G समबाहु त्रिभुज, `DeltaABC` का केन्द्रक हे| यदि `AB=10cm` , तब AG की लम्बाई ज्ञात करे|

Options

A. `5sqrt(3)/3` cm
B. `(10sqrt(3))/3` cm
C. `5sqrt(3)` cm
D. `10sqrt(3)` cm

Answer

D

Explanation

According to question,

Given, AB=BC=CA=10cm
G=centroid
AG=2 units
GD=1unit
`therefore` AD=3 units =Height
As we know that the height of the equilateral triangle is `=sqrt(3)/2xx 10`
`=5sqrt(3)`
`therefore 3 units =5sqrt(3)`
1 unit = `(5sqrt(3))/3 xx 2 =(10sqrt(3))/3`
`2units =(5sqrt(3))/2 xx 2 =(10sqrt(3))/3`
`therefore AG = (10sqrt(3))/3` cm
Alternate:
`AG(r_(c)) = (AB(a))/(sqrt(3))`
`AG=10/sqrt(3)`
`=sqrt(10sqrt(3))/3` cm