S.chand publication New Learning Composite mathematics solution of class 7 Chapter 11 Congruence of Triangles Exercise 11C

 Exercise 11C

Question 1

In the given figure AB and CD bisect each other at K.Find that AC=BD

Fig to be added

AK=KB and CK=KD [AB and CD bisect each other]

∠AKC=∠DKB [vertically opp. ∠s]

∴ΔACK≅DKB

∴AC=BD [∵Corresponding parts of congruent Δs]

Question 2

In the given figure the sides BA and CA have been produced such that BA=AD and CA=AE. Prove that DE||BC

Figure to be added

BA=AD and CA=AE [given]

∠DAE=∠BAC [vertically opp ∠s]

∴ΔADE≅ΔABC

∠ADE=∠ABC [∵corresponding parts of congruent Δs]

∴DE||BC

Question 3

In the given figure OA=OB, OC=OD ∠AOB=∠COD. Proved AC=BD

Figure to be added

∠AOB=∠COD

∠AOB-∠COB=∠COD-∠COB

∠AOC=∠BOD

OA=OB and OC=OD [given]

∴ΔAOC≅ΔBOD

∴AC=BD [∵Corresponding part of congruent Δs]

Question 4

In the given figure, if AB and MN bisect each other at O and AM and BN are perpendicular on XY, prove that Δs, OAM and OBN are congruent and hence, prove that AM=BN

Figure to be added

∠AMO=∠OBN  [∵AM and BN ⊥ XY]

∠AOM=∠NOB  [∵Vertically opp.∠s]

AO=ON  [∵AB and MN bisect each other]

∴ΔAMO≅ΔBNO  [∵Corresponding part of congruent Δs]

∴AM=BN

Question 5

∠XYZ is bisected by YP , L is any point on YP and MLN is perpendicular to YP. Prove that LM=LN

Figure to be added

∠MYL=∠LYN  [∵YP bisect ∠XYZ]

∠YLM=∠YLN  [∵ML⊥YP]

YL common arm of ΔMYL and ΔLYN

∴ΔMYL≅ΔLYN

∴LM=LN  [∵corresponding part of congruent Δs]

Question 6

In the figure , PM=PN, PM⊥AB and PN⊥AC. Show that AM=AN

Figure to be added

In ΔAPN and ΔAPM

PM=PN  [given]

∠ANP=∠AMP [both are 90°]

∴ΔAPN≅ΔAPM

∴AM=AN

Question 7

In the figure , AD=BC and AD||BC. Prove that AB=DC

Figure to be added

In ΔADC and ΔABC

AD=BC  [given]

AC is common arm

∠DAC=∠ACB  [corresponding ∠s, AD||BC]

∴ΔADC≅ΔABC

∴AB=DC

Question 8

In the given figure, triangles ΔABC and ΔDCB are right angled at A and D respectively and AC=DB. Prove that ΔABC≅ΔDCB

Figure to be added

In ΔABC and ΔDCB,

∴AC=DB  [given]

∠BAC=∠BDC  [∵both are right angled]

BC common arm

∴ΔABC≅ΔDCB