Answer:
The area of △AON is [tex]\frac{1}{6}[/tex]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 [tex]\frac{x}{2}[/tex].
The point O divide the median CN is 2:1. So the line AO divides the area in 2:1.
[tex]\frac{\text{Area of triangle AON}}{\text{Area of triangle ANC}}=\frac{\frac{1}{2}\times ON\times h}{\frac{1}{2}\times NC\times h}=\frac{1}{3}[/tex]
Area of triangle AON is [tex]\frac{1}{3}[/tex]rd of area of ANC. Area of AON is
[tex]\frac{1}{3}\text{ of }\frac{x}{2}=\frac{x}{2}\times \frac{1}{3}=\frac{x}{6}[/tex]
The area of ABC is x and the area of AON is [tex]\frac{x}{6}[/tex].
[tex]\frac{\text{Area of triangle AON}}{\text{Area of triangle ABC}}=\frac{\frac{x}{6}}{x}=\frac{1}{6}[/tex]
Therefore the area of △AON is [tex]\frac{1}{6}[/tex]th part of the area of △ABC.
