If the medians of a ^@\Delta ABC^@ intersect at ^@G^@. Prove th | Math Question and Answer

Answer:

Step by Step Explanation:

We are given that $AD$ , $BE$ and $CF$ are the medians of $\Delta ABC$ intersecting at $G$ .

Now, we have to find the area of $\Delta BGC$ .
We know that a median of a triangle divides it into two triangles of equal area.

Now, in $\Delta ABC$ , $AD$ is the median. $\begin{aligned} \therefore ar(\Delta ABD)= ar(\Delta ACD) &&\ldots\text{(i)} \end{aligned}$ Similarly, in $\Delta GBC$ , $GD$ is the median. $\begin{aligned} \therefore ar(\Delta GBD)= ar(\Delta GCD) &&\ldots\text{(ii)} \end{aligned}$
From $\text{(i)}$ and $\text{(ii)}$ , we get: $\begin{aligned} &ar(\Delta ABD) - ar(\Delta GBD)= ar(\Delta ACD) - ar(\Delta GCD) \\ \implies& ar(\Delta AGB) = ar(\Delta AGC) \end{aligned}$ Similarly, $\begin{aligned} & ar(\Delta AGB) = ar(\Delta BGC) \\ \therefore \space & ar(\Delta AGB) = ar(\Delta AGC) = ar(\Delta BGC) &&\ldots\text{(iii)} \end{aligned}$
$\begin{aligned} & \text{But,}\\ & ar(\Delta ABC) = ar(\Delta AGB) + ar(\Delta AGC) + ar(\Delta BGC) = 3 \space ar(\Delta BGC) &&[ \text{Using eq (iii)} ] \\ &\therefore ar(\Delta BGC) = \dfrac { 1 } { 3 } ar(\Delta ABC) \end{aligned}$

You can reuse this answer
Creative Commons License

Chat with us