD is lamppost.

(i) In rt. ∆ABD, = tan 60°

= BD

BD =

BD = 20 × 1.732 = 34.64 m

(ii) In rt. ∆ AEC, = tan 30°

[ EC = BD]

AE = 20 mand EB = AB – AE = 60 – 20 = 40 mAlso EB = CD

CD = 40 m Height of lamp post = 40 m.

The angle of elevation of an aeroplane from a point on the ground is 60°. After a flight of 30seconds the angle of elevation becomes 30°. If the aeroplane is flying at a constant height of 3000

m, find the speed of the aeroplane.

From the point of observation (O), plane is at A, AL = 3000 m and AOL = 60°.

After 30 seconds, plane is at B, therefore,

BM = 3000 m and BOM = 30°.

Distance AB is covered in 30 seconds.In right-angled triangle OLA,

= cot 60°

OL = = 3000 m ...(i)

In right-angled triangle OMB,

= cot 30°

OM = 3000 × = 9000 m ...(ii)

AB = LM = OM – OL = (9000 – 3000) m = 6000 m [from (i) and (ii)]Now in 30 s, distance covered = 6000 m

In 1 hour (3600 s), distance covered = × km = 720 km

Speed of the aeroplane = 720 km/h.

The angles of elevation and depression of the top and bottom of a light-house from the top of a60 m high building are 30° and 60° respectively. Find(i) the difference between the heights of the light-house and the building.(ii) the distance between the light-house and the building.

CREATIVE

Rectangle

CREATIVE

Rectangle

CREATIVE

Typewriter

3

CREATIVE

Rectangle

CREATIVE

Typewriter

44

Ans :

46.

Let AB is the building.

AB = 60 m and CD is the light house.EAC = 30° and EAD = 60°

ADB = 60° AE || BD

In right ABD,

BD = m = m.

BD = AE AE = 20 m

Now In right CEA, tan 30° =

CE = 20 m

(i) Difference between the heights of the light house and the building = CE = 20 m.(ii) The distance between the light house and the building = BD = 20 m

From the top of a building 15 m high, the angle of elevation of the top of a tower is found to be30°. From the bottom of the same building, the angle of elevation of the top of the tower is foundto be 45°. Determine the height of the tower and the distance between the tower and the building.

CREATIVE

Rectangle

CREATIVE

Rectangle

CREATIVE

Typewriter

45

Ans :

47.

Given: A building AB 15 m high and tower CD

Angle of elevation DAE = 30°Angle of elevation DBC = 45°To find: BC and CDSolution: In right ∆DEA,

= tan 30° [ AE = BC = x m]

h – 15 = ...(i)

In right ∆DCB, = tan 45°

h = x m ... (ii)

Putting the value of h from equation (ii) in equation (i)

x – 15 = [From (i)]

x = 35.49 m

Putting in (ii) h = 35.49 m

At a point A, 20 metres above the level of water in a lake, the angle of elevation of a cloud is30°. The angle of depression of the reflection of the cloud in the lake, at A is 60°. Find thedistance of the cloud from A.

CREATIVE

Rectangle

CREATIVE

Rectangle

CREATIVE

Typewriter

46

Ans :

48.

Ans :

Let C is cloud and R is its reflection.

DAC = 30°, DAR = 60°, let CD = x m Height of the cloud above the lake

= (x + 20) m

ER = (20 + x) m.Now In right ADC,

40 + x = 3x

x = 20 m

Now In right ADC, = cosec 30°

= 2 AC = 40 m

Distance of the cloud from A = 40 m

Two poles of equal heights are standing opposite to each other on either side of the road whichis 80 m wide. From a point P between them on the road, the angle of elevation of the top of a poleis 60° and the angle of depression from the top of another pole at point P is 30°. Find the heightsof the poles and the distance of the point P from the poles.

Let AB and CD are two poles

Let BP = x m

PD = (80 – x) m

In right PBA, = tan 60°

=

AB = ...(i)

In right CDP, = tan 30°

CD = ...(ii)

AB = CD

=

3x = 80 – x

4x = 80 x = 20Now AB = AB = 20 m

Height of poles = 20 m and distance of point P from the pole with angle ofelevation 60° is 20 m

CREATIVE

Rectangle

CREATIVE

Typewriter

47


MensurationMensurationMensurationMensuration1.1.1.1. AAAA pendulumpendulumpendulumpendulum swingsswingsswingsswings throughthroughthroughthrough anananan angleangleangleangle ofofofof 30303030 andandandand describesdescribesdescribesdescribes anananan arcarcarcarc 8.88.88.88.8 cmcmcmcm inininin length.length.length.length. FindFindFindFind thethethethe

lengthlengthlengthlength ofofofof thethethethe pendulum.pendulum.pendulum.pendulum. [Use[Use[Use[Use ==== 22/7].22/7].22/7].22/7].

SolutionSolutionSolutionSolution :::: Her, = 30 , l = arc = 8 .8 cm

l =360

2r 8 .8 =36030

2 722

r

r =22

768.8 cm = 16 .8 cm .

2.2.2.2. TheTheTheThe minuteminuteminuteminute handhandhandhand ofofofof aaaa clockclockclockclock isisisis 10101010 cmcmcmcm long.long.long.long. FindFindFindFind thethethethe areaareaareaarea ofofofof thethethethe facefacefaceface ofofofof thethethethe clockclockclockclock describeddescribeddescribeddescribed bybybyby

thethethethe minuteminuteminuteminute handhandhandhand betweenbetweenbetweenbetween 9999 A.M.A.M.A.M.A.M. andandandand 9.359.359.359.35 A.M.A.M.A.M.A.M.

