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I'm terrible at geometry!

Expert replies
by 480ocean » Sat Jan 24, 2009 11:00 am
Please explain the following from GMAT prep:

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

Answer: 48 sq root 3cm

Thanks a bunch!
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Source: — Problem Solving |

by DanaJ » Sat Jan 24, 2009 11:17 am
Well your star is made up of 6 equilateral triangles and a regular hexagon. According to "geometry classics" (my math teacher), a regular hexagon is made up of another 6 equilateral triangles, similar to the other 6. So you've basically got 12 equilateral triangles of 4 cm for each side. So this means total area will be 12(4^2*sqrt(3)/3) = 48sqrt(3).
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by 480ocean » Sat Jan 24, 2009 11:41 am
Thanks so much DanaJ! I totally missed the hexagon!
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by gaggleofgirls » Sat Jan 24, 2009 12:23 pm
I think I have an easier way ( being raised Jewish helped here, :lol: )

A six pointed star can be drawn with two large, equilateral triangles that overlap (doodling at temple school). Their overlap is the hexagon.

See the attachment for a picture

The side of each large triangle is 12, so the total area of the 6-point star (star of david) is the area of one of the large triangles plus the area of the remaining 3 small triangles (with sides of 4).

Height for an equilateral triangle is 1/2b * sqrt3 (from x : x *sqrt3 : 2X ratio of the sides of a right triangle that bisects the equilateral).

So, area of the large triangle is11/2*12 *( 1/2*12 *sqrt3) = 6 *6*sqrt3 = 36*sqrt3
The area of the small triangles is 1/2*4 *(1/2*4 *sqrt3) = 2 * 2*sqrt 3 = 4 *sqrt3 and there are 3 of these so 3 * 4*sqrt3 = 12 * sqrt3

36 * sqrt3 + 12 * sqrt3 = 48 * sqrt3, which is the answer.

HTH

-Carrie
Attachments
star of david question.pdf
(27.42 KiB) Downloaded 237 times
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by 480ocean » Sat Jan 24, 2009 12:58 pm
Yes, that helps too! Thank you!
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by kanha81 » Thu Jun 04, 2009 6:41 am
DanaJ wrote:Well your star is made up of 6 equilateral triangles and a regular hexagon. According to "geometry classics" (my math teacher), a regular hexagon is made up of another 6 equilateral triangles, similar to the other 6. So you've basically got 12 equilateral triangles of 4 cm for each side. So this means total area will be 12(4^2*sqrt(3)/3) = 48sqrt(3).
DanaJ,

I understand the part of how it's 12 equilateral triangles, but why is it
4^2 * sqrt(3) / 3

I also understand that you're dividing by 3 because the 3 sides are equal, but by squaring 4?

Can you please elaborate?

Thanks a bunch
Want to Beat GMAT.
Always do what you're afraid to do. Whoooop GMAT
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by odod » Thu Jun 04, 2009 7:42 am
Hi Same question..I dont get the where the (4^2*sqr3/3) comes from. I thought area of a triangle was 1/2b*h??


Also, even if I do the calculation for 12(4^2*3sqrt/3) I'm getting 64 rt 3 cm.

help?
ODOD
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by odod » Thu Jun 04, 2009 7:55 am
Hi Same question..I dont get the where the (4^2*sqr3/3) comes from. I thought area of a triangle was 1/2b*h??


Also, even if I do the calculation for 12(4^2*3sqrt/3) I'm getting 64 rt 3 cm.

help?
ODOD
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by muna_m » Thu Jun 04, 2009 11:49 am
The formula for area of equilateral triangle = side^2 * sqrt(3)/4

Hope this helps..
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by Osirus@VeritasPrep » Thu Jun 04, 2009 12:04 pm
To solve you add the area of the 6 equilateral triangles and the area of the regular octagon.

Area of equilateral tri = s^2 (srt(3)/4)

Area of regular hexagon = (3 sqrt(3)/2) a^2 where a is the apothem

in this case the apothem would equal four.
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by DanaJ » Fri Jun 05, 2009 12:44 am
Sorry guys... I wrote:

12(4^2*sqrt(3)/3) = 48sqrt(3)

but it should have been

12(4^2*sqrt(3)/4) = 48sqrt(3)

You can easily tell that the first version is not correct, since 12*4^2/3 is NOT 48.

To take it from the top: you've got 12 equilateral triangles, each with sides equal to 4. The area of one triangle will be 4^2sqrt(3)/4, following the well known formula l^2sqrt(3)/4.
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