Please answer 20th question Q.20. If ABCDEF is a regular hexagon, then AD + EB + FC equals (a) zero (b) 2 AB (c) 4 AB (d) 3 AB

Dear student

$4\stackrel{\to }{AB}$ is the answer.

$\begin{array}{l}\stackrel{\to }{\mathrm{AD}}=2\stackrel{\to }{\mathrm{BC}}\\ \stackrel{\to }{\mathrm{EB}}=2\stackrel{\to }{\mathrm{FA}}\\ \stackrel{\to }{\mathrm{FC}}=2\stackrel{\to }{\mathrm{AB}}\end{array}$

In triangle AOF,

$\begin{array}{l}\stackrel{\to }{FA}+\stackrel{\to }{AO}+\stackrel{\to }{FO}=0\\ \therefore \stackrel{\to }{FA}+\stackrel{\to }{AO}=-\stackrel{\to }{FO}\\ \therefore \stackrel{\to }{AD}+\stackrel{\to }{EB}=-2\stackrel{\to }{FO}\end{array}$

And $\stackrel{\to }{AB}=-\stackrel{\to }{FO}$

$\begin{array}{l}\therefore \stackrel{\to }{AD}+\stackrel{\to }{EB}=2\stackrel{\to }{AB}\\ \therefore \stackrel{\to }{AD}+\stackrel{\to }{EB}+\stackrel{\to }{FC}=2\stackrel{\to }{AB}+2\stackrel{\to }{AB}=4\stackrel{\to }{AB}\end{array}$
Regards

• 1
What are you looking for?