Question

if G is centroid of triangle ABC then the area of triangle AGB

Answer 1

Given: G is centroid of triangle ABC

To prove: $ar(AGB) = \dfrac{1}{3}ar(ABC)$

Const: Join AG, BG and CG

Proof:

The median of triangle divides into two equal parts.

In ΔABC, AD is median of triangle intersect BC at D.

Therefore, ar(ADB) = ar(ADC)

In ΔGBC, GD is median of triangle intersect BC at D.

Therefore, ar(GDB) = ar(GDC)

Similarly, ar(GFB) = ar(GFA) and ar(GEC) = ar(GEA)

ar(ABC) = ar(AGB) + ar(BGC) + ar(CGA)

ar(ABC) = 3ar(AGB)          [ ∵ ar(AGB) = ar(BGC) = ar(CGA)]

$ar(AGB) = \dfrac{1}{3}ar(ABC)$

Hence proved

Answer 2

Step-by-step explanation:

the 3 shaded triangles are equal so

if we knew area of one of the shaded triangle we can find area of triangle AGB

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