IIT JEE , AIEEE Forum - Mathematics , Algebra

nick

well ya if that is the case then

my method HOLDS FALSE.

Mr.IITIAN007

I think it is better for the experts to give the answer as fast as possible.
What do you say prathima and nick........????

prathima

do u have the answer?

prathima

i m getting the same answer as i had already given u

Mr.IITIAN007

That is the problem.I don't have the answer.Thanx for your effort prathima .I am rating you but please don't unwatch this topic . We will discuss until we get a final 100 % correct answer ,.O.K??How did you get the answer ...please explain , then I will also put my brain into that solution.
But as I was saying......"'Where in the hail are the so called experts???".

prathima

hey now im getting the other answer
but not sure
i will post the solution soon if i get it

prathima

we can divide the dodecagon into 12 congruent triangles
the base becomes 1/3rd of the side of hexagon
and we get the side as 1unit
so base is 1/3
the angles of the triangle r 30 , 75 , 75
we can find the height as 2 +root 3 / 6
area 1/2 ab
1/2 x 1/3 x 2 + root3 / 6
as there r 12 such triangles the total area is
12 x 1/2 x 1/3 x 2+ root3 / 6 = 2 + root 3 / 6
plz corrct me if i m wrong
i m not sure abt my answer

Mr.IITIAN007

O.K yeah........I support that logic.You are right.Lets see what the damn experts say.

nick

i have as similar Q with two SQUARES INSRIBED INSTEAD OF HEXAGON,

ALSO I HAVE THE ANS., PRATHIMA N THE OTHER PEOPLE FIRST TRY THIS AND THEN PROCEED TO THE "HEXAGON"........

Lolita

Ok nick.Thanks keeping constant watch on the topic.

edison

Well the area which s common to both the hexagons will be minimum if the angle between the corresponding intersecting sides of two hexagons is 120 degrees. Now draw figure and using simple geometry and elementary trigonometry the minimum area can easily be worked out.

Similarly, are will be maximum if the sides overlap that is angle between the corresponding sides of two hexagons is 0 degrees.

So as we rotate one hexagon keeping other fixed the area will be mimnimum if the earliar overlapping sides form angle 120 degree.

navi_mb

when u say min. area then both the hexaogans coincide.So, the ans is 3rt.3

reply me if it is correct.

(0)

if the area is minimum the the hexagons should coincide. the area is 3rt3

reply me me if i'm crt!!!!!!!!!!!1

pirate1_from_jee

Yr, its me agn, tho ihve nt bn able 2 slv d prob. bt hp it wd hlp u smhw.

Look, by formla,

R=a/2cosec(

/n)

Hr, R=1, n=6=no. f sds

so, a=1

If u draw an apprx. fig. ul fnd dt der r adjacent similr tringls in pairs.

Their areas vl b prop. 2 square f der corr.sides.

Nw u hv 2 mk d area f d smlr tringls max.

Use mx.& minma knwldge 2 fnd out hw.

IF psble i vl try 2 mail u the soln. its diffclt to gve derivatvs out here.

Bt surly i'l try.

Vle slvng do remembr dat any reg. hexagon inscrbd in a crcle f unit radius wid unit side hs an area of 1.5(3)^0.5

Ask & Discuss Questions with Community & Experts