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Q2. State giving reasons, whether the following pairs of triangles are congruent or Not.

i) \ \  \Delta ABC  \ in \ which   \angle A = 50^{\circ}, \  \angle B=60^{\circ} \ and \ BC = 4.5 \  cm \  and \ \Delta DEF \ in \ which \  \angle E = 60^{\circ}, \ EF = 4.5 cm, \  \angle F = 70^{\circ}

Answer

\angle C = 70^{\circ}, \angle D^{\circ} = 50

\angle B = \angle E,\  \angle C = \angle F               

BC = EF = 4.5

Congruent by ASA

 

ii)\ \ \Delta DEF \ in\  which\  \angle E=48^{\circ},\  DE=6cm\  and\  EF=8cm\  and \ \Delta MNR \ in \ which\  \angle R=48^{\circ},\  MN = 6 cm\  and\  MR = 8cm

Answer

DE = MN,\  EF = MR \ but \angle E \ne \angle M   \ Hence \ Not \ Congruent.

 

iii)\ \  \Delta KLM \ in \ which \ KM=4 cm, \ \angle K=75^{\circ}, \ \angle M=40^{\circ} \ and\  \Delta PQR \ in \ which \ PR=4 cm,\  \angle Q=65^{\circ},\  \angle R=40^{\circ}  

Answer

 \angle L = 65^{\circ} and \angle P = 75^{\circ} \angle K = \angle P,

\angle M = \angle R \ and \  KM = PR 

ASA Theorem applied, Triangles are Congruent.

 

iv) \\ \Delta ABC\ in\ which\ AB=3 cm,\ \angle A=90^{\circ},\ BC=5 cm\ and\ \Delta  KLM \ in\ which\ KM = 3 cm,\  \angle K=90^{\circ},\  \angle M=5cm

Answer

Applying Pythagoras theorem,

AC = 4 cm \    KL = 4 cm   

AB=KM,\ AC=KL\ and\ \angle A=\angle K

Hence Triangles are congruent by SAS

 

 

Q3. In the adjoining figure, P in the mid point of AB and \angle PAC = \angle PBD . Prove that: \Delta PAC \cong  \Delta PBD

Answer:p4

\angle PAC = \angle PBD , (given)

AP = PB  (given)

\angle APC = \angle DPB   (opposite angles)

Hence  \Delta PAC \cong \Delta ADC

 

Q4. In the adjoining figure,  DC \parallel AB \ and \ \angle B = \angle D . Prove that  \Delta ABC \cong \Delta ADC

Answerp5

Given  \angle B = \angle D

 \angle DCA = \angle CAB (alternate \ angles)

 \angle DAC = \angle ACB (alternate \ angles)

 AC = Common

Therefore, by ASA  \Delta ABC \cong \Delta ADC

 

Q5. In the adjoining figure,  BC = AD, \angle CAB = \angle ABD \ and\  \angle ACB = \angle BDA.

 Prove \Delta ABC \cong \Delta BAD

Answer

In  \Delta ABD \ and\  \Delta ABC

 \angle CAB = \angle DBA p5

 \angle ADB = \angle ACB 

and DA = CB

Therefore, by ASA,  \Delta ABC \cong \Delta BAD

 

Q6. In the adjoining figure, ABC is a triangle in which AB=AC,\ BL \perp AC \ and \ CM \perp AB. \ Prove\ that\ BL=CM  .

Answerp6

Take \Delta ABL \ and\  \Delta ACM  

AB = AC  

\angle A   is common and

\angle ABL = \angle ACM  

Therefore, by ASA, \Delta ABL \cong \Delta ACM \ Hence \ BL = CM  

 

Q7. In \ \Delta ABC AB=AC. \ If \ D \ in \ the \ mid-point \ of \ BC,

Prove \ that: \ AD is \ bisector \ of \angle A \ and \ AD \perp BC  

Answerscreen-shot-2017-01-07-at-5-28-52-pm

In\  \Delta ABD \ and\  \Delta ADC  

AB=AC  

BD = DC  

and AD \ is \ Common  

By SSS,  \Delta ABD \cong \Delta ADC  

Therefore \angle BAD = \angle DAC  

i) AD \ in \ bisector \ of \angle A  

ii) \angle ADB + \angle ADC = 180^{\circ}\ \rightarrow  \angle ADB = \angle ADC = 90^{\circ}  

 

Q8. In  \Delta ABC  , it in given that AB = AC \ and\  AD   in bisector of \angle A  , meeting BC at D. Prove: i) \Delta ABD \cong \Delta ACD \ ii) AD \perp BC  

Answerp8

Given \ AB = AC

 \angle ABD = \angle ACD   (angles opposite equal sides of a triangle)

 \angle BAD = \angle DAC   (angle bisector)

Therefore by ASA:   \Delta ABD \cong  \Delta ACD  

Hence   \angle ADB = \angle ADC = 90^{\circ}\  Therefore,\  AD \perp BC  

 

