If G is the centroid of `triangle` ABC and BC=3,CA=4 AB=5 then BG=
`(sqrt(13))/(3)`
`(2sqrt(13))/(3)`
`sqrt(13)`
`(4sqrt(13))/(3)`

sqrt(13^(2))-sqrt(36) times sqrt(4)=

Evaluate: sqrt(3(13)/(36))

In ABC, the coordinates of B are (0,0),AB=2,/_ABC=(pi)/(3), and the middle point of BC has coordinates (2,0) .The centroid o the triangle is ((1)/(2),(sqrt(3))/(2))(b)((5)/(3),(1)/(sqrt(3)))(4+(sqrt(3))/(3),(1)/(3))( d) none of these

(13+2sqrt(5))^(2)=?sqrt(5)+189

If A(1,-1,2),B(2,1,-1)C(3,-1,2) are the vertices of a triangle then the area of triangle ABC is (A) sqrt(12) (B) sqrt(3) (C) sqrt(5) (D) sqrt(13)

lim_(x rarr1)(13sqrt(x)-7sqrt(x))/(5sqrt(x)-3sqrt(x))

tan(sin^(-1)((3)/(5))+cos^(-1)((3)/(sqrt(13)))=

Which of the following is greater? sqrt(13)-sqrt(12),sqrt(14)-sqrt(13)

cos^(-1)((3cosx-2sinx)/(sqrt(13)))

ln Delta ABC , tan A=3 , tan B=(2)/(3) c=sqrt(10) then circum radius of Delta is (1) 11sqrt((13)/(10)) (2) (10sqrt(13))/(11) (3) (sqrt(130))/(11) (4) (5sqrt(13))/(11)