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

The area of △AON is one sixth part of ABC.

Step-by-step explanation:

Let the area of triangle ABC be x.

It is given that AM and CN are median and they are intersecting at point O. So point and is centroid of the triangle.

We know that centroid divides the medians in 2:1. Therefore the CO:ON is 2:1.

Since CN is median, therefore it divides the area of triangle in two equal parts.

Since the point O divides the line CN is 2:1, therefore the line AO divides the area of triangle ACN is 2:1.

Draw a perpendicular from A to CN. The length of AH is h.

Therefore the area of triangle AON is one-third of area of triangle ANC.

Therefore, the area of △AON is one sixth part of ABC.

1260=3*6*7*10=3^2*2^2*7*5 perfect cude = 3^3*2^

step-by-step explanation: