TWO THIRD CENTRIOD THEOREM OF A TRIANGLE

The centroid theorem states that the centroid of the triangle is at 2/3 of the distance from the vertex to the mid-point of the sides.

centriodtheorem
AG = 23 AEBG = 23 BFCG = 23 CD

Problem 1 :

Given that AG = 14 , FG = 10 and EB = 36, find 

i)  GD     ii)  AD   iii)  CG     iv)  CF    v)  EG  and  vi)  BG

twothirdcentheoremq1

Solution :

Finding GD and AD :

AG = 23 A = 23 A= 3×142

Finding CG and CF :

CG = 23 CFCG = 23 (CG+GF) CG = 23 (CG+10)CG - 23 CG =23×103CG - 2CG3 = 1CG = CF = CG+GF= 20+10CF =

Finding EG and BG :

= 23 = 23 ×= 2×363

i)  GD = 7

ii)  AD = 21

iii)  CG = 20

iv)  CF = 30

v)  EG = 12

vi)  BG = 24

Problem 2 :

Given that AD = 15, GE = 19 and FG = 6, find 

i)  AG     ii) GD   iii) BG     iv)  BE    v)  CF  and  vi)  CG

twothirdcentheoremq1

Solution :

Finding AG and GD :

= 23 = 23 = =

Finding BD and BE :

= 23 = 23 = 23 23 23 BG323 23

Finding CG and CF :

= 23 = 23 = 23 = 23 23 -23 CG=CG3=4

i)  AG = 10

ii) GD = 5

iii) BG = 38

iv) BE = 57

v)  CF = 18

vi)  CG = 12

Problem 3 :

Given that CF = 9, GD = 2 and BG = 16, find 

i)  CG     ii) GF   iii) AG     iv)  AD    v)  GE  and  vi)  BE

twothirdcentheoremq1

Solution :

Finding CG and GF :

= 23 = 23× =

Finding AG and AD :

=23 =23 =232323AG343

Finding GE and BE :

=23 =23 32

i)  CG = 6

ii) GF = 3

iii) AG = 4

iv)  AD = 6

v)  GE = 8

vi)  BE = 24

Problem 4 :

Given that BE = 27, DG = 4 and CG = 10, find 

i) BG     ii) GE   iii) GA     iv)  DA    v)  CF  and  vi)  GF

twothirdcentheoremq1

Solution :

Finding BG and GE :

=23 23

Finding GA and AD :

=23 23232323AG383

Finding CF and GF :

=23 2323232330-2032 GF3

i) BG = 18

ii) GE = 9

iii) GA = 8

iv)  DA = 12

v)  CF = 15

vi)  GF = 5

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