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gluon
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PS: Set 12, Q8 - Geometry 28 Sep 2007, 08:20
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You guys will love this one. I needed 8 minutes to solve it. I would like to find out if there is an easier way. The general steps I followed:

1> Find length of one side of the equilateral triangle using its area.
2> Draw diagonals from the center of the circle to each vertex of triangle. Then find the 'height' of each of one of the sub-triangles.
3> Find the length of one side of the sub-triangle. This gives you the radius of the circle.



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PS: Set 12, Q8 - Geometry 28 Sep 2007, 08:39
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area of eq. triangle is sqrt(3)/4 *s^2 where S is one of the sides..

if eq. triangle inscb inside a cricle then radius of circle = s/sqrt(3)

area =9sqrt(3)

s^2=9*4 i.e s=6

area of circle = pi (r)^2

r=6/sqrt(3) r^2=36/3 =12

area of circle = 12pi
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Bluebird
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PS: Set 12, Q8 - Geometry 28 Sep 2007, 09:24
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fresinha12, you set the radius = 6/sqrt3 - where did you get the sqrt3?

Also, I thought I could take half the height of the triangle and set it equal to the radius why does this not work?

Thanks!
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gluon
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PS: Set 12, Q8 - Geometry 28 Sep 2007, 09:42
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fresinha12
area of eq. triangle is sqrt(3)/4 *s^2 where S is one of the sides..

if eq. triangle inscb inside a cricle then radius of circle = s/sqrt(3)

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Jeez... does GMAT need anything special other than learning formulae? I mean seriously... everytime I need more than 3 - 4 minutes to do any problem there is a formula somewhere that I dont know.

What would I do without you guys?
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Ferihere
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PS: Set 12, Q8 - Geometry 28 Sep 2007, 09:43
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Bluebird
fresinha12, you set the radius = 6/sqrt3 - where did you get the sqrt3?

Also, I thought I could take half the height of the triangle and set it equal to the radius why does this not work?

Thanks!
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There is a such kind of formula,

regarding second question, you should take 2/3 of height not 1/2. It is better if you draw yourself and see then you will memorize it forever.

Ans: C
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gluon
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PS: Set 12, Q8 - Geometry 28 Sep 2007, 09:44
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Bluebird
fresinha12, you set the radius = 6/sqrt3 - where did you get the sqrt3?

Also, I thought I could take half the height of the triangle and set it equal to the radius why does this not work?

Thanks!
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Bluebird, radius = 6/sqrt(3) is just the formula. Use it and save yourself 4 minutes.

Also, height of the equilateral triangle is NOT the diameter. So the half of the height is not going to be the radius. Just draw a picture of it and see.
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GK_Gmat
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PS: Set 12, Q8 - Geometry 28 Sep 2007, 10:29
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Ferihere
Bluebird
fresinha12, you set the radius = 6/sqrt3 - where did you get the sqrt3?

Also, I thought I could take half the height of the triangle and set it equal to the radius why does this not work?

Thanks!

There is a such kind of formula,

regarding second question, you should take 2/3 of height not 1/2. It is better if you draw yourself and see then you will memorize it forever.

Ans: C
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Ferihere,
Can you pls. explain the logic behind taking 2/3rd of the height of the eq. triangle in order to get the r of the circle? How'd you figure that out? Thanks.
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Ravshonbek
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PS: Set 12, Q8 - Geometry 28 Sep 2007, 11:08
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fresinha12 has a nice approach, but i hope gmat doesnt throw ps questions that require special formulae.
even this question is simply solved by a certain procedure which only takes max 2 mins. or less. i have attached it.

Drawing took me ages!
Attachments

circletriangel.JPG
circletriangel.JPG [ 16.73 KiB | Viewed 1361 times ]

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KillerSquirrel
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PS: Set 12, Q8 - Geometry 28 Sep 2007, 11:34
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Ravshonbek
fresinha12 has a nice approach, but i hope gmat doesnt throw ps questions that require special formulae.
even this question is simply solved by a certain procedure which only takes max 2 mins. or less. i have attached it.

Drawing took me ages!
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nice !

part 1

x:x*sqrt(3):2*x

height = x*sqrt(3)

area of triangle = 9*sqrt(3)

(x*x*sqrt(3))/2 = 9*sqrt(3)

x^2 = 9

x = 3

part 2

another triangle 30:60:90 where 3 = y*sqrt(3) and y:y*sqrt(3):2*y

3 = y*sqrt(3)

y = sqrt(3)

r = 3*sqrt(3) - sqrt(3) = 2*sqrt(3) ---> see part 1 for 3*sqrt(3) and part 2 for sqrt(3).

after you have r = 2*sqrt(3)

pi*2*sqrt(3)^2 = pi*4*3 = 12*pi

no special formula is needed.

the answer is (C)

Well done Ravshonbek - Excellent work.

:)
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b14kumar
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PS: Set 12, Q8 - Geometry 28 Sep 2007, 16:24
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Ravshonbek
fresinha12 has a nice approach, but i hope gmat doesnt throw ps questions that require special formulae.
even this question is simply solved by a certain procedure which only takes max 2 mins. or less. i have attached it.

Drawing took me ages!
Show more


Good work and a Good question....

Well, this problem can also be solved by using two formulae of Area of Triangle:

Area of a Triangle = 1/2 (Base * Height)

Area of a Triangle = Sqrt( s (s-a) (s-b) (s-c) ) --------- (1)

Where s = [(a+b+c)/2] , a,b and c are the sides of the triangle.

For equilateral Triangle a = b = c
Let's x = a = b = c then just put the values in Eq (1) and you will get following formula:

Area of an Equilateral Triange = [sqrt(3)/4] * x^2 where x is the length of the side.

Using above formula, you will get the length of the side.
=> x = 6 = Base.

Now use formula:

Area of a Triangle = 1/2 (Base * Height)
=> 1/2 (Base * Height) = 9sqrt(3)
=> Height = 3 sqrt(3)

We can use above values to get the Radius of the Circle and then we can find the Area of the circle.

Area of the circle , thus, comes = 12*pi

- Brajesh
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FN
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PS: Set 12, Q8 - Geometry 28 Sep 2007, 20:41
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word to the wise...first practice and understand the concepts then learn the formulae to save time..

it is imperitive that you know how to solve a problem without a formula..then only use a formula if it can save you 1-2 mins..but you first need to realize that a question will take you more than 2 mins to solve..once you have got an idea how to solve the problem then by all means use a formula to save time.
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PS: Set 12, Q8 - Geometry 04 Oct 2007, 14:54
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Ravshonbek
fresinha12 has a nice approach, but i hope gmat doesnt throw ps questions that require special formulae.
even this question is simply solved by a certain procedure which only takes max 2 mins. or less. i have attached it.

Drawing took me ages!
Show more


thanks. this problem took me forever and i got stuck on step 20 lol



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