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ABCDEF is a regular hexagon, prove that AB+AC+AD+AE+AF=3AD ABCDEF is a regular hexagon, prove that AB+AC+AD+AE+AF=3AD
ABCDEF is a regular hexagon, prove that AB+AC+AD+AE+AF=3AD
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.
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
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 =
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
AB+AC+AD+AE+AF =(AB+AE)+(AC+AF)+AD =AD+AD+AD+=3AD=3(2AO)=6AO therefore,AB+AC+AD+AE+AF=3(2AD)=6AO.
As we know all side of the regular hexagon are equal 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 #prove
Hello studentsDivide thehexagoninto 4 triangles by joiningAC,AD,AE. Now,AF= CD, because its aregular hexagon, therefore it has equal and parallel opposide sides. Add one moreADboth the sides and v get the desired result. therefore,AB+AC+AD+AE+AF=3(2AD)=6AO.Hope it helpsthankyou
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