SolutionSolutionSolutionSolution :::: We have ,Ang le described by the m inu te hand in one m inu te = 6

Ang le described by the m inu te hand in 35 m inu tes = (6 35 ) = 210 Area swep t by the m inu te hand in 35 m inu tes

= Area of a secto r of ang le 210 in a circle of rad iu s 10 cm

=

210

722

360210 cm 2 = 183 .3 cm 2

2r360

A:gs inU

3.3.3.3. AAAA carcarcarcar hashashashas wiperswiperswiperswipers whichwhichwhichwhich dodododo notnotnotnot overlap.overlap.overlap.overlap. EachEachEachEach wiperwiperwiperwiper hashashashas aaaa bladebladebladeblade ofofofof lengthlengthlengthlength 25252525 cmcmcmcm sweepingsweepingsweepingsweepingthroughthroughthroughthrough anananan angleangleangleangle ofofofof 115115115115.... FindFindFindFind thethethethe totaltotaltotaltotal areaareaareaarea cleanedcleanedcleanedcleaned atatatat eacheacheacheach sweepsweepsweepsweep ofofofof thethethethe blades.blades.blades.blades.

SolutionSolutionSolutionSolution :::: Clea rly, each wip er sweep s a secto r of a circle of rad iu s 25 cm and sector ang le 115 . The re fo re ,a rea A cleaned at each sweep is g iven by

A =360

r2

A =360115

722

25 25cm 2 = 617 .48 cm 2.

4.4.4.4. ToToToTo warmwarmwarmwarm shipsshipsshipsships forforforfor underwaterunderwaterunderwaterunderwater rocks,rocks,rocks,rocks, aaaa lightlightlightlight househousehousehouse throwsthrowsthrowsthrows aaaa redredredred colouredcolouredcolouredcoloured lightlightlightlight overoveroverover aaaa sectorsectorsectorsectorofofofof 80808080 angleangleangleangle totototo aaaa distancedistancedistancedistance ofofofof 16.516.516.516.5 km.km.km.km. FindFindFindFind thethethethe areaareaareaarea ofofofof thethethethe seaseaseasea overoveroverover whichwhichwhichwhich thethethethe shipsshipsshipsships areaareaareaareawarmed.warmed.warmed.warmed. [Use[Use[Use[Use ==== 3.14]3.14]3.14]3.14]

SolutionSolutionSolutionSolution :::: We have ,r = 16 .5 km and = 80

Area of the sea over wh ich the sh ip s are warm ed

=360

r2 =36080

3 .14 16 .5 16 .5 km 2 = 189 .97 km 2 .

5.5.5.5. DetermineDetermineDetermineDetermine thethethethe ratioratioratioratio ofofofof thethethethe volumevolumevolumevolume ofofofof aaaa cubecubecubecube thatthatthatthat ofofofof spherespherespheresphere whichwhichwhichwhich willwillwillwill exactlyexactlyexactlyexactly fitfitfitfit insideinsideinsideinside thethethethecube.cube.cube.cube.

SolutionSolutionSolutionSolution :::: Let the rad iu s of the sphere wh ich fits exactly in to a cube be r un its .Then , Leng th of each edge of the cube = 2 r un itsLet V1 and V2 be the vo lumes of the cube and sphe re respective ly. Then ,


V1 = (2 r)3 and V2 =34r2

2

1

VV

=3

2

r34r8

=

6

V1 : V2 = 6 : .

6.6.6.6. FindFindFindFind thethethethe maximummaximummaximummaximum volumevolumevolumevolume ofofofof aaaa coneconeconecone thatthatthatthat cancancancan bebebebe carvedcarvedcarvedcarved outoutoutout ofofofof aaaa solidsolidsolidsolid hemispherehemispherehemispherehemisphere ofofofof radiusradiusradiusradius r.r.r.r.

SolutionSolutionSolutionSolution :::: Clearly,Rad iu s of the base of cone = Rad iu s of the hem isphe re = rHeigh t of the cone = Rad iu s of the hem isphe re

Volum e of the cone =31r2 r =

31r3

7777.... TheTheTheThe circumferencecircumferencecircumferencecircumference ofofofof aaaa circlecirclecirclecircle exceedsexceedsexceedsexceeds thethethethe diameterdiameterdiameterdiameter bybybyby 16.816.816.816.8 cm.cm.cm.cm. FindFindFindFind thethethethe radiusradiusradiusradius ofofofof thethethethe

circle.circle.circle.circle.

SolutionSolutionSolutionSolution :::: Let the rad iu s of the circle be r cm . Then ,Diam ete r = 2 r cm and Circum ference = 2r cmIt is g iven tha t the circum ference exceed s the d iam ete r by 16 .8 cm

Circum ference = Diam ete r + 16 .8 2r = 2 r + 16 .8

2 722

r = 2 r + 16 .8 [Q =722 ]

44 r = 14 r + 16 .8 7

44 r 14 r = 117 .6 30 r = 117 .6 r =30

6.117 = 3 .92

Hence , rad iu s = 3 .92 cm .

8888.... AAAA wirewirewirewire isisisis loopedloopedloopedlooped inininin thethethethe formformformform ofofofof aaaa circlecirclecirclecircle ofofofof radiusradiusradiusradius 28282828 cm.cm.cm.cm. ItItItIt isisisis re-bentre-bentre-bentre-bent intointointointo aaaa squaresquaresquaresquare form.form.form.form.DetermineDetermineDetermineDetermine thethethethe lengthlengthlengthlength ofofofof thethethethe sidesidesideside ofofofof thethethethe square.square.square.square.

SolutionSolutionSolutionSolution :::: We have ,Leng th of the wire = Circum ference of the circle

Leng th of the wire =

28

7222 cm [Us ing C = 2r]

Leng th of the wire = 176 cm …….(i)Let the s ides of the square be x cm . Then ,

Perim eter of the square = Leng th of the wire 4x = 176 [Us ing (i)] x = 44 cm

Hence , the leng th of the s id e of the square is 44 cm .


9999.... AAAA raceraceracerace tracktracktracktrack isisisis inininin thethethethe formformformform ofofofof aaaa ringringringring whosewhosewhosewhose innerinnerinnerinner circumferencecircumferencecircumferencecircumference isisisis 352352352352 m,m,m,m, andandandand thethethethe outerouterouteroutercircumferencecircumferencecircumferencecircumference isisisis 396396396396 mmmm .... FindFindFindFind thethethethe widthwidthwidthwidth ofofofof thethethethe track.track.track.track.

SolutionSolutionSolutionSolution :::: Let the ou ter and inner rad ii o f the ring be R metre s and r me tres respective ly. Then ,2R = 396 and 2r = 352

2 722

R = 396 and 2 722

r = 352

R = 396 227

21 and r = 352

227

21

R = 63 m and r = 56mHence , wid th of the track = (R r) me tres

= (63 56 ) me tres = 7 metres .

