OG Geometry - Find the external angles of the 5 pointed star - The Beat The GMAT Forum

OG Geometry - Find the external angles of the 5 pointed star

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Lasve: Senior | Next Rank: 100 Posts; Posts: 30; Joined: Tue Dec 20, 2011 2:11 pmFollowed by: 1 member; Thanked: 4 times

by Lasve » Sun Jan 08, 2012 6:37 am

Hi Guys I'm writing about a tricky question that took me quite a while to solve.

I wanted to ask your opinion about the solve method given by the OG, because I find it quite confusing and time consuming.

I think that a faster and easyer way to solve it would be:
i. Realize the center is a pentagon
ii. That means that the total internal angles are 180*(n-2) = 180 * (5-2) = 540
iii. That means that the Average internal angle is 108
iv. Realize that each internal angle is part of a 180-degrees Straight angle, That means that the complementary one (the Base of the triangle) is 180-108= 72
v. Since every triangle is 180 degree, the external angle must be 180-(72*2) = 36
vi. Multiply 36 for the number of external angles: 36*5=180

What do you think?
Did I get something wrong and solve it just for sheer luck or is it ok?
Thanks![/img]

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LalaB: Master | Next Rank: 500 Posts; Posts: 425; Joined: Wed Dec 08, 2010 9:00 amFollowed by: 7 members; Thanked: 56 times; GMAT Score:690

by LalaB » Sun Jan 08, 2012 11:15 am

hm, have doubts about step III (u divide 540 by 5 ,assuming that all angles are equal. r they?)

my approach-

a+b+c+d+e=180(5-2)=540
remember-the measure of an exterior angle equals the sum of its two non-adjacent interior angles

lets assume that the triangle with the angle z has also angles m (near c) and p(near b)
then c=p+z b=m+z c+b=p+m+2z=180+z
same with the rest angles of the polygon abcde.

now pay attention to the fact that the angles of the polygon counted twice (so we need to divide the result by 2). having this in the mind, we got the following-

(180+z+180+v+180+w+180+x+180+y)/2=540

5*180+(z+w+v+x+y)=540*2
z+w+v+x+y=180

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rijul007: Legendary Member; Posts: 588; Joined: Sun Oct 16, 2011 9:42 amFollowed by: 9 members; Location: New Delhi, India; Thanked: 130 times; GMAT Score:720

by rijul007 » Sun Jan 08, 2012 11:40 am

Lasve wrote:Hi Guys I'm writing about a tricky question that took me quite a while to solve.

I wanted to ask your opinion about the solve method given by the OG, because I find it quite confusing and time consuming.

I think that a faster and easyer way to solve it would be:
i. Realize the center is a pentagon
ii. That means that the total internal angles are 180*(n-2) = 180 * (5-2) = 540
iii. That means that the Average internal angle is 108
iv. Realize that each internal angle is part of a 180-degrees Straight angle, That means that the complementary one (the Base of the triangle) is 180-108= 72
v. Since every triangle is 180 degree, the external angle must be 180-(72*2) = 36
vi. Multiply 36 for the number of external angles: 36*5=180

What do you think?
Did I get something wrong and solve it just for sheer luck or is it ok?
Thanks![/img]

This approach would have been correct if the pentagon was regualar, but we dont know that.
You got the correct answer because you were asked the sum of all angles v,w,x,y,z.
If the ques would have been, find x+y+z.
you would have got 36*3 = 108, which is wrong.
the data would be insufficient fot that summation.

check ou the figure

a = v+m
e = v+n
a+e = v+m+n+v = 180+v
similarly similarly you can get expressions for other pairs of adjacent angles
a+b+b+c+c+d+d+e+e+a = 5(180) +v+w+x+y+z
2(a+b+c+d+e) = 900 + v+w+x+y+z
2(540) = 900 + v+w+x+y+z
1080-900 = v+w+x+y+z
v+w+x+y+z = 180

Option C

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Lasve: Senior | Next Rank: 100 Posts; Posts: 30; Joined: Tue Dec 20, 2011 2:11 pmFollowed by: 1 member; Thanked: 4 times

by Lasve » Mon Jan 09, 2012 4:32 am

Thank's for the answers!

Anyway, It is obvious that this method would not have worked if I had been asked to find only 3 of the sides!
But in that case it would have been impossible to determine them with no further information.

Moreover I treat the penthagon as regular because, since I'm asked to find out the sum of all angles, lesser and larger angles would compensate each other.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon Jan 09, 2012 5:07 am

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ArunangsuSahu: Master | Next Rank: 500 Posts; Posts: 382; Joined: Thu Mar 31, 2011 5:47 pm; Thanked: 15 times

by ArunangsuSahu » Mon Jan 09, 2012 1:34 pm

180 degree

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Lasve: Senior | Next Rank: 100 Posts; Posts: 30; Joined: Tue Dec 20, 2011 2:11 pmFollowed by: 1 member; Thanked: 4 times

by Lasve » Thu Jan 12, 2012 3:44 am

Thank's Guru, I did exactly as in the thread you post!
(althought I took me more than 10 minutes to do it!)

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