Home
Class 12
MATHS
If ABCDEF is a regular hexagon and AB+AC...

If ABCDEF is a regular hexagon and AB+AC+AD+AE+AF=`lamdaAD`, then `lamda` is equal to

Text Solution

Verified by Experts

We have,
`AD+EB+FC=(AB+BC+CD)+(ED+DC+CB)+FC`
`=AB+(BC+CB)+(CD+DC)+ED+FC`

`=AB+O+O+AB+2AB=4AB`
`(becauseED=AB,FC=2AB)` hence proved.
Promotional Banner

Topper's Solved these Questions

  • VECTOR ALGEBRA

    ARIHANT MATHS|Exercise Exercise For Session 1|7 Videos

  • VECTOR ALGEBRA

    ARIHANT MATHS|Exercise Exercise For Session 2|17 Videos

  • TRIGONOMETRIC FUNCTIONS AND IDENTITIES

    ARIHANT MATHS|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|18 Videos

Similar Questions

Explore conceptually related problems

In a regular hexagon ABCDEF, AB=a,BC=b and CD=c. Then, vec(AE) is equal to

In a regular hexagon ABCDEF, prove that AB+AC+AD+AE+AF=3AD.

If ABCDEF is a regular hexagon then vec(AD)+vec(EB)+vec(FC) equals :

If ABCDEF is a regular hexagon with vec(AB) = vec(a) ,vec(BC ) = vec(b) then vec(CE) equals :

In the given figure,if AB= AC, angle BAD=30^(@)and AE=AD, then x is equal to

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

In the given figure, ABCDEF is a regular hexagon, which vectors are: (i) Collinear (ii) Equal (iii) Coinitial (iv) Collinear but not equal.

ABCDEF is a regular hexagon in the x-y plance with vertices in the anticlockwise direction. If vecAB=2hati , then vecCD is

Consider the regular hexagon ABCDEF with centre at O (origin). Q. Five forces AB,AC,AD,AE,AF act at the vertex A of a regular hexagon ABCDEF. Then, their resultant is