10101010.... AAAA wheelwheelwheelwheel hashashashas diameterdiameterdiameterdiameter 84848484 cm.cm.cm.cm. FindFindFindFind howhowhowhow manymanymanymany completecompletecompletecomplete revolutionsrevolutionsrevolutionsrevolutions mustmustmustmust itititit taketaketaketake totototo covercovercovercover 792792792792meters.meters.meters.meters.

Solution:Solution:Solution:Solution: Let r be rad iu s of the wheel. Then ,Diam ete r = 84 cm [Given ]

2 r = 84 r = 42 cm

Circum ference of the wheel = 2r cm = 2 722

42 cm = 264 cm = 2 .64 m

So, the wheel covers 2 .64 me ters in one comp le te revo lu t ion . Tota l number of revo lu t ions in covering 792 mete rs .

Hence , the wheel takes 300 revo lu tions in covering 792 mete rs .

11.... FindFindFindFind thethethethe areasareasareasareas ofofofof thethethethe shadedshadedshadedshaded regionregionregionregion inininin thethethethe fig.fig.fig.fig.

SolutionSolutionSolutionSolution :::: We have ,Rad iu s of the b igger sem i- circle = 14 cm

Area of the b igger sem i- circle

=21r2 =

21

722

(14 )2 cm 2 = 308 cm 2

Rad iu s of each of the sm alle r circles = 7 cm

Area of 2 sm alle r sem i- circles = 2

27

722

21 cm 2 = 154 cm 2

Hence , Requ ired area = (308 + 154 ) cm 2 = 462 cm 2.

12121212.... TheTheTheThe radiiradiiradiiradii ofofofof thethethethe basesbasesbasesbases ofofofof twotwotwotwo rightrightrightright circularcircularcircularcircular solidsolidsolidsolid conesconesconescones ofofofof samesamesamesame heightheightheightheight areareareare rrrr1111 andandandand rrrr2222respectively.respectively.respectively.respectively. TheTheTheThe conesconesconescones areareareare meltedmeltedmeltedmelted andandandand recastrecastrecastrecast intointointointo aaaa solidsolidsolidsolid spherespherespheresphere ofofofof radiusradiusradiusradius R.R.R.R. ShowShowShowShow thatthatthatthat thethethethe

heightsheightsheightsheights ofofofof eacheacheacheach coneconeconecone isisisis givengivengivengiven bebebebe hhhh ==== 22

21

3

rrR4

....

SolutionSolutionSolutionSolution :::: Let h be the heigh t of each cone . Then ,Sum of the vo lumes of two cones = Volum e of the sphere

31r1 2 h+

31r22h =

34R3

(r 12 + r22) h = 4R3

h = 22

21

3

rrR4

....


13131313.... MayankMayankMayankMayank mademademademade aaaa bird-bathbird-bathbird-bathbird-bath forforforfor hishishishis gardengardengardengarden inininin thethethethe shapeshapeshapeshape ofofofof aaaa cylindercylindercylindercylinder withwithwithwith aaaa hemisphericalhemisphericalhemisphericalhemispherical

dipressiondipressiondipressiondipression atatatat oneoneoneone endendendend asasasas shownshownshownshown inininin fig.fig.fig.fig. TheTheTheThe heightheightheightheight ofofofof thethethethe cylindercylindercylindercylinder isisisis 1.451.451.451.45 mmmm andandandand itsitsitsits radiusradiusradiusradius isisisis

30303030 cm.cm.cm.cm. FindFindFindFind thethethethe totaltotaltotaltotal surfacesurfacesurfacesurface areaareaareaarea ofofofof thethethethe bird-bath.bird-bath.bird-bath.bird-bath. (Take(Take(Take(Take ==== 22/7).22/7).22/7).22/7).

SolutionSolutionSolutionSolution :::: Let r be the common rad iu s of the cylinderand hem isphere and h be the heigh t of the cylinder. Then ,r = 30 cm and h = 1 .45 m = 145 cm .Tota l su rface area of the b ird - ba th= Curved su rface area of the cylinder +

cu rved su rface area of the hem isphe re= 2rh + 2r2 = 2r (h + r)

= 2 722

30 (145 + 30 ) cm 2 = 33000 cm 2 = 3 .3 m 2.

11114444.... TwoTwoTwoTwo circlecirclecirclecircle touchtouchtouchtouch externally.externally.externally.externally. TheTheTheThe sumsumsumsum ofofofof theirtheirtheirtheir areasareasareasareas isisisis 130130130130 sq.sq.sq.sq. cm.cm.cm.cm. andandandand thethethethe distancedistancedistancedistance betweenbetweenbetweenbetween

theirtheirtheirtheir centrescentrescentrescentres isisisis 14141414 cm.cm.cm.cm. FindFindFindFind thethethethe radiiradiiradiiradii ofofofof thethethethe circles.circles.circles.circles.

Solution:Solution:Solution:Solution: If two circles touch ex te rna lly, then the d is tance between the ir cen tres is equal to the sum of

the ir rad ii.

Le t the rad ii o f the two circles be r1 cm and r2 cm respective ly.

Let C1 and C2 be the cen tres of the g iven circles . Then ,

C1C2 = r 1 + r2

14 = r1 + r2 [Q C1C2 = 14cm (g iven )]

r1 + r2 = 14 ……(i)

It is g iven tha t the sum of the areas of two circles is equal to 130 cm 2 .

r1 2 + r22 = 130

r1 2 + r2 2 = 130 ……(ii)

Now, (r1 + r2)2 = r1 2 + r2 2 + 2r 1r2

14 2 = 130 + 2r 1r2 [Us ing (i) and (ii)]

196 130 = 2r 1r2 …..(iii)

r1r2 = 33

Now,

(r1 r2)2 = r12 + r22 2 r1r2

(r1 r2)2 = 130 2 33 [Us ing (ii) and (iii)]

(r1 r2)2 = 64

r1 r2 = 8

Solving (i) and (iv), we get r 1 = 11 cm and r2 = 3 cm .

