Question

In the given figure, AD, CE and BF are the medians. O is the point at which the medians of the $\Delta ABC$ intersect. Area of the $\Delta BOC$ is $24c{m}^2$. Then the area of the $\Delta ABC$ is:

• A. $48c{m}^2$
• B. $72c{m}^2$
• C. $76c{m}^2$
• D. $60c{m}^2$

Answer 1

The correct option is B

$72c{m}^2$

Area of $\Delta ADB$ and $\Delta ADC$ are same (Since BD = DC, and AD is the common height)
Similarly, Area of $\Delta ODB$ = Area of$\Delta ODC$
$\Rightarrow$ (Area of $\Delta ADB$ - Area of $\Delta ODB)$ = (Area of $\Delta ADC$ - Area of $\Delta ODC)$
$\Rightarrow$ Area of $\Delta AOB$ = Area of $\Delta AOC$
Similarly, Area of $\Delta AOB$ = Area of $\Delta BOC$
So, Area of $\Delta AOB$ = Area of $\Delta AOC$ = Area of $\Delta BOC=\frac{1}{3}\times$ Area of $\Delta ABC$
Area of $\Delta ABC=3$ (Area of $\Delta BOC)=3\times 24=72c{m}^2$.