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Question
If G is the centroid of a ΔABC, prove that (area of ΔGAB) = (area of ΔGBC) = (area of ΔGCA) = `1/3` (area of ΔABC)
Solution
W.K.T the median of a triangle divides it into two triangles of equal area.
In ΔABC, AD is the median
Area(ΔABD) = Area (ΔACD) ........(1)
In ΔGBC, GD is die median
Area(ΔGBD) = Area (ΔGCD) ........(2)
Sub (2) from (1) we get
Area(ΔABD) – Area (ΔGBD)
= Area(ΔACD) – Area (ΔGCD)
Area(ΔAGB) = Area(ΔAGC) ........(3)
Similarly
Area(ΔAGB) = Area(ΔBGC) ........(4)
From (3) and (4) we get
Area(ΔAGB) = Area(ΔAGC) = Area(ΔBGC) ........(5)
Now
Area(ΔAGB) + Area(ΔAGC) + Area(ΔBGC) = Area(ΔABC)
⇒ Area(ΔAGB) + Area(ΔAGB) + Area(ΔAGB)
= Area(ΔABC) .......(Using 5)
⇒ 3Area(ΔAGB) = Area(ΔABC)
⇒ Area(ΔAGB) = `1/3` area(ΔABC) ........(6)
From (5) and (6) we get
Area(ΔAGB) = Area(ΔAGB) = Area(ΔBGC)
= `1/3` area(ΔABC)
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