Question

The four lines drawing from the vertices of any tetrahedron to the centroid of the opposite faces meet in a point whose distance from each vertex is ‘k’ times the distance from each vertex to the opposite face, where k is

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{3}{4}$
• D. $\frac{5}{4}$

Answer 1

The correct option is C $\frac{3}{4}$

Let A$(x_1,y_1,z_1)$ B$(x_2,y_2,z_2)$ C$(x_3,y_3,z_3)$ D $(x_4,y_4,z_4)$ be the vertices of tetrahedron. If E is the centroid of face BCD and G is the centroid of A B C D then $AG=\frac{3}{4}\left(AE\right)\therefore K=\frac{3}{4}$