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

Question

https://gmatclub.com/forum/download/file.php?id=27112 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} 2015-07-01_1424.png

GMATinsight · Answer

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

Answer: Option D

noTh1ng · Answer

How did you get to the part with sqrt 3?

Is there any other way to solve this?

GMATinsight · Answer

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!

noTh1ng · Answer

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...

chetan2u · Answer

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

balamoon · Answer

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

adityadon · Answer

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

So answer = D

reto · Answer

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}

Answer D

mejia401 · Answer

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

Bunuel · Answer

MANHATTAN GMAT OFFICIAL SOLUTION: https://gmatclub.com/forum/download/file.php?id=27160 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} . Answer: D. 2015-07-06_1435.png

Sirakri · Answer

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

DinoPen · Answer

Couldn't you assume that the hexagon in the middle would be equivalent to 6 equilateral triangles? Thus, you would just find the area of 1 equilateral triangle. Then multiply that by 12? -- You get the same answer. I'm wondering if this solution would be sufficient.
 Bunuel

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

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