RS Aggarwal 2019 2020 Solutions for Class 7 Math Chapter 16 Congruence are provided here with simple step-by-step explanations. These solutions for Congruence are extremely popular among class 7 students for Math Congruence Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal 2019 2020 Book of class 7 Math Chapter 16 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal 2019 2020 Solutions. All RS Aggarwal 2019 2020 Solutions for class 7 Math are prepared by experts and are 100% accurate.

Page No 199:

Question 1:

Answer:

We have to state the correspondence between the vertices, sides and angles of the following pairs of congruent triangles.(i) ABCEFDCorrespondence between vertices:   AE, BF, CDCorrespondence between sides:  AB=EF, BC=FD,CA=DECorrespondence between angles:  A=E,B=F,C=D(ii)CABQRPCorrespondence between vertices:CQ, AR, BPCorrespondence between sides:CA=QR, AB=RP, BC=PQCorrespondence between angles:C=Q, A=R, B=P(iii) XZY QPRCorrespondence between vertices:XQ, ZP, YRCorrespondence between sides:XZ=QP, ZY=PR, YX=RQCorrespondence between angles:X=Q, Z=P, Y=R(iv) MPNSQRCorrespondence between vertices:MS, PQ, NRCorrespondence between sides:MP=SQ, PN=QR, NM=RSCorrespondence between angles:M=S, P=Q, N=R



Page No 200:

Question 2:

We have to state the correspondence between the vertices, sides and angles of the following pairs of congruent triangles.(i) ABCEFDCorrespondence between vertices:   AE, BF, CDCorrespondence between sides:  AB=EF, BC=FD,CA=DECorrespondence between angles:  A=E,B=F,C=D(ii)CABQRPCorrespondence between vertices:CQ, AR, BPCorrespondence between sides:CA=QR, AB=RP, BC=PQCorrespondence between angles:C=Q, A=R, B=P(iii) XZY QPRCorrespondence between vertices:XQ, ZP, YRCorrespondence between sides:XZ=QP, ZY=PR, YX=RQCorrespondence between angles:X=Q, Z=P, Y=R(iv) MPNSQRCorrespondence between vertices:MS, PQ, NRCorrespondence between sides:MP=SQ, PN=QR, NM=RSCorrespondence between angles:M=S, P=Q, N=R

Answer:

(i) ACB DEF(SAS congruence property)(ii) RPQ LNM(RHS congruence property)(iii) YXZ TRS(SSS congruence property)(iv)DEF PNM(ASA congruence property)(v) ACB ACD(ASA congruence property)

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Question 3:

(i) ACB DEF(SAS congruence property)(ii) RPQ LNM(RHS congruence property)(iii) YXZ TRS(SSS congruence property)(iv)DEF PNM(ASA congruence property)(v) ACB ACD(ASA congruence property)

Answer:

Given:      PLOA      PM OB       PL=PMTo prove:PLOPMOProof:In PLO andPMO:PLO=PMO    (90°each)PO=PO                   (common)PL=PM                    (given)By RHS congruence property:PLO PMO

Page No 200:

Question 4:

Given:      PLOA      PM OB       PL=PMTo prove:PLOPMOProof:In PLO andPMO:PLO=PMO    (90°each)PO=PO                   (common)PL=PM                    (given)By RHS congruence property:PLO PMO

Answer:

Given:            AD=BC          ADBCWe have to show that AB=DC.Proof:ADBCBCA= DAC   (alternate angles) In ABC andCDA: BC =DA                     (given )BCA= DAC       (proved above) AC= AC                     (common)By SAS congruence property:  ABCCDA=> AB=CD                            (corresponding parts of the congruent triangles)

Page No 200:

Question 5:

Given:            AD=BC          ADBCWe have to show that AB=DC.Proof:ADBCBCA= DAC   (alternate angles) In ABC andCDA: BC =DA                     (given )BCA= DAC       (proved above) AC= AC                     (common)By SAS congruence property:  ABCCDA=> AB=CD                            (corresponding parts of the congruent triangles)

Answer:

Given:AB=AC, BD =DCTo prove: ADBADCProof:(i) In ADB andADC:AB=AC          (given)BD=DC          (given)DA=DA      (common)By SSS congruence property:ADB ADCADB=ADC     (corresponding parts of the congruent triangles)       ...(1)ADB and ADC are on the straight line.  ADB+ADC=180°ADB+ADB=180°=>2ADB=180°=>ADB=90°From (1): ADB=ADC  =90°(ii)BAD=CAD     (corresponding parts of the congruent triangles)

