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If ABCDEF is a regular hexagon then `vec(AD)+vec(EB)+vec(FC)` equals : |
Answer» Correct Answer - D ABCDEF is a regular hexogen. We know from the hexagon that AD is parallel to BC `implies AD=2BC` Similarly, EB is parallel to FA `implies EB = 2FA` and FC is parallel to AB `impliesFC =2AB` Thus, `AD+EB +FC =2BC+2FA+2AB` `=2(FA+AB+BC)` `=2(FC=2(2AB)=4AB)` |
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