Suppose that the medians BD, CE and AF of a triangle ABC meet at G. Th – McqOptions

MCQOPTIONS

1.	Suppose that the medians BD, CE and AF of a triangle ABC meet at G. Then AG : GF is
A.	1 : 2
B.	2 : 1
C.	1 : 3
D.	2 : 3
Answer» C. 1 : 3

Discussion

No Comment Found

Related MCQs

In the adjoining figure, AB || CD, <b>t</b> is the traversal, EG and FG are the bisectors of BEE and DFE respectively, then EGF is equal to :
In the given figure, <b>DE || BC</b> and <b>DE : BC</b> = 3 : 5 the ratio of the areas of <b> ADE</b> and the trapezium <b>BCED</b>.
If <b>H</b> is the orthocentre of <b> ABC</b>, then the orthocentre of <b> HBC</b> is (fig. given) :
In a triangle <b>ABC</b>, the length of the sides <b>AB, AC</b> and <b>BC</b> are <b>3, 5</b> and <b>6 cm</b> respectively. If a point <b>D</b> on <b>BC</b> is drawn such that the line <b>AD</b> bisects the angle <b>A</b> internally, then what is the length of <b>BD</b>?
<b>X, Y</b> are the mid-points of opposite sides <b>AB</b> and <b>DC</b> of a parallelogram <b>ABCD</b>. <b>AY</b> and <b>DX</b> are joined intersecting in <b>P; CX </b>and <b>BY</b> are joined intersecting in <b>Q</b>. Then <b>PXQY</b> is a :
<b>ABCD</b> is a rhombus with <b> ABC</b> = 56 , then <b> ACD </b>is equal to :
In the given figure, In a <b> ABC , B</b> = <b> C</b>. If <b>AM</b> is the bisector of <b> BAC</b> and <b>AN BC</b>, then <b> MAN</b> is equal to :
ABCD is a square, F is mid point of AB and E is a point on BC such that BE is one-third of BC. If area of FBE = 108 m<sup>2</sup>, then the length of AC is:
The diagonals of a rectangle <b>ABCD</b> meet at <b>0</b>. If <b> BOC</b> = 44 , then <b> OAD</b> is equal to :
ABCD is a parallelogram P, Q, R and S are points on sides AB, BC, CD and DA respectively such that AP = DR. If the area of the parallelogram ABCD is 16 cm<sup>2</sup>, then the area of the quadrilateral PQRS is: