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# The points of a six-pointed star consist of six identical equilateral

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Math Expert
Joined: 02 Sep 2009
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The points of a six-pointed star consist of six identical equilateral  [#permalink]

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01 Jul 2015, 02:28
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Difficulty:

45% (medium)

Question Stats:

69% (01:56) correct 31% (02:17) wrong based on 191 sessions

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The points of a six-pointed star consist of six identical equilateral triangles, with each side 4 cm (see figure). What is the area of the entire star, including the center?

A. $$36\sqrt{3}$$
B. $$40\sqrt{3}$$
C. $$44\sqrt{3}$$
D. $$48\sqrt{3}$$
E. $$72\sqrt{3}$$

Attachment:

2015-07-01_1424.png [ 5.27 KiB | Viewed 6806 times ]

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Re: The points of a six-pointed star consist of six identical equilateral  [#permalink]

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01 Jul 2015, 02:44
1
Bunuel wrote:

The points of a six-pointed star consist of six identical equilateral triangles, with each side 4 cm (see figure). What is the area of the entire star, including the center?

A. $$36\sqrt{3}$$
B. $$40\sqrt{3}$$
C. $$44\sqrt{3}$$
D. $$48\sqrt{3}$$
E. $$72\sqrt{3}$$

Attachment:
The attachment 2015-07-01_1424.png is no longer available

Joining Diagonal of Inner Hexagon gives another set of six equilateral triangle making it a figure of a total 12 equilateral triangle of side 4 each

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Solution 12111.jpg [ 117.07 KiB | Viewed 5281 times ]

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Re: The points of a six-pointed star consist of six identical equilateral  [#permalink]

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01 Jul 2015, 05:42
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How did you get to the part with sqrt 3?

Is there any other way to solve this?
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The points of a six-pointed star consist of six identical equilateral  [#permalink]

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01 Jul 2015, 05:48
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noTh1ng wrote:
How did you get to the part with sqrt 3?

Is there any other way to solve this?

Point 1) $$\sqrt{3}$$ comes from the property of Area of Equilateral Triangle which is $$(\sqrt{3}/4)*side^2$$

Point 2) Another method is to Find Area of Bigger Equilateral Triangle of side (4+4+4 = 12) and adding three small Equilateral Triangles with it

i.e. Total Area = $$(\sqrt{3}/4)*12^2$$ $$+ 3*(\sqrt{3}/4)*4^2$$
i.e. Total Area = $$(\sqrt{3}/4)*(144+3*16)$$
i.e. Total Area = $$(\sqrt{3}/4)*(192)$$
i.e. Total Area = $$(\sqrt{3})*(48)$$

I hope it helps!
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The points of a six-pointed star consist of six identical equilateral  [#permalink]

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01 Jul 2015, 06:10
Ok, thank you!

I approached it in the same way you did it in your second choice, by figuring out the area of the big triangle and the one of the three smaller ones, but couldn't remember the Area formula for equilateral triangles so calculated the sides using Pythagorean Theorem.

Numerically I got to the same solution, but it obviously wasn't among the answer choices...
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Re: The points of a six-pointed star consist of six identical equilateral  [#permalink]

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01 Jul 2015, 06:34
Bunuel wrote:

The points of a six-pointed star consist of six identical equilateral triangles, with each side 4 cm (see figure). What is the area of the entire star, including the center?

A. $$36\sqrt{3}$$
B. $$40\sqrt{3}$$
C. $$44\sqrt{3}$$
D. $$48\sqrt{3}$$
E. $$72\sqrt{3}$$

Attachment:
2015-07-01_1424.png

we have two inverted equilateral triangle with sides 12(4+4+4)...
so the area of one of them is $$12^2*\sqrt{3}/4=36\sqrt{3}$$..

apart from this there are three smaller equilateral triangles of sides 4.. so area of these 3 is $$3*4^2*\sqrt{3}/4$$=$$12\sqrt{3}$$..
total=$$48\sqrt{3}$$

ans D
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Re: The points of a six-pointed star consist of six identical equilateral  [#permalink]

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01 Jul 2015, 07:04
Bunuel wrote:

The points of a six-pointed star consist of six identical equilateral triangles, with each side 4 cm (see figure). What is the area of the entire star, including the center?

