Median am and cn of △abc intersect at point o. what part of area of △abc is the area of △aon?

1/6 part of the area of △ABC is the area of △AON

Step-by-step explanation:

Here, ABC is a triangle.

In which,  AM and AN are the median of the Δ ABC,

where, and .

Since, By the property of the median,

Area of Δ ANC = Area of Δ BNC ( because median divide the triangle in two equal parts)

⇒ Area of Δ ANC = 1/2× Area of Δ ABC ------------ (1)

Now,  Median AM and CN of △ABC intersect at point O.  ( given)

That is, O is the centroid of the Δ ABC.

And, we know that, The centroid divides each median in a ratio of 2:1.

Thus, OC : ON = 2:1

Let OC = 2x and ON= x where x is any number.

Now, In triangle ANO,

Let h shows the height of Δ ANO,

Then the area of Δ ANO = 1/2 × x × h

And, area of Δ AOC = 1/2 × 2x × h ( because both triangles must have the same height)

But, the area of Δ AOC / area of Δ ANO = 2/1

⇒ AO divides the area of Δ ANC in the ratio of 2:1

That is, the area of Δ ANO = 1/3 × area of ANC

From equation (1) the area of Δ ANO = 1/3 × ( 1/2 × area of Δ ABC) = 1/6 × area of Δ ABC.

Thus, Area of Δ ANO is 1/6 th part of the area of Δ ABC.

i think is g(1)=-1.

hope that i can