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# In the figure above an equilateral triangle with a perimeter

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Director
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In the figure above an equilateral triangle with a perimeter [#permalink]

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23 Feb 2005, 11:59
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In the figure above an equilateral triangle with a perimeter of 18 inches is surrounded by a border inches thick. What is the area of the border?
(A) 18xsqrt(3)
(B) 21xsqrt(3)
(C) 24xsqrt(3)
(D) 27xsqrt(3)
(E) 36xsqrt(3)

Tell me if you do this one under 2 mins.
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triangle.jpg [ 4.67 KiB | Viewed 2390 times ]

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Manager
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23 Feb 2005, 12:14
x+x+x=18
side=6 and with border is 6+(3^1/2)
area1=s^2*(3^(1/2)/4)=12.98
area2=38.94
border=18*3sqrt

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SVP
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23 Feb 2005, 12:50
Hmmm I got D
(1/2*3*sqrt(3)*2+6*sqrt(3))*3=27sqrt(3)

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23 Feb 2005, 14:10
"D"....27*3^1/2.....

Divide the area into 3 rectangles and 6 triangles.

Area of 3 rectangles = 3* 6*3^1/2 = 18*3^1/2

Area of 6 triangles = 6 * 1/2 *{(3^1/2)/tan 30} *3^1/2 = 9*3^1/2

Total = 27*3^1/2

Can't do anything under 2 mins while at work

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Senior Manager
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23 Feb 2005, 14:33
Actually I have taken about ten and really have gotten no where. All I can figure out is that it is more than 18*sqrt3 because that is the summ of the four rectangle you make when you draw lines down from the edge of the inner triangle to the outer one...so that leave D or E...but in terms of how to calculate it I'm not quie sure...baneerjeas method seem legit....but just conceptually you can say that it must be a multiple of 9sqrt3 because that is the area of the inner triangle and since the outer and inner traiangle are equivalent their areas are related by ratio...therefore if you subtract the area of the outer triagle and inner triangle you should get a 9sqrt3 or a multiple ruling out B and C. So within two minutes I was able to get rid of A, B and C but not produce an answer.
_________________

"No! Try not. Do. Or do not. There is no try.

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VP
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23 Feb 2005, 14:41
banerjeea_98 wrote:
"D"....27*3^1/2.....

Divide the area into 3 rectangles and 6 triangles.

Area of 3 rectangles = 3* 6*3^1/2 = 18*3^1/2

Area of 6 triangles = 6 * 1/2 *{(3^1/2)/tan 30} *3^1/2 = 9*3^1/2

Total = 27*3^1/2

Can't do anything under 2 mins while at work

can u plz explain how you solved the area of 6 triangles ? what is tan 30 ?

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Manager
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23 Feb 2005, 14:46
Hey guys how do i see picture/diagram in such questions. I cannot c the diagram??????

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23 Feb 2005, 14:52
christoph wrote:
banerjeea_98 wrote:
"D"....27*3^1/2.....

Divide the area into 3 rectangles and 6 triangles.

Area of 3 rectangles = 3* 6*3^1/2 = 18*3^1/2

Area of 6 triangles = 6 * 1/2 *{(3^1/2)/tan 30} *3^1/2 = 9*3^1/2

Total = 27*3^1/2

Can't do anything under 2 mins while at work

can u plz explain how you solved the area of 6 triangles ? what is tan 30 ?

U already know the height which is 3^1/2.....also when u draw the triangles the top angle is divided in 30 each.....in trigonometry tan30 = height/base.....u need the base....and tan 30 = 1/3^1/2....so base = height/tan 30 = {3^1/2}^2 = 3

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23 Feb 2005, 15:35
You don't have to memorize tan30, as long as you know that the side that is across from the 30 degree angel in a right triangle is half of the longest side (forgot what it is called in English). Then you can work out the other side. eg. a=sqrt(3), c=2sqrt(3). => b=3.

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23 Feb 2005, 16:20
HongHu wrote:
You don't have to memorize tan30, as long as you know that the side that is across from the 30 degree angel in a right triangle is half of the longest side (forgot what it is called in English). Then you can work out the other side. eg. a=sqrt(3), c=2sqrt(3). => b=3.

hm ? can u plz explain your equation ?

Quote:
(1/2*3*sqrt(3)*2+6*sqrt(3))*3=27sqrt(3)

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Director
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23 Feb 2005, 16:42
Does the GMAT actually test for stuff like this? I cvan't see to figure this out let alone under 2 minutes

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Director
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23 Feb 2005, 16:57
Yes, OA is (D) guys
Thanks all.