A. $$36\sqrt{3}$$
B. $$40\sqrt{3}$$
C. $$44\sqrt{3}$$
D. $$48\sqrt{3}$$
E. $$72\sqrt{3}$$

Attachment:
2015-07-01_1424.png

We can do in different way -

Area of large triangle of side 4+4+4 = 12 is (√3/4)*12^2 = 36√3 ---------1

Then the area of remaining 3 small triangles not covered = 3*(√3/4)*4^2 = 12√3 ----------2

So total area 1+2 = 36√3 + 12√3 = 48√3
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Re: The points of a six-pointed star consist of six identical equilateral  [#permalink]

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01 Jul 2015, 08:07
total area of star = area of one vertical big triangle + area of 3 small remaining traingles
= 36 root 3 + 12 root 3
= 48 root 3

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Re: The points of a six-pointed star consist of six identical equilateral  [#permalink]

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01 Jul 2015, 08:23
1
Bunuel wrote:

The points of a six-pointed star consist of six identical equilateral triangles, with each side 4 cm (see figure). What is the area of the entire star, including the center?

A. $$36\sqrt{3}$$
B. $$40\sqrt{3}$$
C. $$44\sqrt{3}$$
D. $$48\sqrt{3}$$
E. $$72\sqrt{3}$$

Attachment:
2015-07-01_1424.png

First calculate the area of one big equiliteral triangle with 12^2*$$\sqrt{3}$$/4 = 36*$$\sqrt{3}$$
Then the smaller outer equiliteral triangles of which are in total 3: 3 * 4^2*$$\sqrt{3}$$/4 = 12*$$\sqrt{3}$$

In total we have 48 $$\sqrt{3}$$

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The points of a six-pointed star consist of six identical equilateral  [#permalink]

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01 Jul 2015, 17:32
Since there are twelves equilateral triangles, the answer must be divisible by 12, and therefore some of the answer choices can be excluded, which ones that are not divisible by twelve.

A. $$36\sqrt{3}$$ The area of 3, the area for each small triangle, is too small for an equaliteral triangle of side length 4.
B. $$40\sqrt{3}$$ Not divisible by 12. Out
C. $$44\sqrt{3}$$ Not divisible by 12. Out
D. $$48\sqrt{3}$$ The area of $$4\sqrt{3}$$ is the area of the triangle with a side lenght of four. Correct answer.
E. $$72\sqrt{3}$$ If answer is above, E is out.

Thanks,
A
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Posts: 52274
Re: The points of a six-pointed star consist of six identical equilateral  [#permalink]

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06 Jul 2015, 02:43
Bunuel wrote:

The points of a six-pointed star consist of six identical equilateral triangles, with each side 4 cm (see figure). What is the area of the entire star, including the center?

A. $$36\sqrt{3}$$
B. $$40\sqrt{3}$$
C. $$44\sqrt{3}$$
D. $$48\sqrt{3}$$
E. $$72\sqrt{3}$$

Attachment:
The attachment 2015-07-01_1424.png is no longer available

MANHATTAN GMAT OFFICIAL SOLUTION:

You can think of this star as a large equilateral triangle with sides 12 cm long, and three additional smaller equilateral triangles with sides 4 inches long. Using the same 3 0 -60-90 logic you applied in problem #13, you can see that the height of the larger equilateral triangle is $$6\sqrt{3}$$, and the height of the smaller equilateral triangle is $$2\sqrt{3}$$. Therefore, the areas of the triangles are as follows:

Large triangle: $$A = \frac{bh}{2} = \frac{12*6\sqrt{3}}{2}=36\sqrt{3}$$

Small triangles: $$A = \frac{bh}{2} = \frac{4*2\sqrt{3}}{2}=4\sqrt{3}$$

The total area of three smaller triangles and one large triangle is:

$$36\sqrt{3}+3(4\sqrt{3})=48\sqrt{3}$$.

Attachment:

2015-07-06_1435.png [ 8.44 KiB | Viewed 6409 times ]

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Re: The points of a six-pointed star consist of six identical equilateral  [#permalink]

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27 Dec 2016, 09:59
Bunuel wrote:

The points of a six-pointed star consist of six identical equilateral triangles, with each side 4 cm (see figure). What is the area of the entire star, including the center?

A. $$36\sqrt{3}$$
B. $$40\sqrt{3}$$
C. $$44\sqrt{3}$$
D. $$48\sqrt{3}$$
E. $$72\sqrt{3}$$

Attachment:
2015-07-01_1424.png

Area of 1 big triangle = $$12^2$$$$\frac{srt3}{4}$$=$$36\sqrt{3}$$
Area of 3 small triangles= $$3*4^2$$$$\frac{srt3}{4}$$=$$12\sqrt{3}$$
Area of total figure=$$48\sqrt{3}$$

D
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Re: The points of a six-pointed star consist of six identical equilateral  [#permalink]

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