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In a regular hexagon ABCDEF, bar(AB) + b...

In a regular hexagon `ABCDEF, bar(AB) + bar(AC)+bar(AD)+ bar(AE) + bar(AF)=k bar(AD)` then k is equal to

A

2

B

3

C

4

D

6

Text Solution

Verified by Experts

The correct Answer is:
B
By triangle law, `AB=AD-BD,AC=AD-CD`

Therefore, `AB+AC+AD+AE+AF`
`=3AD+(AE-BD)+(AF-CD)=3AD`
hence, `lamda=3" "(becauseAE=BD,AF=CD)`
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