Question

In the figure, point $D$ is the mid point of side $BC$ and point $G$ is the centroid of $△ABC$.
Find $\frac{A\left(△AGB\right)}{A\left(△ABD\right)}$.

Answer 1

R.E.F image

Given,

'D' is the midpoint of side BC.

'G' is the centroid of

$△$ ABC.

also given,

to find out,

$\frac{ar(△AGB)}{ar(△ABD)}$

as 'D' is the midpoint of BC

we get,BD=DC

'G' is the centroid of $△$ABC.

since we know that,

centroid divides a median in 2:1 ratio

we get,

AG:GD=2:1

as in the given figure ,

the two triangles $△$ ABD and $△$AGB

have the same base of line and the

common verier ,so their ratio of

the triangles area will be equal to their

base ratio

so we get,

$\frac{ar(△AGB)}{ar(△ABD)}=\frac{AG}{AD}\Rightarrow \frac{AG}{AG+GD}$

$\Rightarrow \frac{2}{2+1}\Rightarrow \frac{2}{3}$

$\frac{(△AGB)}{ar(△ABD)}=\frac{2}{3}$