Answer:
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., [tex]\frac{x}{2}[/tex].
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.
[tex]\frac{\text{Area of }\triangle ANO}{\text{Area of }\triangle ACN}=\frac{\frac{1}{2}\times ON\times h}{\frac{1}{2}\times CN\times h}=\frac{1}{3}[/tex]
Therefore the area of △AON is one-third of △ACN.
[tex]\text{Area of }\triangle ANO=\frac{1}{3}\times\text{Area of }\triangle ACN[/tex]
[tex]\text{Area of }\triangle ANO=\frac{1}{3}\times\frac{x}{2}[/tex]
[tex]\text{Area of }\triangle ANO=\frac{x}{6}[/tex]
The area of ANO is [tex]\frac{x}{6}[/tex]. Therefore the area of △AON is one-sixth part of △ABC.
