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# Geometry problem

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Intern
Joined: 05 Feb 2009
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22 Feb 2009, 02:43
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

What is the area of the triangle?
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Intern
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Re: Geometry problem [#permalink]

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22 Feb 2009, 03:25
Area of the traingle is 25 sqrt3

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Senior Manager
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Re: Geometry problem [#permalink]

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22 Feb 2009, 04:24
Let A B and C be the vertices of the triangle.

three circles are of equal area so the the tangent from the vertices to circle are of same length = x

Length AB = BC = CA = 2x + 4 ( Equlateral tringle so all the three interior angles are 60)

so we can find x using l = X *60*( 2 *3.14/360) where l = 4*3.14/3

substituting x = 4.

so side of triangle = 12

Area of equilateral triangle = 36(3)^1/2.

hope so i am correct.

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Re: Geometry problem [#permalink]

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24 Feb 2009, 08:11
-------.---------------.-------
---x---|---2---|--2----|---x---

Draw line joining center of the circle to the nearest vertex. Draw another line perpendicular down which makes small
triangle (30,60,90) with one side =2

triangle (30,60,90) sides are in the ratio 1:sqrt(3):2
smallest side radius of the circle=2
other sides are 2:2sqrt(3):4

x= 2sqrt(3)

Side of Big triangle = 2+2+2x = 4(1+sqrt(3))

Area of equilatrial traingle = sqrt(3)/4 * s^2 = sqrt(3)/4 * (4(1+sqrt(3)))^2
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Intern
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Re: Geometry problem [#permalink]

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12 May 2009, 07:43
i think the answer is 6 + 16*(3)^.5.

This is how i get there... look at the top of the trinagle (call it ABC). So we are looking at angle B and the "top" circle with centre O.
join two D and E points where the circle touches the two angles.
join OD and OE. they are equal (= 2). also note that angle ODE and OED are equal to each other and equal to 30. They are equal to 30 becouse angle between a tangent and a radius is 90 and angle BDE = BED = 60 due to eqilateral trinagle DBE.
all this means that DOE = 120 and BD = 2(2-cos120)^(.5) = 2*(3)^.5. (This is the formula for a chord) and since DBE is equilatral BD is also 2*(3)^.5.
This all means that a side of ABC is equal to 2 times sides like BD and two times a radius of 2 or (4+4*(3)^.5)
which means that the area of the triangle is

(4+4*(3)^.5)^(2) times ((3)^.5) divided by 4 or (4*4 + 2*4*(3)^.5+4*4*3) times ((3)^.5) divided by 4

= (16 +2*(3)^.5)(((3)^.5) = 6 + 16*(3)^.5.

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Intern
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Re: Geometry problem [#permalink]

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13 May 2009, 15:20
Hello!
I completely agree with x2suresh; I think his answer is correct, that the area is $$16sqrt3+24$$. What I've done here is a series of illustrations that go along with his explanation.

The original:

The diagram seems to imply the necessary symmetry to get that it's equilateral.

Bisecting angle BCA. By extending the line, I convinced myself that we truly were bisecting it.

Thus, angle DCR must be 30 degrees, which leaves angle CRD to be 60 degrees. Recall that the short side, that opposite the 30, was of length 2.

Here's a handy reminder of the 30-60-90 triangle:

Therefore:

And finally, the middle part is clearly 4.

And so, the side length must be $$4+4sqrt3$$

Now, a bit of algebra to clean it up, to match what would probably be listed as answer choices:

Thus, $$Area=16sqrt3+24$$. And yes, I worked out x2suresh's answer, and the result is the same.

Patrykgb: Your explanation also works; you clearly get the side length correct. The only reason your final answer is not right must be just due to some little computation error. However, that formula with cosine in it shouldn't be necessary, since all GMAT problems must be solvable using "fundamental" math. Also, dude, check out the main forum page; there's a stickied topic that talks about how to use the fun math formatting on this forum. It makes it all look pretty, and it's very easy to lose a little symbol or sign somewhere without it!

tkarthi4u: What formulas are you using? It seems like you're trying to convert to radians.

Vemuri: How did you arrive at your answer?

Enjoy!

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Re: Geometry problem [#permalink]

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13 May 2009, 16:06
Ahh yes... there is a 4 missing before that 6... yes it is 24 + 16*(3)^.5 sorry...
though problem

would this ever come up on the real GMAT?

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Re: Geometry problem [#permalink]

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30 May 2009, 13:04
@patrykgb:

I think it could possibly appear, although the work is a bit tedious. I think this is a fair upper level question, since it is answerable using fundamental math.

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Re: Geometry problem   [#permalink] 30 May 2009, 13:04
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# Geometry problem

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