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 th part of the area of △ABC.

Step-by-step explanation:

It is given that AM and CN are medians. The intersection point of medians is centroid. Median divides the triangle is two equal parts.

The centroid divides each median in a ratio of 2:1.

Therefore CO:ON is 2:1.

Let the area of △ABC be x.

Since CN is median therefore the area of triangle ANC and BNC are same and half of the total area of the triangle ABC.

Area of triangle ANC is .

The point O divide the median CN is 2:1. So the line AO divides the area in 2:1.

Area of triangle AON is rd of area of ANC. Area of AON is

The area of ABC is x and the area of AON is .

Therefore the area of △AON is th part of the area of △ABC.

the given 6 = w/p is equivalent to 6p = w, where w is the total area of the walls and p is the area covered by a single can of paint.

knowing that 6 cans of paint will be needed, 6p = w becomes 6(6) = w = 36.   this tells us that the total area to be painted is 36 square units.