Hence , the rad ii o f the two circles are 11 cm and 3 cm .

15151515.... AAAA coppercoppercoppercopper wire,wire,wire,wire, whenwhenwhenwhen bentbentbentbent inininin thethethethe formformformform ofofofof aaaa square,square,square,square, enclosesenclosesenclosesencloses anananan areaareaareaarea ofofofof

r1C1 C2

r 2

1 .45 M

30 cm


484484484484 cmcmcmcm2222.... IfIfIfIf thethethethe samesamesamesame wirewirewirewire isisisis bentbentbentbent inininin thethethethe formformformform ofofofof aaaa circle,circle,circle,circle, findfindfindfind thethethethe areareareare enclosedenclosedenclosedenclosed bybybyby it.it.it.it.

(Use(Use(Use(Use ==== 22/7).22/7).22/7).22/7).

SolutionSolutionSolutionSolution :::: We have ,

Area of the square = 484 cm 2

Side of the square 484 cm = 22 cm

AreaSid e

Sid eAreaQ

So, Perim eter of the square = 4 (s id e) = (4 22 ) cm = 88cm

Let r be the rad iu s of the circle . Then ,

Circum ference of the circle = Perim eter of the square .

2r = 88

2 722

r = 88 r = 14 cm

Area of the circle = r 2 =

214722 cm 2 = 616 cm 2.

16161616.... AAAA carcarcarcar hashashashas wheelswheelswheelswheels whichwhichwhichwhich areareareare 80808080 cmcmcmcm inininin diameter.diameter.diameter.diameter. HowHowHowHow manymanymanymany completecompletecompletecomplete revolutionsrevolutionsrevolutionsrevolutions doesdoesdoesdoes eacheacheacheach

wheelwheelwheelwheel makemakemakemake inininin 10101010 minutesminutesminutesminutes whenwhenwhenwhen thethethethe carcarcarcar isisisis travelingtravelingtravelingtraveling atatatat aaaa speedspeedspeedspeed ofofofof 66666666 kmkmkmkm perperperper hour?hour?hour?hour?


Speed of the car = 66 km / h r

Dis tance trave lled by the car in 1 hou r = 66 km

Dis tance trave lled by the car in 10m in .

=6066

10 km = 11 km = 11 1000 100 cm

We have ,

Rad iu s of car wheels = 40 cm

Circum ference of the wheels = 2 722

40 cm

Dis tance trave lled by the car when its wheels take one comp le te revo lu t ion

= 2 722

40cm

Number of revo lu t ions made by the wheels in 10 m inu tes .

=revo lu t ioncomp le teonem akeswheelsitswhencarthebyeredcovcetanDis

u tesm in10incarthebyeredcovcetanDis

=40

7222

100100011

=40222

7100100011

= 4375


Hence , each wheel makes 4375 revo lu tions in 10 m inu tes .

17171717.... Fig.,Fig.,Fig.,Fig., depictsdepictsdepictsdepicts anananan archeryarcheryarcheryarchery targettargettargettarget marketmarketmarketmarket withwithwithwith itsitsitsits fivefivefivefive scoringscoringscoringscoring areasareasareasareas fromfromfromfrom thethethethe centrecentrecentrecentre outwardsoutwardsoutwardsoutwards asasasas

Gold,Gold,Gold,Gold, Red,Red,Red,Red, BlueBlueBlueBlue BlackBlackBlackBlack andandandand White.White.White.White. TheTheTheThe diameterdiameterdiameterdiameter ofofofof thethethethe regionregionregionregion representingrepresentingrepresentingrepresenting GoldGoldGoldGold scorescorescorescore isisisis 21212121 cmcmcmcm

andandandand eacheacheacheach ofofofof thethethethe otherotherotherother bandsbandsbandsbands isisisis 10.510.510.510.5 cmcmcmcm wide.wide.wide.wide. FindFindFindFind thethethethe areaareaareaarea ofofofof eacheacheacheach ofofofof thethethethe fivefivefivefive scoringscoringscoringscoring regions.regions.regions.regions.


R = Rad iu s of the reg ion rep re sen ting Gold score = 10 .5 cm

r1 = Rad iu s of the reg ion rep resen ting Gold and Red scoring areas

= (10 .5 + 10 .5 ) cm = 21 cm = 2r cm

r2 = Rad iu s of the reg ion rep resen ting Gold , Red and Blue scoring areas

= (21 + 10 .5 ) cm = 31 .5 cm = 3r cm

r3 = Rad iu s of the reg ion rep res ing Gold , Red , Blue and Black scoring areas

= (31 .5 + 10 .5 ) cm = 42 cm = 4r cm

r4 = Rad iu s of the reg ion rep res en ting Go ld , Red , Blu e , Black and wh ite sco ring areas = (42 + 10 .5 )

cm = 52 .5 cm = 5r cm

Now,

A1 = Area of the reg ion rep resen ting Gold scoring area

= r2 =722

(10 .5 )2 =722

10 .5 10 .5 = 22 1 .5 10 .5 = 346 .5 cm 2

A2 = Area of the reg ion rep resen ting Red scoring area

= (2 r)2 r 2 = 3r2 = 3A1 = 3 346 .5 cm 2 = 1039 .5 cm 2

A3 = Area of the reg ion rep resen ting Blue scoring are

= (3 r)2 (2 r)2 = 9r2 4r2 = 5r2 = 5 1 = 5 346 .5 cm 2 = 1732 .5 cm 2

A4 = Area of the reg ion rep resen ting Black scoring area

= (4 r)2 (3 r)2 = 7r2 = 7A1 = 7 346 .5 cm 2 = 2425 .5 cm 2

A5 = Area of the reg ion rep resen ting White scoring area


= (5 r)2 (4 r)2 = 9r 2 = 9A1 = 9 346 .5 cm 2 = 3118 .5 cm 2.