Page No 200:

Question 6:

Given:AB=AC, BD =DCTo prove: ADBADCProof:(i) In ADB andADC:AB=AC          (given)BD=DC          (given)DA=DA      (common)By SSS congruence property:ADB ADCADB=ADC     (corresponding parts of the congruent triangles)       ...(1)ADB and ADC are on the straight line.  ADB+ADC=180°ADB+ADB=180°=>2ADB=180°=>ADB=90°From (1): ADB=ADC  =90°(ii)BAD=CAD     (corresponding parts of the congruent triangles)

Answer:

Given:AD is a bisector of A.=>DAB=DAC          ...(1)ADBC=>BDA=CDA         (90° each)To prove:ABC is isosceles.Proof:InDAB and DAC:BDA=CDA         (90° each)DA=DA                      (common)DAB=DAC           (from 1)  By ASA congruence property:    DAB  DAC=>AB=AC     (corresponding parts of the congruent triangles)Therefore, ABC is isosceles.



Page No 201:

Question 7:

Given:AD is a bisector of A.=>DAB=DAC          ...(1)ADBC=>BDA=CDA         (90° each)To prove:ABC is isosceles.Proof:InDAB and DAC:BDA=CDA         (90° each)DA=DA                      (common)DAB=DAC           (from 1)  By ASA congruence property:    DAB  DAC=>AB=AC     (corresponding parts of the congruent triangles)Therefore, ABC is isosceles.

Answer:

Given:          AB=AD          CB=CDTo prove:ABC ADCProof:In ABC and ADC:AB=AD         (given)BC=DC         (given)AC=AC         (common)ABC ADC                                  (by SSS congruence property)

Page No 201:

Question 8:

Given:          AB=AD          CB=CDTo prove:ABC ADCProof:In ABC and ADC:AB=AD         (given)BC=DC         (given)AC=AC         (common)ABC ADC                                  (by SSS congruence property)

Answer:

Given:             PAAB             QBAB             PA = QBTo prove: OAPOBQFind whether OA=OB.Proof: InOAP andOBQ:POA=QOB           (vertically opposite angles)OAP =OBQ            (90° each)PA=QB                          (given)By AAS congruence property: OAPOBQ=>OA=OB   (corresponding parts of the congruent triangles)

Page No 201:

Question 9:

Given:             PAAB             QBAB             PA = QBTo prove: OAPOBQFind whether OA=OB.Proof: InOAP andOBQ:POA=QOB           (vertically opposite angles)OAP =OBQ            (90° each)PA=QB                          (given)By AAS congruence property: OAPOBQ=>OA=OB   (corresponding parts of the congruent triangles)

Answer:

Given:Triangles ABC and DCB are right angled at A and D, respectively.AC=DBTo prove: ABC DCBIn ABC and DCB:CAB=BDC         (90° each)  BC=BC                      (common)AC= DB                         (given)By R.H.S. congruence property:    ABC DCB                                          

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Question 10:

Given:Triangles ABC and DCB are right angled at A and D, respectively.AC=DBTo prove: ABC DCBIn ABC and DCB:CAB=BDC         (90° each)  BC=BC                      (common)AC= DB                         (given)By R.H.S. congruence property:    ABC DCB                                          

Answer:

Given:ABC is an isosceles triangle in which AB=AC.E and F are midpoints of AC and AB, respectively.To prove:BE=CFProof:E and F are midpoints of AC and AB, respectively.=>AF=FB, AE=ECAB=AC=>12AB=12AC=>FB=ECABC=ACB       (angle opposite to equal sides are equal )=>FBC=ECBConsider BCF and CBE:BC=BC                  (common)FBC=ECB      (proved above)FB=EC                   (proved above)By SAS congruence property:BCF  CBEBE =CF        (corresponding parts of the congruent triangles)

Page No 201:

Question 11:

Given:ABC is an isosceles triangle in which AB=AC.E and F are midpoints of AC and AB, respectively.To prove:BE=CFProof:E and F are midpoints of AC and AB, respectively.=>AF=FB, AE=ECAB=AC=>12AB=12AC=>FB=ECABC=ACB       (angle opposite to equal sides are equal )=>FBC=ECBConsider BCF and CBE:BC=BC                  (common)FBC=ECB      (proved above)FB=EC                   (proved above)By SAS congruence property:BCF  CBEBE =CF        (corresponding parts of the congruent triangles)

Answer:

