Question

In the given figure, $AC=3AD,3BC=4EC,BF=\frac{1}{5}BD,$ if FG is the median of $\Delta EFD,thenar\left(\Delta FGD\right)$ is equal to

• A. $\frac{1}{27}ar\left(\Delta ABC\right)$
• B. $\frac{1}{15}ar\left(\Delta ABC\right)$
• C. $\frac{3}{40}ar\left(\Delta ABC\right)$
• D. $\frac{4}{5}ar\left(\Delta ABC\right)$

Answer 1

The correct option is B

$\frac{1}{15}ar\left(\Delta ABC\right)$

Let $BJ\perp AC.$ Given that AC = 3AD. $\Rightarrow DC=AC-AD=3AD-AD\left(Given\right)=2AD\Rightarrow DC=\frac{2}{3}AC...\left(i\right)(\because AD=\frac{1}{3}AC)Also,3BC=4EC\Rightarrow 3BC=4(BC-BE)\Rightarrow 4BE=BC\Rightarrow BE=\frac{1}{4}BCAnd,BF=\frac{1}{5}BD\Rightarrow FD=\frac{4}{5}BDGE=GD=\frac{1}{2}ED(\because FGisthemedian)Ar\left(\Delta BDC\right)=\frac{1}{2}\times DC\times BJand,Ar\left(\Delta ABC\right)=\frac{1}{2}\times AC\times BJ\therefore \frac{Ar\left(\Delta BDC\right)}{Ar\left(\Delta ABC\right)}=\frac{DC}{AC}=\frac{2}{3}\left[from\right(i\left)\right]\Rightarrow Ar\left(\Delta BDC\right)=\frac{2}{3}Ar(\Delta ABC).\left(ii\right)Similarly,BE=\frac{1}{4}BC\therefore Ar\left(\Delta BED\right)=\frac{1}{4}Ar\left(\Delta BDC\right)..\left(iii\right)and,FD=\frac{4}{5}BD\therefore Ar\left(\Delta FED\right)=\frac{4}{5}Ar\left(\Delta BED\right).\left(iv\right)and,GD=\frac{1}{2}ED\therefore Ar\left(\Delta FGD\right)=Ar\left(\Delta FED\right)\dots ..\left(v\right)\text{From (ii), (iii), (iv) and (v), we get}Ar\left(\Delta FGD\right)=(\frac{1}{2}\times \frac{4}{5}\times \frac{1}{4}\times \frac{2}{3})Ar\left(\Delta ABC\right)=\frac{1}{15}Ar\left(\Delta ABC\right)$

Hence, the correct answer is option (b).