18181818.... PQRSPQRSPQRSPQRS isisisis aaaa diameterdiameterdiameterdiameter ofofofof aaaa circlecirclecirclecircle ofofofof radiusradiusradiusradius 6666 cm.cm.cm.cm. TheTheTheThe lengthslengthslengthslengths PQ,PQ,PQ,PQ, QRQRQRQR andandandand RSRSRSRS areareareare equal.equal.equal.equal. Semi-Semi-Semi-Semi-

circlescirclescirclescircles areareareare drawndrawndrawndrawn onononon PQPQPQPQ andandandand QSQSQSQS asasasas diametersdiametersdiametersdiameters asasasas shownshownshownshown inininin fig.,fig.,fig.,fig., FindFindFindFind thethethethe perimeterperimeterperimeterperimeter andandandand areaareaareaarea ofofofof

thethethethe shadedshadedshadedshaded region.region.region.region.


PS = Diam ete r of a circle of rad iu s 6 cm = 12 cm

PQ = QR = RS =312 = 4 cm

QS = QR + RS = (4 + 4) cm = 8 cm

Hence , requ ired perim e ter

= Arc of sem i- circle of rad iu s 6 cm

+ Arc of sem i- circle of rad iu s 4 cm

+ Arc of sem i- circle of rad iu s 2 cm

= ( 6 + 4 + 2 ) cm = 12 cm

Requ ired area = Area of sem i- circle with PS as d iame ter

+ Area of sem i- circle with PQ d iam ete r

Area of sem i- circle with QS d iame ter.

=21

722

(6 )2 +21

722

2 2 21

722

(4 )2

=21

722 (6 2 + 2 2 4 2)

=21

722

24 =7

264 cm 2 = 37 .71 cm 2 .

19191919.... OnOnOnOn aaaa circularcircularcircularcircular tabletabletabletable covercovercovercover ofofofof radiusradiusradiusradius 32323232 cm,cm,cm,cm, aaaa designdesigndesigndesign isisisis formedformedformedformed leavingleavingleavingleaving anananan equilateralequilateralequilateralequilateral triangletriangletriangletriangle ABCABCABCABC

inininin thethethethe middlemiddlemiddlemiddle asasasas shownshownshownshown inininin fig.fig.fig.fig. FindFindFindFind thethethethe areaareaareaarea ofofofof thethethethe designdesigndesigndesign (shaded(shaded(shaded(shaded region).region).region).region).

SolutionSolutionSolutionSolution :::: In OBD, we have

cos 60 =OBOD and s in 60 =

OBBD

21 =

32OD and

23 =

32BD

OD = 16 and BD = 16 3

BC = 2 BD = 32 3

Area of the shaded reg ion

= Area of the circle Area of ABC

=

22 332

4332 cm 2


=

37683232722 cm 2

=

3768

722528 cm 2.

20202020.... IfIfIfIf thethethethe diameterdiameterdiameterdiameter ofofofof cross-sectioncross-sectioncross-sectioncross-section ofofofof aaaa wirewirewirewire isisisis decreaseddecreaseddecreaseddecreased bybybyby 5555 %%%% howhowhowhow muchmuchmuchmuch percentpercentpercentpercent willwillwillwill thethethethe

lengthlengthlengthlength bebebebe increasedincreasedincreasedincreased sosososo thatthatthatthat thethethethe volumevolumevolumevolume remainsremainsremainsremains thethethethe same?same?same?same?

SolutionSolutionSolutionSolution :::: Let r be the rad iu s of cross - s ect ion of wire and h be its leng th . Then ,

Volum e = r2h ……(i)

5%diam ete r of cross - sect ion =1005

2 r =10r

New diam ete r = 2 r 10r =

10r19

New rad iu s =20r19

Let the new leng th be h 1. Then ,

Volum e = 2

20r19

h 1

From (i) and (ii), we have

r 2h = 2

20r19

h 1

h = 1h400361

h 1 = h361400

Increas e in leng th = h 1 h =361

h400 h =

361h39

Percen tage increase in leng th =h

hh1 100 =

hh39

100 =3613900 = 10 .8 %

Hence , the leng th of the wire increase by 10 .8 %.

21212121.... AAAA cylindricalcylindricalcylindricalcylindrical pipepipepipepipe hashashashas innerinnerinnerinner diameterdiameterdiameterdiameter ofofofof 7777 cmcmcmcm andandandand waterwaterwaterwater flowsflowsflowsflows throughthroughthroughthrough itititit atatatat 192.5192.5192.5192.5 litreslitreslitreslitres perperperper

minute.minute.minute.minute. FindFindFindFind thethethethe raterateraterate ofofofof flowflowflowflow inininin kilometerskilometerskilometerskilometers perperperper hour.hour.hour.hour.


Volume of water tha t flows per hou r = (192 .50 60 ) lit res

= (192 .50 60 1000 ) cm 2 …(i)


Inner d iam ete r of the p ipe = 7cm

Inner rad iu s of the p ipe27 cm = 3 .5 cm

Let h cm be the leng th of the co lumn of water tha t flows in one hou r.

Clearly, water co lumn form s a cylinder of rad iu s 3 .5 cm and leng th h cm .

Volume of water tha t flows in one hou r

= Volum e of the cylinder of rad iu s 3 .5 cm and leng th h cm

=

h5.3722 2 cm 2

From (i) and (ii), we have

722

3 .5 3 .5 h = 192 .50 60 1000

h =5.35.322

710006050.192

cm = 300000 cm = 3 km

Hence , the ra te of flow of water is 3 km per hou r.

22222222.... TheTheTheThe radiiradiiradiiradii ofofofof thethethethe internalinternalinternalinternal andandandand externalexternalexternalexternal surfacesurfacesurfacesurface ofofofof aaaa metallicmetallicmetallicmetallic sphericalsphericalsphericalspherical shellshellshellshell areareareare 3333 cmcmcmcm andandandand 5555 cmcmcmcm

respectively.respectively.respectively.respectively. ItItItIt isisisis meltedmeltedmeltedmelted andandandand recastrecastrecastrecast intointointointo aaaa solidsolidsolidsolid rightrightrightright circularcircularcircularcircular cylindercylindercylindercylinder ofofofof heightheightheightheight3210 cm.cm.cm.cm.

FindFindFindFind thethethethe diameterdiameterdiameterdiameter ofofofof thethethethe basebasebasebase ofofofof thethethethe cylinder.cylinder.cylinder.cylinder.

