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area of shaded region

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by sibbineni » Sun Aug 10, 2008 11:10 am
Find the area of hexagon

it has 6 equilateral triangles

area of equilateral triangle is =sqrt(3)/4*side^2
=sqrt(3)/4*6^2
=sqrt(3)/4*36
=sqrt(3)*9

for six equilateral triangles =6*sqrt(3)*9
=54sqrt(3)----(1)


area of circle is =pi *r^2

=pi*3^2

=9pi-----(2)


area of portion of the circle is 120/720* 9pi =3/*pi

then area of all portions =9pi----(3)

(1)-(2)-(3)

gives 54sqrt(3)-9pi-9pi

C
Last edited by sibbineni on Sun Aug 10, 2008 11:42 am, edited 1 time in total.
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by parallel_chase » Sun Aug 10, 2008 11:18 am
I think the answer C, not too sure coz I make lot of silly mistakes.

The sum of all the angles = (n-2)*180 = 4*180 = 720

perimeter = 36, therefore the length of AB, BC, CD,DE, EF, FA=6

draw a line from each of the points ABCDEF to the center, you will have 6 equilateral triangles.

area of first triangle = a^2sqrt3/4= 36sqrt3/4= 9sqrt3

area of the hexagon = 6*(9sqrt3) = 54sqrt3

since all sides of the hexagon are equal, all the angles will also be equal


Therefore 720/6 = 120 is each individual central angle.

area of portion of the circle = 120/720 * 9pi = 3/2 pi

area of all the portions of the circle = 3/2pi * 6 = 9 pi


area of the circle inside = 9pi


area of the shaded region = 54sqrt3 -9pi-9pi = 54sqrt3 -18pi.

Hence C is the answer.

Whats the OA?
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by kshin78 » Sun Aug 10, 2008 12:24 pm
sibbineni wrote:Find the area of hexagon

it has 6 equilateral triangles

area of equilateral triangle is =sqrt(3)/4*side^2
=sqrt(3)/4*6^2
=sqrt(3)/4*36
=sqrt(3)*9

for six equilateral triangles =6*sqrt(3)*9
=54sqrt(3)----(1)


area of circle is =pi *r^2

=pi*3^2

=9pi-----(2)


area of portion of the circle is 120/720* 9pi =3/*pi

then area of all portions =9pi----(3)

(1)-(2)-(3)

gives 54sqrt(3)-9pi-9pi

C

I did the same exact up until the 3rd part. why do you divide 120/720? shouldn't it be 120/360 which is the portion of the circle and multiply by the area?

I have 120/360 x 9pi = 3pi x # of portions (6) = 18pi

Thus, 54sqr3-9pi-18pi = 54sqr3-27pi (E)

What do you think?
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by parallel_chase » Sun Aug 10, 2008 12:27 pm
Good work Kshin78. Absolutely Perfect.

Cant argue with that. Answer has to be E.

Like I said in my post I make lot of silly mistakes.:)
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by pepeprepa » Sun Aug 10, 2008 2:59 pm
Let's find the areas of the 7 circles.
The radius of the circles is 3 given side of the hegaxon is 6 and we have to radius for each side.
We have one total circle so its area is 9pi
The area of the other 6 circles is 120/360*9pi*6=18pi (proportion of the circle (we know each angle of an hexagon is 120° and a circle is 360°) times the area of each circle times the number of circles)

Now the hegaxon:
The area of the hegaxon is 3sqrt(3)/2 * 6^2 = 54sqrt(3)
(General formula for regular hegaxon is 3sqrt(3)/2*a^2 with a one side, useful but hard to remember)

Now we can subtract all that and we have 54sqrt(3)-27pi

That is a nice question
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by sudhir3127 » Sun Aug 10, 2008 7:51 pm
Thnks all fro the response.. yes the OA is E.

thank u!!
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