Question

$\text{In fig. BD || CA, E is mid-point of CA}$
$\text{and BD=}\frac{1}{2}\text{AC. Then}$.

• A. $ar\left(ABC\right)=3ar\left(DBC\right)$
• B. $ar\left(ABC\right)=ar\left(DBC\right)$
• C. $ar\left(ABC\right)=2ar\left(DBC\right)$

Answer 1

The correct option is C $ar\left(ABC\right)=2ar\left(DBC\right)$


$\text{Here, BCED is a parallelogram,}\because BD=CE\text{and}BD||CE$
$\text{Hence, ar(DBC) = ar(EBC)...(i)}$

$\text{E is the midpoint of CA.}\text{Hence, CE = AE.}$

$\text{In}\Delta ABC,\text{BE is the median,}\Rightarrow ar\left(EBC\right)=ar\left(EBA\right)=\frac{1}{2}ar\left(ABC\right)$

$\text{Now, ar(ABC) = ar(EBC)+ar(ABE)}$

$\text{Also, ar(ABC) = 2 ar(EBC)}$

$\text{From (i),}\text{ar(ABC) = 2ar(DBC).}$