SolutionSolutionSolutionSolution :::: Let the rad iu s of the base of the cylinder be r cm . Then ,

Volume of the meta llic so lid cylinder of he igh t3210 cm

= Volum e of the me ta l in the spherica l she ll

r2 332 =

34(5 3 3 3)

r2 =323

34 98

r2 =449

r =27 cm

Hence , d iam ete r of the base of the cylinder = 7 cm .

23232323.... AAAA conicalconicalconicalconical vesselvesselvesselvessel ofofofof radiusradiusradiusradius 6666 cmcmcmcm andandandand heightheightheightheight 8888 cmcmcmcm isisisis completelycompletelycompletelycompletely filledfilledfilledfilled withwithwithwith water.water.water.water. AAAA spherespherespheresphere isisisis

loweredloweredloweredlowered intointointointo thethethethe waterwaterwaterwater andandandand itsitsitsits sizesizesizesize isisisis suchsuchsuchsuch thatthatthatthat whenwhenwhenwhen itititit touchestouchestouchestouches thethethethe sides,sides,sides,sides, itititit isisisis justjustjustjust immersedimmersedimmersedimmersed

asasasas shownshownshownshown inininin fig.,fig.,fig.,fig., WhatWhatWhatWhat fractionfractionfractionfraction ofofofof waterwaterwaterwater overoveroverover flows?flows?flows?flows?


SolutionSolutionSolutionSolution :::: Let the rad iu s of the sphere be r cm .

In VOA, we have

tan =86 =

43

s in =53

In VPO, We have

Sin =VOr

53 =

r8r

24 3 r = 5 r

8 r = 24

r = 3 cm

V1 = Volum e of the sphere =34 3 3 cm 3 = 36 cm 3

V2 = Volum e of the water = Volum e of the cone =31 6 2 8 cm 3 = 96 cm 3

Clearly, vo lume of the water tha t flows ou t of the cone is sam e as the vo lume of the sphe re i.e ., V1.

Fraction of the water tha t flows ou t = V1 : V2 = 36 : 96 = 3 : 8 .

22224444.... ABCDABCDABCDABCD isisisis aaaa fieldfieldfieldfield inininin thethethethe shapeshapeshapeshape ofofofof aaaa trapezium.trapezium.trapezium.trapezium. ABABABAB |||||||| DCDCDCDC andandandand ABCABCABCABC ==== 90909090,,,, DABDABDABDAB 60606060.... FourFourFourFour sectorsectorsectorsector

areareareare formedformedformedformed withwithwithwith centrescentrescentrescentres A,A,A,A, B,B,B,B, CCCC andandandand D.D.D.D. TheTheTheThe radiusradiusradiusradius ofofofof eacheacheacheach sectorsectorsectorsector isisisis 17.517.517.517.5 m.m.m.m. FindFindFindFind thethethethe

(i)(i)(i)(i) totaltotaltotaltotal areaareaareaarea ofofofof thethethethe fourfourfourfour sectors.sectors.sectors.sectors.

(ii)(ii)(ii)(ii) areaareaareaarea ofofofof remainingremainingremainingremaining portionportionportionportion givengivengivengiven thatthatthatthat ABABABAB ==== 75757575 mmmm andandandand CDCDCDCD ==== 50m.50m.50m.50m.

SolutionSolutionSolutionSolution :::: Since AB | | CD and ABC = 90 . There fore BCD = 90 . Also ,

BAD = 60

CDA = 180 60 = 120 [Co- in te rio r ang les ]

(i) We have ,

Tota l a rea of the fou r secto rs

= Area of secto r a t A + Area of secto r a t B + Area of secto r a t C

+ Area of secto r D

=36060

(17 .5 )2 +36090

(17 .5 )2 +36090

(17 .5 )2 +360120

(17 .5 )2

=

25.17

31

41

41

21 m 2

= 2

235

m 2 =722

235

235 m 2 = 962 .5 m 2


(ii) Le t DL be perp end icu lar d rawn from D on AB. Then ,

AL = AB BL = AB CD = (75 50 ) m = 25 m

In ALD, we have

tan 60 =ALDL

3 =25DL

DL = 25 3 m

Area of trapez ium ABCD =21 (AB + CD) DL

=21 (75 + 50 ) 25 3 m 2

= 1562 .5 1 .732m 2 = 2706 .25 m 2

Hence ,

Area of the rem ain ing port ion

= Area of trapez ium ABCD Area of 4 secto rs

= 2706 .25 m 2 962 .5 m 2 = 1743 .75 m 2

25252525.... HowHowHowHow manymanymanymany sphericalsphericalsphericalspherical bulletsbulletsbulletsbullets cancancancan bebebebe mademademademade outoutoutout ofofofof aaaa solidsolidsolidsolid cubecubecubecube ofofofof lendlendlendlend whosewhosewhosewhose edgeedgeedgeedge measuresmeasuresmeasuresmeasures

44444444 cm,cm,cm,cm, eacheacheacheach bulletbulletbulletbullet beingbeingbeingbeing 4444 cmcmcmcm inininin diameter.diameter.diameter.diameter.

SolutionSolutionSolutionSolution :::: Let the to ta l number of bu lle ts be x .

Rad iu s of a sphe rica l bu lle t =24 cm = 2 cm

Now, Volum e of a sphe rica l bu lle t =24 (2 )3 cm 3 =

8

722

34 cm 2

Volum e of x spherica l bu lle ts =

x8

722

34 cm 3

Volum e of the so lid cube = (44 )3 cm 3

Clearly, Volum e of x sphe rica l bu lle ts = Volum e of cube

34

722

8 x = (44 )3

34

722

8 x = 44 44 44

x =8224

73444444

= 2541

Hence , to ta l number of sphe rica l bu lle ts = 2541 .

26262626.... AAAA solidsolidsolidsolid ironironironiron rectangularrectangularrectangularrectangular blockblockblockblock ofofofof dimensionsdimensionsdimensionsdimensions 4.44.44.44.4 m,m,m,m, 2.62.62.62.6 mmmm andandandand 1111 mmmm isisisis castcastcastcast intointointointo aaaa hollowhollowhollowhollow

cylindricalcylindricalcylindricalcylindrical pipepipepipepipe ofofofof internalinternalinternalinternal radiusradiusradiusradius 30303030 cmcmcmcm andandandand thicknessthicknessthicknessthickness 5555 cm.cm.cm.cm. FindFindFindFind thethethethe lengthlengthlengthlength ofofofof thethethethe pipe.pipe.pipe.pipe.



