Question

In the given figure D is a point on BC such that BD : DC = $1:3$. If O is the midpoint of AD, then the ratio of areas of $\Delta$AOB and $\Delta$ ABC will be

• A. $1:8$
• B. $1:3$
• C. $1:4$
• D. $1:6$

Answer 1

The correct option is A $1:8$

R.E.F image

Let A be origin

& $B\left(\overline{b}\right)$ & $C\left(\overline{c}\right)$

By section formula,

$D=\frac{3\overrightarrow{b}+\overrightarrow{c}}{3+1}=\frac{3\overrightarrow{b}+\overrightarrow{c}}{4}$

$ar\left(AOB\right)=\frac{1}{2}(\overrightarrow{AB}\times \overrightarrow{AO})=\frac{1}{2}(\overline{b}\times \frac{3\overrightarrow{b}+\overrightarrow{c}}{8})$

$=\frac{1}{6}(\overrightarrow{b}\times \overrightarrow{c})\{\overrightarrow{b}\times \overrightarrow{b}=0\}$

$ar\left(△ABC\right)=\frac{1}{2}(\overrightarrow{AB}\times \overrightarrow{AC})=\frac{1}{2}(\overrightarrow{b}\times \overrightarrow{c})$

$\frac{ar\left(AOB\right)}{ar\left(ABC\right)}=\frac{1}{16}\times 2=\frac{1}{8}\Rightarrow \left(A\right)$