help me

[madhu_infos]

frends solve this problem give the solution for it?


A Hexagon(H1) with all sides of length 'a', If Midpoints of the sides of Hexagon are Joined we can get another Hexagon(H2), What is the ratio of Areas of H2 and H2?

[madhu_infos]

consider this ratio between h1 and h2.

[Tan]

1:1
i think..
but not confident

[keerthi]

No its cant be 1:1 sharma.Just try it once. I am working on that .


How to find the length of side of inner hexagon?Any one help me.[img]smileys/smiley1.gif[/img]Edited by: keerthi

[tns.rao]

Originally Posted by madhu_infos

consider this ratio between h1 and h2.

The area ration between the External hexagon and Internal one is


2:1.732

[Anusha_j]

the answer may be H1:H2 = 4:3

[keerthi]

[img]smileys/smiley5.gif[/img]Please post ur explanations anusha.How did u arrive at the answer.I am quite confused[img]smileys/smiley9.gif[/img]

[sowmya571]

hello



please explain ya even i got 1:1 aS answer



bye

[madhu_infos]

plz explain the entire procedure.

[aayushkaistha]

consider the hexagon as shown above...


each angle at the center is 60 degrees.


and since it is a regular hexagon, each triangle is an equilateral triangle.


area of 1 equilateral triangle = ((a pow 2) * root3)/4
area of hexagon = 6 * area of 1 triangle...
= 3*((a pow 2) * root3)/2


since we only need to find the ratio,
we can only consider that...


area of hexagon = K * (side pow 2)
where K consists of all other constants...


for outer hexagon...
side = a


for inner hexagon we need to find the length of side...


consider the following fig..





in order to find the side, we can consider the triangle highlighted by the circle...





in this triangle...
the base is the side of the inner hexagon,
and other two sides r equal of length a/2


each angle in a regular hexagon is 120 degrees, so other two angles will be 30 degrees each


if we drop a perpendicular on the base, it will bisect the base..


so, length of half of the base will simply be = (a/2)*cos30
so, side of inner triangle = a*cos30


square of this will be = (a*cos30) pow 2
= 3*(a pow 2)/4


now,
area of H1/H2 = (K*(a pow 2)) / (k*(3*(a pow 2)/4))
= 4/3 Ans

[keerthi]

[img]smileys/smiley32.gif[/img]Good explanation aayush.

[Tan]

[img]smileys/smiley32.gif[/img]

[Anusha_j]

good explanation aayush......


We can do this way also but bit lenghthy procedure


Area of hexagon1 = 6 * area of six equilateral triangles


Base of triangle = a


Since diagonal = 2a ,other 2 sides length = a


Height of the triangle by pythogoras th = root3a/2


area of triangle = 1/2* a *root3a/2


area of hexagon = 6( 1/2 * a * root3a/2 )


similarly for inner hexagon, here the previous triangle height becomes the side and find the height by pythogoras therom


H2: base oftriangle = root 3a/2


Height = 3a/4


area of hexagon = 6 ( 1/2 * root3a/2 * 3a/4 )


On solving we get 4:3

[Varun Kumar]

just try to find out the ratio of the sides of the hexagons h1 and h2 then just square the the ratio of lengths u will get the ratio of areas.

[narainash25]

[img]smileys/smiley32.gif[/img][img]smileys/smiley32.gif[/img][img]smileys/smiley32.gif[/img]vry vry good work ayush my kisses for u[img]smileys/smiley32.gif[/img][img]smileys/smiley32.gif[/img][img]smileys/smiley32.gif[/img]