Q9. In the adjoining figure,   \Delta ABC   in such that   AB = AC and \angle OBC = \angle OCB  . Prove that:   i) \Delta ABO \cong \Delta ACO \ ii) AO \ in \ the \ bisector \ of \ \angle A  

Answerp9

  \angle ABC = \angle ACB  \ given \ AB = AC  

  \angle ABO + \angle OBC = \angle ACO+\angle OCB  

Given   \angle OBC = \angle OCB  

Therefore,

  \angle ABO = \angle ACO  

AB = AC (given)

AO is Common

Because  \angle OBC=\angle OCB \rightarrow OB=OC

Therefore, SAS applies, Hence \Delta ABO \cong \Delta ACO  

Since \Delta ABO \cong \Delta ACO, \ i) \angle BAO = \angle CAO \ ii) AO \ in \ Bisector \ of \ \angle A  

 

Q10. In the adjoining figure, AB \parallel GF \ and \ AC \parallel DE \ and \ BF = CE.

Prove \ that \ \Delta BDE \cong \Delta FGC

Answerp10

In \Delta BDE \ and\  \Delta FGC

\angle DBE = \angle GFC  (alternate angles)

\angle DEB = \angle GCF  (alternate angles)

BF + FE = FE + EC  

Hence,  BE = FC  

Given BF=EC \ and \  FE    is common.

Therefore,  BE = FC  

Applying ASA, \Delta BDE \cong \Delta FGC  

 

Q11. In the adjoining figure, ABCD as a square and CEB is an isosceles triangle in which EC = EB show: \ i) \Delta DCE \cong \Delta ABE \ ii) AE = DE

Answer p11

Given DC = AB (Sides of a square)

CE = EB (Sides of an isosceles triangle)

\angle ECB = \angle EBC

Add \ 90^{\circ} on \ both \ side, \angle ECB + 90^{\circ} = \angle EBC + 90^{\circ} or \angle DCE = \angle ABE

Hence SAS applies,

Therefore, i)\  \Delta DCE \cong \Delta ABE \ ii) \ Because \ of i)\  AE = DE

 

Q12. Find the vales of x and y in each of the following cases:

Answers

i)p12-1

Given PN \parallel LM,  

PQ = KM  

\angle PQN = \angle MKL  

\angle PNL = \angle NLM   (alternate angles)

Therefore, AAS applies Hence, \Delta PQN \cong \Delta LMK  

Therefore 19=3x+4 \Rightarrow x=5 2y-3=55 \Rightarrow y=29

 

ii)p12-2

In \Delta ABC \ \& \  \Delta ADC 

AC = Common

AB = AD \ and \ BC = DC

Therefore by SSS, \Delta ABC \cong \Delta ADC

Hence, 7y + 1 = 180-103-34  

7y+1=43 \Rightarrow y=6 \ and\ 2x-5=103 \Rightarrow x=54

 

iii)screen-shot-2017-01-07-at-5-53-55-pm

In \Delta PQS \ \&\  \Delta PQR

PR = SQ

PS = QR

And PQ in common

Therefore, by SSS, \Delta PQS \cong \Delta PQR

Hence 2y-5=71 \Rightarrow y = 38

\Delta SPO \cong  \Delta ORQ  

Therefore \ \angle SPO = \angle RQO = 34 \ \& \ \angle SOP = 75^{\circ} \ \&\  \angle POQ = 105^{\circ}

Since, \angle SPQ = \angle RQP \angle OPQ = (3x+3)

Now Calculate, 105 + 2(3x+3) = 180 \Rightarrow x = 11.5

 

iv)p12

In  \Delta RMP \ and\  \Delta RQN

 \angle RPQ = \angle RQP

 \angle RPM = \angle RQN

 RP = RQ \ \&\  \angle MRP = \angle QRN

Hence, ASA applies, Therefore,   \Delta RPM \cong \Delta RQN  

Hence x -3=14 \Rightarrow x = 15 \

And 5y-3=3x+1 \ or \ 5y = 55 \Rightarrow y=11

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Q13. In adjoining figure, the sides BA and CA of  have been produced to D and E such that BA = AD and CA = AE Prove ED ∥ BC

Answerp13

In \Delta AED \ and\  \Delta ABC

\angle EAD = \angle BAC(Opposite Angles)

BA = AD \ and\  EA = AC  

Hence \Delta AED \cong \Delta ABC  

\Rightarrow \angle ABC = \angle ADE \ and\  \angle ACB = \angle AED  

Hence ED \parallel BC  

 

Q14. Equilateral triangle ABC and ACE have been drawn on the sides AB and AC respectively of \Delta ABC , as shown in the adjoining figure, prove: i) \angle DAC = \angle EAB \ ii) \ DC = BE

Answerp14

In \Delta DAC \ \&\ \Delta EAB  

DA = AB 

AC = AE 

\angle DAC = 60^{\circ} +\angle BAC =\angle BAE = 60^{\circ} +\angle BAC 

Hence \Delta DAC \cong \Delta EAB   Therefore,

i)\ \angle DAC =\angle BAE 

ii) \ DC = BE 

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Q15. In a regular pentagon ABCDE, prove that \Delta ABD \ \ is isosceles.

