(4) If ABCDEF is a regular hexagon, show that vectors
AB + AC + AD + AE + AF = 6 AO, where O is the center of the hexagon.
Answers
Answer:
AB+AC+AD+AE+AF=AB+(AC+AF)+AD+AE
=AB+(AC+CD)+AD+AE [since AF=CD]
=AB+AD+AD+AE
=2AD+(AB+AE)
=2AD+(ED+AE) [since AB=ED]
=2AD+AD
=3AD
=3*(2AO) [since O is the center and AO=OD]
=6AO
Thus, AB+AC+AD+AE+AF=6AO
Step-by-step explanation:
Answer: AB+AC+AD+AE+AF=AB+(AC+AF)+AD+AE
=AB+(AC+CD)+AD+AE [since AF=CD]
=AB+AD+AD+AE
=2AD+(AB+AE)
=2AD+(ED+AE) [since AB=ED]
=2AD+AD
=3AD
=3*(2AO) [since O is the center and AO=OD]
=6AO
Thus, AB+AC+AD+AE+AF=6AO
Or The I have Attached.
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