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Manager
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23 Feb 2005, 18:26
mine will be a bit more complicated in term of imagination. Took 2 minutes to do it

we can split the boarder in
i. 6 small rectangle triangle
ii. 3 rectangles
iii. 3 parallelogram

i. aera of each : 0.5 * (Square root 3 *1)
ii. aera of each : square root 3 * 6
iii. aera of each : 2square root of 3 * 1

i. aera : 3 square root of 3
ii. aera : 18 square root 3
iii. aera : 6 square root of 3

total : 27 square root of 3
D

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23 Feb 2005, 20:41
christoph wrote:
HongHu wrote:
You don't have to memorize tan30, as long as you know that the side that is across from the 30 degree angel in a right triangle is half of the longest side (forgot what it is called in English). Then you can work out the other side. eg. a=sqrt(3), c=2sqrt(3). => b=3.

hm ? can u plz explain your equation ?

Quote:
(1/2*3*sqrt(3)*2+6*sqrt(3))*3=27sqrt(3)

My method really is the same with baner. It involves three steps. First is to divide the area into three parts, each part contains two side triangles and the rectangle in the middle. Second step is to figure out the length of the base of the little triangles using its heighth (sqrt(3)). Third, knowing the base and height of the two triangles, and the two sides for the rectangle, we can then calculate each respective area, and then add them up, and multiply by three.

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24 Feb 2005, 10:51
YES BY DIVIDNG IT IN TO 6 TRIANGLES AND 3 RECT

WE GET: 18.SQRT(3) + 9. SQRT(3)
= 27.SQRT(3)

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24 Feb 2005, 13:05
1 hour...
but I solved it

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Intern
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24 Feb 2005, 18:09
HongHu wrote:
christoph wrote:
HongHu wrote:
You don't have to memorize tan30, as long as you know that the side that is across from the 30 degree angel in a right triangle is half of the longest side (forgot what it is called in English). Then you can work out the other side. eg. a=sqrt(3), c=2sqrt(3). => b=3.

hm ? can u plz explain your equation ?

Quote:
(1/2*3*sqrt(3)*2+6*sqrt(3))*3=27sqrt(3)

My method really is the same with baner. It involves three steps. First is to divide the area into three parts, each part contains two side triangles and the rectangle in the middle. Second step is to figure out the length of the base of the little triangles using its heighth (sqrt(3)). Third, knowing the base and height of the two triangles, and the two sides for the rectangle, we can then calculate each respective area, and then add them up, and multiply by three.

"Second step is to figure out the length of the base of the little triangles using its heighth (sqrt(3)). " Could u please explain this.... here we know height which is given (sqrt(3)) how did u get the base ? I just dont get it.

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24 Feb 2005, 19:20
Answer should be D by symmetry.

the angle subtended by the border thickness (sqrt of 3 ) at the vertex of the bigger triangle should be half of 60 deg = 30 deg.

So the extra half - length of the bigger trainge should be tan30 deg * sqrt 3 = 3. So the lenght of the bigger equialteral triangle will be 6+3+3 = 12.

Hence border area = (sqrt 3)/4 * (12*12 - 6*6) = 27sqrt(3)

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Director
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24 Feb 2005, 20:29
Well, here is the hint:
the border area = big triangle's area - small triangle's area

Can you find small triangle's area?
Yes, easy!!
You need to find big triangle's area?
How?
Look at the pictures.
Attachments

2.JPG [ 13.77 KiB | Viewed 2293 times ]

image001.gif [ 1.97 KiB | Viewed 2298 times ]

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25 Feb 2005, 00:53
HongHu wrote:
christoph wrote:
HongHu wrote:
You don't have to memorize tan30, as long as you know that the side that is across from the 30 degree angel in a right triangle is half of the longest side (forgot what it is called in English). Then you can work out the other side. eg. a=sqrt(3), c=2sqrt(3). => b=3.

hm ? can u plz explain your equation ?

Quote:
(1/2*3*sqrt(3)*2+6*sqrt(3))*3=27sqrt(3)

My method really is the same with baner. It involves three steps. First is to divide the area into three parts, each part contains two side triangles and the rectangle in the middle. Second step is to figure out the length of the base of the little triangles using its heighth (sqrt(3)). Third, knowing the base and height of the two triangles, and the two sides for the rectangle, we can then calculate each respective area, and then add them up, and multiply by three.

thx !

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25 Feb 2005, 00:53
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