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

How did you get to the part with sqrt 3?

Is there any other way to solve this?

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

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

Answer D

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:

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

Answer: D.

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...

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

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

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

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

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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