1.

If ABCDEF is regular hexagon, then AD+EB+FC is equal to

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2AB
3AB
4AB

Solution :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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