Given: AB=AC ABC is an isosceles triangle.AP= AQTo prove:BQ=CPProof:AB=AC   (given)AP=AQ   (given)AB-AP=AC-AQ =>BP=CQABC=ACB     (angle opposite to the equal sides are equal)=>PBC=QCBIn PBC andQCB:PB=QC       (proved above)PBC=QCB    (proved above)BC=BC        (common)By SAS congruence property:PBC QCBBQ=CP         (corresponding parts of the congruent triangles)

Page No 201:

Question 12:

Given: AB=AC ABC is an isosceles triangle.AP= AQTo prove:BQ=CPProof:AB=AC   (given)AP=AQ   (given)AB-AP=AC-AQ =>BP=CQABC=ACB     (angle opposite to the equal sides are equal)=>PBC=QCBIn PBC andQCB:PB=QC       (proved above)PBC=QCB    (proved above)BC=BC        (common)By SAS congruence property:PBC QCBBQ=CP         (corresponding parts of the congruent triangles)

Answer:

Given:ABC is an isosceles triangle. AB=ACBD=CETo prove:BE=CD Proof:AB+BD=AC+CE         (As, AB=AC, BD=CE)=>AD=AEConsider ACD and ABE: AC=AB       (given)CAD=BAE     (common)AD=AE           (proved above)By SAS congruence property: ACD ABE =>CD =BE      (corresponding parts of the congruent triangles)



Page No 202:

Question 13:

Given:ABC is an isosceles triangle. AB=ACBD=CETo prove:BE=CD Proof:AB+BD=AC+CE         (As, AB=AC, BD=CE)=>AD=AEConsider ACD and ABE: AC=AB       (given)CAD=BAE     (common)AD=AE           (proved above)By SAS congruence property: ACD ABE =>CD =BE      (corresponding parts of the congruent triangles)

Answer:

.Given:ABC is an isosceles triangle.AB=ACBD =CD To prove:AD bisects A and D.Proof:Consider ABD and ACD:AB=AC    (given)BD=CD    (given)AD=AD     (common) By SSS congruence property:ABD ACD =>BAD=CAD     (by cpct)=>BDA=CDA      (by cpct)

Page No 202:

Question 14:

.Given:ABC is an isosceles triangle.AB=ACBD =CD To prove:AD bisects A and D.Proof:Consider ABD and ACD:AB=AC    (given)BD=CD    (given)AD=AD     (common) By SSS congruence property:ABD ACD =>BAD=CAD     (by cpct)=>BDA=CDA      (by cpct)

Answer:

No, its not necessary. If the corresponding angles of two triangles are equal, then they may or may not be congruent.
They may have proportional sides as shown in the following figure:

Page No 202:

Question 15:

No, its not necessary. If the corresponding angles of two triangles are equal, then they may or may not be congruent.
They may have proportional sides as shown in the following figure:

Answer:

No, two triangles are not congruent if their two corresponding sides and one angle are equal. They will be congruent only if the said angle is the included angle between the sides.

Page No 202:

Question 16:

No, two triangles are not congruent if their two corresponding sides and one angle are equal. They will be congruent only if the said angle is the included angle between the sides.

Answer:


Both triangles have equal area due to the the same product of height and base. But they are not congruent.

Page No 202:

Question 17:


Both triangles have equal area due to the the same product of height and base. But they are not congruent.

Answer:

(i) the same length

(ii) the same measure

(iii)the same side length

(iv) the same radius

(v) the same length and the same breadth

(vi) equal parts

Page No 202:

Question 18:

(i) the same length

(ii) the same measure

(iii)the same side length

(iv) the same radius

(v) the same length and the same breadth

(vi) equal parts

Answer:

(i) False
This is because they can be equal only if they have equal sides.

(ii) True
This is because if squares have equal areas, then their sides must be of equal length.

(iii) False
For example, if a triangle and a square have equal area, they cannot be congruent.

(iv) False
For example, an isosceles triangle and an equilateral triangle having equal area cannot be congruent.

(v) False
They can be congruent if two sides and the included angle of a triangle are equal to the corresponding two sides and the included corresponding angle of another triangle.

(vi) True
This is because of the AAS criterion of congruency.

(vii) False
Their sides are not necessarily equal.


(viii)  True
This is because of the AAS criterion of congruency.

(ix) False
This is because two right triangles are congruent if the hypotenuse and one side of the first triangle are respectively equal to the hypotenuse and the corresponding side of the second triangle.

(x) True



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