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 total area of △ABC be x.

Median divides the area of a triangle in two equal parts.

Since AM are CN are medians, therefore the area of △ACN, △BCN, △ABM and △ATM are equal, i.e., .

The intersection point of medians is called centroid of the triangle. A centroid divides the median in 2:1.

Since CN is median and O is the centroid of the triangle, therefore CO:ON is 2:1.

Draw a perpendicular on CN from A.

Therefore the area of △AON is one-third of △ACN.

The area of ANO is . Therefore the area of △AON is one-sixth part of △ABC.

show us the picture

step-by-step explanation:

show the !

here you hope this .

step-by-step explanation:

hypotenuse = √(15² + 8²)

=  √(225 +64)

=  √(289) = 17

sin(x) = cos(y) = 8/17

they are identical so long as you are doing the sine of one angle and the cosine of the only other angle that isn't the right angle.