Request a call back

Join NOW to get access to exclusive study material for best results

Class 9 SELINA Solutions Maths Chapter 9 - Triangles [Congruency in Triangles]

Triangles [Congruency in Triangles] Exercise Ex. 9(A)

Solution 1

 

 

 

 

 

 








 




 

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

 

Solution 11

 

 

Solution 12

  

  

Solution 13

  

Solution 14

  

Solution 15

  

Triangles [Congruency in Triangles] Exercise Ex. 9(B)

Solution 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 3

 

 

Solution 4

 

 

 

 

 

 

 

 

Solution 5

 

 

 

 

 

Solution 6

Solution 7

 

Solution 8

Solution 9

 

Solution 10

Solution 11

 

 

 

 

Solution 12

 

  

  

Solution 13

 

  

Solution 14

  

Solution 15

 

  

 

 

In triangles AOE and COD,

A = C  (given)

AOE = COD  (vertically opposite angles)  

A + AOE = C + COD

180° - AEO = 180° - CDO

AEO = CDO  ….(i)

Now, AEO + OEB = 180° (linear pair)

And, CDO + ODB = 180° (linear pair)

AEO + OEB = CDO + ODB

OEB = ODB [Using (i)]

CEB = ADB ….(ii)

Now, in ΔABD and ΔCBE,

A = C  (given)

ADB = CEB  [From (ii)]

AB = BC  (given)

ΔABD ΔCBE  (by AAS congruence criterion) 

Solution 16

  

 

 

In ΔAOD and ΔBOC,

AOD = BOC (vertically opposite angles)

DAO = CBO (each 90°)

AD = BC (given)

ΔAOD ΔBOC (by AAS congruence criterion) 

AO = BO [cpct]

O is the mid-point of AB.

Hence, CD bisects AB.

Solution 17

  

 

 

In ΔABC,

AB = AC

B = C (angles opposite to equal sides are equal)

 

Solution 18

  

 

 

Solution 19

  

 

 

 

Solution 20

  

In ΔABD and ΔBAC,

AD = BC (given)

BD = CA (given)

AB = AB (common)

ΔABD ΔBAC (by SSS congruence criterion)

 ADB = BCA and DAB = CBA (cpct)

Get FREE Sample Mock Papers for Final Exams
Download Now
×