I'm terrible at geometry!

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480ocean: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Mon Oct 27, 2008 1:01 pm; Thanked: 1 times

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: Beat The GMAT — Problem Solving | Sat Jan 24, 2009 11:00 am

DanaJ: Site Admin; Posts: 2567; Joined: Thu Jan 01, 2009 10:05 amFollowed by: 550 members; Thanked: 712 times; GMAT Score:770

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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480ocean: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Mon Oct 27, 2008 1:01 pm; Thanked: 1 times

by 480ocean » Sat Jan 24, 2009 11:41 am

Thanks so much DanaJ! I totally missed the hexagon!

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gaggleofgirls: Master | Next Rank: 500 Posts; Posts: 138; Joined: Thu Jan 15, 2009 7:52 am; Location: Steamboat Springs, CO; Thanked: 15 times

by gaggleofgirls » Sat Jan 24, 2009 12:23 pm

I think I have an easier way ( being raised Jewish helped here,

)

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 208 times

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480ocean: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Mon Oct 27, 2008 1:01 pm; Thanked: 1 times

by 480ocean » Sat Jan 24, 2009 12:58 pm

Yes, that helps too! Thank you!

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kanha81: Master | Next Rank: 500 Posts; Posts: 431; Joined: Sat Jan 10, 2009 9:32 amFollowed by: 1 member; Thanked: 16 times

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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odod: Senior | Next Rank: 100 Posts; Posts: 56; Joined: Tue Jan 13, 2009 12:59 pm; Location: Toronto; Thanked: 1 times

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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odod: Senior | Next Rank: 100 Posts; Posts: 56; Joined: Tue Jan 13, 2009 12:59 pm; Location: Toronto; Thanked: 1 times

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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muna_m: Junior | Next Rank: 30 Posts; Posts: 22; Joined: Fri Mar 06, 2009 3:36 pm

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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Osirus@VeritasPrep: GMAT Instructor; Posts: 1578; Joined: Thu May 28, 2009 8:02 amFollowed by: 34 members; Thanked: 128 times; GMAT Score:760

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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DanaJ: Site Admin; Posts: 2567; Joined: Thu Jan 01, 2009 10:05 amFollowed by: 550 members; Thanked: 712 times; GMAT Score:770

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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