Question

In the given $\Delta ABC$, Z is the mid-point of the median AD. If the area of $\Delta ABC$ is $18m^2$, find the area of $\Delta BZC$.

• A. $7m^2$
• B. $9m^2$
• C. $6m^2$
• D. $5m^2$

Answer 1

The correct option is B $9m^2$

In $\Delta ABC,$ $Z$ is the mid-point of the median $AD$

Join $Z$ to $B$ and $C$

Therefore, it will divide ΔABC into two triangles of equal area.

$\therefore A\left(\Delta BZD\right)=A\left(\Delta BZA\right)$

$\Rightarrow A\left(\Delta BZD\right)=\frac{1}{2}A\left(\Delta ABD\right)$...............(i)

Also, $A\left(\Delta DZC\right)=A\left(\Delta AZC\right)$

$\Rightarrow A\left(\Delta DZC\right)=\frac{1}{2}A\left(\Delta ADC\right)$......................(ii)

$A\left(\Delta BZD\right)+A\left(\Delta DZC\right)=\frac{1}{2}A\left(\Delta ABD\right)+\frac{1}{2}A\left(\Delta ADC\right)$

$A\left(\Delta BZC\right)=\frac{1}{2}A\left(\Delta ABC\right)=\frac{1}{2}\times 18=9m^2$