Let the leng th of the p ipe b l cm . Then , vo lume of iron in the p ipe is equal to the vo lume of iron in

the b lock.

We have ,

Volum e of the b lock = (4 .4 2 .6 1 ) m 3 = (440 260 100 ) m 3

r = In terna l rad iu s of the p ipe = 30 cm

R = Externa l rad iu s of the p ipe = (30 + 5) cm = 35 cm

Volum e of iron in the p ipe = (Externa l Volum e) (in te rna l Volume)

Volum e of iron in the p ipe = R2h r 2h = (R2 r 2)h

Volum e of iron in the p ipe = (R + r) (R r) h

Volum e of iron in the p ipe = (35 + 30 ) (35 30 ) h cm 3

Volum e of iron in the p ipe= 65 5 h cm 3

Now, Volum e of iron in the p ipe = Volum e of iron in the b lock

65 5 h = 440 260 100

722

65 5 h = 440 260 100

h =

51

651

227100260440 cm

h = 11200 cm = 112 m .

27272727.... AAAA well,well,well,well, whosewhosewhosewhose diameterdiameterdiameterdiameter isisisis 7777 m,m,m,m, hashashashas beenbeenbeenbeen dugdugdugdug 22.522.522.522.5 mmmm deepdeepdeepdeep andandandand thethethethe earthearthearthearth dugoutdugoutdugoutdugout isisisis usedusedusedused totototo

formformformform anananan embankmentembankmentembankmentembankment aroundaroundaroundaround it.it.it.it. IfIfIfIf thethethethe heightheightheightheight ofofofof thethethethe embankmentembankmentembankmentembankment isisisis 1.51.51.51.5 m,m,m,m, findfindfindfind thethethethe widthwidthwidthwidth ofofofof

thethethethe embankment.embankment.embankment.embankment.


Rad iu s of the well =27 m = 3 .5 m

Dep th of the well = 22 .5 m

Volum e of the earth dugou t = (3 .5 )2 22 .5 m 3

= 27

27

245 m 2

Let the wid th of the embankmen t be r me tres .

Clearly, embankmen t form s a cylind rica l she ll whose inner and other rad ii a re 3 .5 m and (r + 3 .5 )

m respective ly and heigh t 1 .5 m .

Volum e of the embankmen t = {(r + 3 .5 )2 (3 .5 )2} 1 .5 m 3

= (r + 7 ) r 23 m 2


But, Volum e of the embankmen t = Volum e of the earth dugou t

r (r + 7 ) 23 =

27

27

245

r(r + 7 ) =449

15

4 r 2 + 28 r = 735

4 r2 + 28r 735 = 0

r =8

1176078428

r =81254428 =

811228 =

884 = 10 .5 [ r > 0 ]

Hence , the wid th of the embankmen t is 10 .5 m .

28282828.... WaterWaterWaterWater isisisis flowingflowingflowingflowing atatatat thethethethe raterateraterate ofofofof 7777 metresmetresmetresmetres perperperper secondsecondsecondsecond throughthroughthroughthrough aaaa circlecirclecirclecircle pipepipepipepipe whosewhosewhosewhose internalinternalinternalinternal

diameterdiameterdiameterdiameter isisisis 2222 cmcmcmcm intointointointo aaaa cylindricalcylindricalcylindricalcylindrical tanktanktanktank thethethethe radiusradiusradiusradius ofofofof whosewhosewhosewhose basebasebasebase isisisis 40404040 cm.cm.cm.cm. DetermineDetermineDetermineDetermine thethethethe

increaseincreaseincreaseincrease inininin thethethethe waterwaterwaterwater levellevellevellevel inininin ½½½½ hour.hour.hour.hour.


Rate of flow of water = 7 m / sec = 7 m / sec = 700 cm / sec.

Leng th of the water co lumn in21 hou rs = (700 30 60 ) cm

In terna l rad iu s of circu lar p ipe = 1 cm .

Clearly, water co lumn form s a cylinder of rad iu s 1 cm and leng th

= (700 30 60 ) cm .

Volum e of the water tha t flows in the tank in21 h r

=

603070011722 cm 2 [Us ing : V = r2h ]

Let h cm be the rise in the leve l of water in the tank. Then ,

Volum e of the water in the tank =722

40 40 h cm 3

From (i) and (ii). We have

722

40 40 h =722

1 1 700 30 60

h =4040

6030700

cm = 787 .5 cm

Hence , the rise in the leve l of water in the tank in21 h r is 787 .5 cm .


22229999.... AAAA rightrightrightright triangle,triangle,triangle,triangle, whosewhosewhosewhose sidessidessidessides areareareare 15151515 cmcmcmcm andandandand 20202020 cm,cm,cm,cm, isisisis mademademademade totototo revolverevolverevolverevolve aboutaboutaboutabout itsitsitsits hypotenuse.hypotenuse.hypotenuse.hypotenuse.

FindFindFindFind thethethethe VolumeVolumeVolumeVolume andandandand surfacesurfacesurfacesurface areaareaareaarea ofofofof thethethethe doubledoubledoubledouble coneconeconecone sosososo formed.formed.formed.formed. (Use(Use(Use(Use ==== 3.14)3.14)3.14)3.14)

SolutionSolutionSolutionSolution :::: Let ABC be the righ t ang led triang le such tha t AB = 15 cm and AC = 20 cm .

Us ing Pythagoras theorem , we have

BC2 = AB2 + AC2

BC2 = 15 2 + 20 2

BC2 = 225 + 400

BC2 = 625

BC = 25 cm

Let OB = x and OA = y.

App lying Pythagoras theorem s in triang les OAB and OAC, we have

AB2 = OB2 + OA2 and AC2 = OA2 + OC2

15 2 = x 2 + y2 and 20 2 = y2 + (25 x)2

x 2 + y2 = 225 and (25 x)2 + y2 = 400

{(25 x)2 + y2} {x 2 + y2} = 400 225

(25 x)2 x 2 = 175

(25 x x) (25 x + x) = 175

(25 2x) 25 = 175

25 2x = 7

2x = 18

x = 9

Pu tt ing x = 9 in x 2 + y2 = 225 , we get

81 + y2 = 225 y2 = 144 y = 12 .

