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sneha sunil raikar Grade: 12
`        ABCDEF is a regular hexagon, prove that AB+AC+AD+AE+AF=3AD`
8 years ago

## Answers : (4)

10 Points
```										HiDivide the hexagon into 4 triangles by joining AC, AD, AE .Considering triangle ACD, v get AD= AC + CDfor Tri ADE , AD= AE + EDAdd the two eq=>2AD= AC + CD + AE + EDNow, AF = CD, because its a regular hexagon, therefore it has equal and parallel opposide sides.Similarly , ED= ABSubstitute in the eq :2AD= AB+ AC + AE + AFAdd one more AD both the sides and v get the desired result.
```
8 years ago
Krishna Deepakrao Kulkarni
33 Points
```										From vector properties,AB=ED,AE+AB=AD,        ...(1)Similarly,AF=CD,AC+AF=AD       ...(2)From (1), (2)AF+AC+AD+AE+AB=AD+AD+AD=3AD
```
8 years ago
Surendra Pandey
13 Points
```										Here,To prove: AB+AC+AD+AE+AF=3ADNow,,  LHS= AB+AC+AD+AE+AF       = AB+AE+AC+AF+AD       = AE+ED+AC+CD+AD ( opposite sides of regular hexagon are equal So AB=ED & AF=CD )       = AD+AD+AD       =3AD #proved              =
```
9 months ago
parth
11 Points
```										given: ABCDEF is a regular hexagon with center O.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=6AOhope this helps you
```
5 months ago
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