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the conked out star

by sanju09 » Sat May 08, 2010 4:38 am
When two congruent equilateral triangles share a common center, their union can be a star, as shown. If their overlap is a regular hexagon with an area of 60, what is the area of one of the original equilateral triangles?
(A) 50
(B) 60
(C) 70
(D) 80
(E) 90
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by shashank.ism » Sat May 08, 2010 4:55 am
sanju09 wrote:When two congruent equilateral triangles share a common center, their union can be a star, as shown. If their overlap is a regular hexagon with an area of 60, what is the area of one of the original equilateral triangles?
(A) 50
(B) 60
(C) 70
(D) 80
(E) 90
what do you mean by overlap here..I can't understand this. Please share figure if u have any.
because overlapping could be of many types. if it is a partial overlap , there could be an infinite hexagon. so in that case we can't solve the problem.
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by sanju09 » Sat May 08, 2010 5:06 am
shashank.ism wrote:
sanju09 wrote:When two congruent equilateral triangles share a common center, their union can be a star, as shown. If their overlap is a regular hexagon with an area of 60, what is the area of one of the original equilateral triangles?
(A) 50
(B) 60
(C) 70
(D) 80
(E) 90
what do you mean by overlap here..I can't understand this. Please share figure if u have any.
because overlapping could be of many types. if it is a partial overlap , there could be an infinite hexagon. so in that case we can't solve the problem.
Is the star.docx not opening?
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
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www.manyagroup.com
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by shashank.ism » Sat May 08, 2010 5:10 am
sanju09 wrote:
shashank.ism wrote:
sanju09 wrote:When two congruent equilateral triangles share a common center, their union can be a star, as shown. If their overlap is a regular hexagon with an area of 60, what is the area of one of the original equilateral triangles?
(A) 50
(B) 60
(C) 70
(D) 80
(E) 90
what do you mean by overlap here..I can't understand this. Please share figure if u have any.
because overlapping could be of many types. if it is a partial overlap , there could be an infinite hexagon. so in that case we can't solve the problem.
Is the star.docx not opening?
Hi sanju its after a long time I am here..and you have done a lot of post here..nice to see your effort. so how's your preparation going on . when is ur paper of GMAT.

Well for the attachment I want to know if the star and regular hexagon is the same thing or different things....??
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by sanju09 » Sat May 08, 2010 5:22 am
shashank.ism wrote:
sanju09 wrote:
shashank.ism wrote:
sanju09 wrote:When two congruent equilateral triangles share a common center, their union can be a star, as shown. If their overlap is a regular hexagon with an area of 60, what is the area of one of the original equilateral triangles?
(A) 50
(B) 60
(C) 70
(D) 80
(E) 90
what do you mean by overlap here..I can't understand this. Please share figure if u have any.
because overlapping could be of many types. if it is a partial overlap , there could be an infinite hexagon. so in that case we can't solve the problem.
Is the star.docx not opening?
Hi sanju its after a long time I am here..and you have done a lot of post here..nice to see your effort. so how's your preparation going on . when is ur paper of GMAT.

Well for the attachment I want to know if the star and regular hexagon is the same thing or different things....??
Welcome back shashank.ism, it's good to see you again and thanks for your kind acknowledgements. I am not planning to write it again for now, are you planning to?

Well, I wished to be another MF Hussain but I couldn't. My drawing is pathetic. Please take the drawing as two identical equilateral triangles with center of one lying on the other, and the corresponding sides appearing to be parallel are parallel. When this is fixed, the hexagon appearing in the middle of the star will be regular and the question talks about that only.
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by shashank.ism » Sat May 08, 2010 5:39 am
Ok sanju I got the problem now. I will give it a try.