Thus , we have OA = 12 cm and OB = 9cm

Now,

Volum e of the doub le cone = Vol. o f cone CAA + Vol. o f BAA

=31(OA2) OC +

31 (OA2) OB

=31 12 2 16 +

31 12 2 9

=31 144 (16 + 9) =

31 3 .14 144 25 cm 2 = 3768 cm 2

Surface area of the doub le cone = Curved su rface area of cone CAA

+ Curved su rface area of cone BAA

A

o

15cm20cm

20cm15cm

y

x25cm

A

B C


= OA AC + OA AB

= ( 12 20 + 12 15 )cm 2

= 420 cm 2

= 420 3 .14 cm 2 = 1318 .8 cm 2.

33330000.... AnAnAnAn ironironironiron pillarpillarpillarpillar hashashashas somesomesomesome partpartpartpart inininin thethethethe formformformform ofofofof aaaa rightrightrightright circularcircularcircularcircular cylindercylindercylindercylinder andandandand remainingremainingremainingremaining inininin thethethethe formformformform

ofofofof aaaa rightrightrightright circularcircularcircularcircular cone.cone.cone.cone. TheTheTheThe radiusradiusradiusradius ofofofof thethethethe basebasebasebase ofofofof eacheacheacheach ofofofof coneconeconecone andandandand cylindercylindercylindercylinder isisisis 8888 cm.cm.cm.cm. TheTheTheThe

cylindricalcylindricalcylindricalcylindrical partpartpartpart isisisis 240240240240 mmmm highhighhighhigh andandandand thethethethe conicalconicalconicalconical partpartpartpart isisisis 36363636 cmcmcmcm high.high.high.high. FindFindFindFind thethethethe weightweightweightweight ofofofof thethethethe pillarpillarpillarpillar

ifififif oneoneoneone cubiccubiccubiccubic cmcmcmcm ofofofof ironironironiron weighsweighsweighsweighs 7.87.87.87.8 grams.grams.grams.grams.

SolutionSolutionSolutionSolution :::: Let r1 cm and r2 cm denote the rad ii o f the base of the cylinder and cone resp ective ly. Then ,

r1 = r2 = 8 cm

Let h 1 cm be and h 2 cm the heigh ts of the cylinder and the cone respective ly. Then ,

h 1 = 240 cm and h 2 = 36 cm

Now,

Volum e of the cylinder = r1 2h 1cm 3

= ( 8 8 240 ) cm 3

= ( 64 240 ) cm 3

Volum e of the cone =31r2 2h 2 cm 3

= (31 8 8 36 ) cm 3

= (31 64 36 ) cm 3

Tota l vo lume of the iron = Volum e of the cylinder + Volum e of the cone

= ( 64 240 +31 64 36 ) cm 3

= 64 (240 + 12 ) cm 3

=722

64 252 cm 3

= 22 64 36 cm 3

Hence , Tota l weigh t of the p illa r = Volume Weigh t per cm 3

= (22 64 36 ) 7 .8 gms

= 395366 .4 gms = 395 .3664 kg .

O

V

AO

B

A

r1= 8cm

B

h 2= 36 cm

h 1= 240 cm


31313131.... AAAA leadleadleadlead pencilpencilpencilpencil consistsconsistsconsistsconsists ofofofof aaaa cylindercylindercylindercylinder ofofofof woodwoodwoodwood withwithwithwith aaaa solidsolidsolidsolid cylindercylindercylindercylinder ofofofof graphitegraphitegraphitegraphite filledfilledfilledfilled intointointointo it.it.it.it. TheTheTheThe

diameterdiameterdiameterdiameter ofofofof thethethethe pencilpencilpencilpencil isisisis 7777 mm,mm,mm,mm, thethethethe diameterdiameterdiameterdiameter ofofofof thethethethe graphitegraphitegraphitegraphite isisisis 1111 mmmmmmmm andandandand thethethethe lengthlengthlengthlength ofofofof thethethethe

pencilpencilpencilpencil isisisis 10101010 cm.cm.cm.cm. CalculateCalculateCalculateCalculate thethethethe weightweightweightweight ofofofof thethethethe wholewholewholewhole pencil,pencil,pencil,pencil, ifififif thethethethe specificspecificspecificspecific gravitygravitygravitygravity ofofofof thethethethe woodwoodwoodwood isisisis

0.70.70.70.7 gm/cmgm/cmgm/cmgm/cm3333 andandandand thatthatthatthat ofofofof thethethethe graphitegraphitegraphitegraphite isisisis 2.12.12.12.1 gm/cmgm/cmgm/cmgm/cm3333....


Diame ter of the g raph ite cylinder = 1 mm =101 cm

Rad iu s =201 cm

Leng th of the g raph ite cylinder = 10 cm

Volum e of the g raph ite cylinder =

10

201

201

722 cm 3

Weigh t of g raph ite = Volum e Specific g ravity

Weigh t of g raph ite =

1.210

201

201

722 gm

Weigh t of g raph ite =

102110

201

201

722 gm = 0 .165 gm

Diam ete r of pencil = 7 mm =107 cm

Rad iu s of pencil =207 cm and , Leng th of pencil = 10 cm

Volum e of pencil =722

207

207

10 cm 3

Volum e of wood =

10

201

201

72210

207

207

722 cm 3

Volum e of wood =722

201

201

10 (7 7 1 )cm 3 =711

201

48 cm 3

Weigh t of wood =

7.048

201

711 gm

Weigh t of wood =

10748

201

711 gm = 2 .64 gm

Hence , Tota l weigh t = (2 .64 + 0 .165 )gm = 2 .805 gm .

��

1 & 2. Similarity & Pythagoras' Theorem · Geometry (Std. X) 1 1 & 2. Similarity & Pythagoras'...

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Transcript of 1 & 2. Similarity & Pythagoras' Theorem · Geometry (Std. X) 1 1 & 2. Similarity & Pythagoras'...