Question

If $ABCDE$ is a pentagon, then $\overrightarrow{AB}+\overrightarrow{AE}+\overrightarrow{BC}+\overrightarrow{DC}+\overrightarrow{ED}+\overrightarrow{AC}$ equals

• A. $3\overrightarrow{AD}$
• B. $3\overrightarrow{AC}$
• C. $3\overrightarrow{BE}$
• D. $3\overrightarrow{CE}$

Answer 1

The correct option is B $3\overrightarrow{AC}$

In the pentagon ABCDE, join AC and AD.

By triangle law,

In $\Delta ABC$,

$\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$

In $\Delta ACD$,

$\overrightarrow{AD}+\overrightarrow{DC}=\overrightarrow{AC}$

In $\Delta AED$,

$\overrightarrow{AE}+\overrightarrow{ED}=\overrightarrow{AD}$

Now,

$\overrightarrow{AB}+\overrightarrow{AE}+\overrightarrow{BC}+\overrightarrow{DC}+\overrightarrow{ED}+\overrightarrow{AC}=\left(\overrightarrow{AB}+\overrightarrow{BC}\right)+\left(\overrightarrow{AE}+\overrightarrow{ED}\right)+\overrightarrow{DC}+\overrightarrow{AC}$

$=\overrightarrow{AC}+\left(\overrightarrow{AD}+\overrightarrow{DC}\right)+\overrightarrow{AC}$

$=\overrightarrow{AC}+\overrightarrow{AC}+\overrightarrow{AC}$

$=3\overrightarrow{AC}$