If G is the centroid of Trinanle ABC. Prove that area of GAB = 1/3 of ABC.
With Figure
Answers
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Given : AN , BN , CL are the medians
To Prove :
ar ∆AGB = ar ∆AGC
ar ∆AGB = ar ∆ BGC
ar ∆ AGB = 1/3 ∆ABC
Proof :
In ∆AGC and ∆AGB
AG is the median
So, ar ∆AGB = ar ∆AGC
Similarly,
BG is the median
ar ∆AGB = ar ∆BGC
We can say,
ar ∆AGB = ar ∆AGC = ar ∆BGC
Now,
ar∆AGB + ar∆ AGC + ar ∆BGC = ar∆ABC
As they are equal in area, we can say,
3 (ar∆AGB + ar∆AGC + ar∆BGC)= ar∆ABC
Hence,
ar ∆AGB = ar ∆AGC = ar ∆BGC = 1/3 ABC
Proved.
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Hope it helps
[ I am so sorry, due to network issues I am not able to upload the figure ]
Lemme tell you how it is,
A ∆ABC
L is the median on AB
M is the median on BC
and
N is the median on AC
And G is the centroid.
Answer:
G is the centroid of triangle ABC. Area of triangle ABC is 72. What is the area of triangle GBC?
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6 Answers

Ved Prakash Sharma, former Lecturer at Sbm Inter College, Rishikesh (1971-2007)
Answered December 9, 2018 · Author has 8.2K answers and 4.1M answer views
Let .A(h,k). , B(0,0). and. C(p,0). then coordinates of G{(h+0+p)/3 ,(k+0+0)/3}
or. G(h+p/3, k/3).
Area of ∆ ABC=1/2.[h.(0–0)+0.(0-k)+p.(k-0)]
=1/2.[pk]. sq.unit. ,Acordingly:-
1/2.p.k=72
p.k=144……………………….(1)
Area of GBC=1/2.[(h+p/3)(0–0)+0(0-k/3)+p.(k/3–0)]
=1/2.[p.k/3]
=1/6.(p.k)
Putting p.k=144 from eqn. (1)
=1/6.(144)
=24 sq.units. Answer.
Second - method:-
We know that :-
In a triangle ABC , if G be the its centroid then
area (∆ GBC) =area(∆ GCA)= area (∆ GAB) = 1/3.area (∆ ABC).
Therefore. Area of ∆GBC=1/3.area of ∆ ABC.
or. Area of triangle GBC=1/3×72 sq.units.= 24 sq.units. Answer.