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# see attachment A 18 sqrt 3 B 21sqrt 3 C 24sqrt 3 D

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Senior Manager
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see attachment A 18 sqrt 3 B 21sqrt 3 C 24sqrt 3 D [#permalink]

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14 Dec 2006, 16:33
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see attachment
A 18 sqrt 3
B 21sqrt 3
C 24sqrt 3
D 27sqrt 3
E 35sqrt 3
Attachments

pr_t2_36.GIF [ 5.14 KiB | Viewed 1060 times ]

Last edited by gk3.14 on 14 Dec 2006, 19:27, edited 1 time in total.

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14 Dec 2006, 18:52
Getting 27sqrt3
and the side of the larger triangle = 12

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VP
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14 Dec 2006, 19:06
anindyat wrote:
Getting 27sqrt3
and the side of the larger triangle = 12

whats the size of the larger triangle you got?

I got it as 6+3*sqrt(3)

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Re: PS - triangle [#permalink]

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14 Dec 2006, 19:16
gk3.14 wrote:
see attachment

The side of the larger triangle is

sqrt(3)*sqrt(3)*2 + 6 = 12

since its an equilateral triangle

(sqrt(3)/4)*(12^2)
(sqrt(3)/4)*144

whatever this is ...

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14 Dec 2006, 19:46
Side of the smaller triangle = 6, and larger triangle is x
Using Pythagoras,
height of smaller triangle = sqrt3 * 3
height of larger triangle = sqrt3 * x/2

OB = 1/3 * height of larger triangle = 1/3 * sqrt3 * x/2
OA = 1/3 * height of smaller triangle = 1/3 * sqrt3 * 3 = sqrt3
OB = OA + AB => x*sqrt3/6 = sqrt3 + sqrt3 => x = 12

Area of the border = Area of larger triangle â€“ area of smalletr triangle = (sqrt3 * 12^2)/4 - (sqrt3 * 6^2)/4 = 27 * sqrt3

Friends ...Couldnâ€™t find a shorter way to do it, looking forward for a smarter response
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pr_t2_36_195.jpg [ 9.33 KiB | Viewed 1026 times ]

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14 Dec 2006, 21:49
sorry that i can't draw... but...

take a vertex of the inner triangle. take a perpendicular line towards an edge of the larger triangle (this line is given to be of length sqrt(3).
connect the same internal vertex to the closest external vertex.
you get a small triangle. one side is sqrt(3). the opposite angle is 30 degrees. now - tan(30)= 1/sqrt(3) = edge opposite to angle/edge near angle. hence, edge near angle (part of the edge of the external traingle) is 3.
but this small 3 is exactly the difference (on each side), between the internal edge and external edge. hence external edge is 12.

since external edge is twice the internal edge, the external area is four times the internal area.

the difference is therefore 3 times the internal area.
the internal area is 9*sqrt(3) (i always forget the formula, so i recalculate it...)

so the final answer is 27*sqrt(3).

sorry for the lengthy answer.

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15 Dec 2006, 07:57
Anindyat..
wonderful explanation..

Can this be done under 3 minutes on the day?

OA is 27 sqrt 3

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15 Dec 2006, 08:31
...I can't
In fact, sometimes much simpler questions are also going wrong in less than 2 mins. This one is way above my bar...

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15 Dec 2006, 10:25
I too got 27 sqrt3
in the same manner - given by hobbit
This looks faster to me.

larger side A = a+2*(sqrt3*cot30)=6+2*3=12
desired area= 1/2*( AH-ah)
= (1/2)* sqrt3/2 * ( A^2-a^2)
...as in a 60:60:60 Trnge, h=sqrt3/2 * a

= 1/2 * 1/2 * 18*6 * sqrt3
= 27sqrt3

now, hopefully, we can solve these types in the exam.

thanks gk3.14 for the post!

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15 Dec 2006, 16:27
Yes, hobbit has the fastest way.

First we know that the long side for the little triangle is 2sqrt(3) (the side faces 30 degree angle is half of the longest side of a straight triangle), then we can calculate the other side is 3.

Now the two triangle on both side can simply be put together to make a rectangle. So the total area for one of the three edges would be sqrt(3)*(3+6) which is 9sqrt(3). The total area for all three edges would be 27sqrt(3).

The calculation itself would be less than 1 minute. The key though, is to be able to form the little triangle on the side.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

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15 Dec 2006, 16:27
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# see attachment A 18 sqrt 3 B 21sqrt 3 C 24sqrt 3 D

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