So when did you gave your paper and you scored 760 thats a nice score btw . So when you are planning to join college and in which university.
Well I haven't given my paper as yet . I really got busy after winning BTG giveaway. I m planning to give it by october. Since you have given ur papers, I would be in contact with you in this regard. It would really help me to fetch good scores. Rather I will PM you or mail you . What's your email id btw??
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by shashank.ism » Sat May 08, 2010 5:47 am
So here is the solution..
Since there are two congruent equilateral triangle sharing a common center. obviously this center is orthocenter which divides altitude in in the ration 2/3
so the base of triangle is also divided in 3 equal parts.
hence area of hexagon = 6 x area of eq. traingles = 60\
area of each triangle = 10
Now if you see the figure this star is a combination of hexagon + 6 equal triangles(which is equivalent to 6 traingles of under hexagon ...since base of triangle is divided in 3 equal parts)
hence area of outer triangles = 10 x6 = 60
area of triangle = area of hexagon + area of 3 smaller triangles =[spoiler] 60 + 30 = 90 Ans E[/spoiler]
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by sanju09 » Sat May 08, 2010 5:47 am
shashank.ism wrote:Ok sanju I got the problem now. I will give it a try.

So when did you gave your paper and you scored 760 thats a nice score btw . So when you are planning to join college and in which university.
Well I haven't given my paper as yet . I really got busy after winning BTG giveaway. I m planning to give it by october. Since you have given ur papers, I would be in contact with you in this regard. It would really help me to fetch good scores. Rather I will PM you or mail you . What's your email id btw??
Isn't it public place for such queries?
The mind is everything. What you think you become. -Lord Buddha



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by harshavardhanc » Sat May 08, 2010 6:24 am
sanju09 wrote:When two congruent equilateral triangles share a common center, their union can be a star, as shown. If their overlap is a regular hexagon with an area of 60, what is the area of one of the original equilateral triangles?
(A) 50
(B) 60
(C) 70
(D) 80
(E) 90
Formula :

Area of a regular polygon with n sides and s as the measure of each side is

n/4 * s^2 * cot (180/n)
now, as the area of hexagon is 60, putting it in the above formula yields

60 = 6/4 * s^2 * cot30 OR s^2 = 40/ V3

coming to the area of smaller equilateral triangles, we find that

area of the eq triangle = 3/4 * s^2 * cot 60

= 3/4 * 40/ V3 * 1/ V3

=10


the area of bigger equilateral triangle is sum of 3 smaller triangles + area of the hexagon

= 3 * 10 + 60

=90
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Harsha
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by kstv » Sun May 09, 2010 1:00 am
Sad I can't see sanju's drawing just yet as I do not have Office 2007.
Please post word doc of earlier versions, if possible. It is easier to open.
I will try origami to rival his drawing skills.
The equilateral triangles have a hexagon (Red!)as the overlap area equal to 60.
As they are equilateral triangles and have a common centre the pointed parts or the six smaller equilateral triangles (Pinky ones) are jutting out from the edges of hexagon.
If these Pinkies are folded along the edges of the hexagon will meet at the common centre and cover the hexagon.
Red star eclipsed by Pink star.
so the area of these six equilateral traingle together = 60
The original equilateral triangle consists of the hexagon and three small equilateral triangles
= 60+60/2 = 90

Colour code ------Hexagon in Red
Smaller equilateral triangles in pink
Image
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by sanju09 » Sun May 09, 2010 9:42 pm
kstv wrote:Sad I can't see sanju's drawing just yet as I do not have Office 2007.
Please post word doc of earlier versions, if possible. It is easier to open.
I will try origami to rival his drawing skills.
The equilateral triangles have a hexagon (Red!)as the overlap area equal to 60.
As they are equilateral triangles and have a common centre the pointed parts or the six smaller equilateral triangles (Pinky ones) are jutting out from the edges of hexagon.
If these Pinkies are folded along the edges of the hexagon will meet at the common centre and cover the hexagon.
Red star eclipsed by Pink star.
so the area of these six equilateral traingle together = 60
The original equilateral triangle consists of the hexagon and three small equilateral triangles
= 60+60/2 = 90

Colour code ------Hexagon in Red
Smaller equilateral triangles in pink
Image
nice drawing, how did you do that?
The mind is everything. What you think you become. -Lord Buddha



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by shashank.ism » Sun May 09, 2010 10:22 pm
nice picture... it really gives a clear idea.. otherwise I was totally confused about the question...
guys please save your document in compatibility mode so that people using WORD 2003 can also open the document..
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