Question

Median AM and CN of △ABC intersect at point O. What part of area of △ABC is the area of △AON?

Answer 1

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., $\frac{x}{2}$.

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.

$\frac{\text{Area of }\triangle ANO}{\text{Area of }\triangle ACN}=\frac{\frac{1}{2}\times ON\times h}{\frac{1}{2}\times CN\times h}=\frac{1}{3}$

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

$\text{Area of }\triangle ANO=\frac{1}{3}\times\text{Area of }\triangle ACN$

$\text{Area of }\triangle ANO=\frac{1}{3}\times\frac{x}{2}$

$\text{Area of }\triangle ANO=\frac{x}{6}$

The area of ANO is $\frac{x}{6}$. Therefore the area of △AON is one-sixth part of △ABC.