Answerp15

Since ABCD in a regular pentagon, all sides are equal and all internal angles are 108^{\circ} .

In \Delta ADE \ and\  \Delta BDC

ED = DC \ \&\  EA = CB \ \&\  \angle EA = \angle DCB =108^{\circ}  

Therefore \Delta AED \cong \Delta DCB

Hence DA = DB    Therefore,\Delta ABD = isosceles triangle

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Q16. In the adjoining figure,  ABCD in a square, and P,\  Q,\  R,\   are points on the side AB,\  BC,\  and\  CD  respectively such that,AP = BQ = CR \ and\  \angle PQR = 90^{\circ}.  

Prove: \ i) \ PB = QC \ ii)\  PQ = QR \ iii) \ \angle QPR = 45^{\circ}  

Answerp16

Given ABCD in a square

\therefore DC = AB 

DR + RC = AP + PB 

RC = AP  (given)

\therefore DR = PB 

Similarly

DC =BC 

DR + RC = CQ + QB 

RC = QB (Given) 

 \therefore DR = CQ 

Now Consider \Delta PQB \ \&\  \Delta RCQ 

\angle PBQ = \angle QCR = 90^{\circ} (Square)

\therefore  RQ = PQ  

RC=QB \ \& \  PB=CQ

\therefore applying \ SSS, \Delta PQB \cong \Delta RCQ, 

PQ = QR \Rightarrow \angle QPR = \angle QRP = 45^{\circ}  

 

Q17. In the adjoining figure, QK \parallel ML, \ QM \parallel KL \ and \ RL=LP.

Prove:\  i)\  \angle MPL=\angle KLR \ ii)\  ML=QK

Answer:p18

Since ML \parallel QK \angle PLM= \angle LRK

And Similarly, KL \parallel QM \angle LKR=\angle PML

PL = LR   Given

Therefore, using AAS, \Delta PML \cong \Delta KLR  

i) \therefore \angle KLR= \angle MPL

ii) \ Since \ QM \parallel KL, \ ML = QK  

 

Q18. In the adjoining figure ABCD is a square, EF \parallel AC  , and R is mid point of EF. Prove:  i) \ AE=CF \ ii) \ DE=DF \ iii)\ DR \ bisect\  \angle EDF  

Answerp17

In \Delta DER \ \&\ \Delta DFR    ,

\angle EDR = \angle FDR   (Median would bisect the angle)

ER = FR \ \& \ DR =   Common

\therefore  \ \Delta DER \cong  \Delta DFR  

\Rightarrow DE = DF  

In \Delta DAE \ \& \ \Delta DCF  

DA = DC   (Side of a square)

angle DAE = \angle DCF = 90^{\circ} DE = DF   (Proved above)

Because \angle EDR=\angle FDR \Rightarrow \angle ADE = \angle CDF  

Applying SAS, Proves that \Delta DAE \cong \Delta DCF  

Therefore, AE = CF  . Hence Proven.

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Q19. In the adjoining figure, \angle TPQ=\angle SQP,\  \angle SRP=\angle TRQ \ and \ R in the Mid point of PQ Prove: \ i) \ \Delta PRT \cong  \Delta QRS \ ii)\  PT = SQ \  iii)\  \angle PTR = \angle QSR  

Answerscreen-shot-2017-01-07-at-5-23-41-pm

In \Delta PRT \ \&\  \Delta QRS  

\angle TPR = \angle SQR  (given)

 PR = QR     (given)

\angle TRP = \angle SRQ  (\angle SRT \ is\  Common)  

 \therefore \ By \ applying \ ASA \ \ \Delta PRT \cong \Delta QRS  

Since \Delta PRT \cong \Delta QRS, PT = SQ \ \&\  \angle PTR = \angle QSR  

 

Q20. In adjoining figure, \angle PON = 90^{\circ} and MO = ON.  

Prove:\  i) \ PM = PN \ ii)\  \angle OMQ = \angle ONQ  

Answerscreen-shot-2017-01-07-at-5-27-10-pm

Consider \Delta MPO \ \&\  \Delta NPO     

PO   is common

MO = ON   (given)

\angle POM = \angle PON  

Hence, \Delta MPO \cong \Delta NPO    

Therefore,  i) \  PM = PN   

Now Consider \Delta MOQ \ \&\  \Delta NOQ  

MO = ON  

\angle MOQ = \angle NOQ  

OQ   is common

Therefore, \Delta MOQ \cong \Delta NOQ \ Hence, \ \angle OMQ = \angle ONQ  

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