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Senior Manager
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ice cream cone [#permalink] New post 16 May 2007, 13:51
try this
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Re: ice cream cone [#permalink] New post 16 May 2007, 17:20
Sergey_is_cool wrote:
try this


(B) it is

C = Cc + Ct

Cc = 3/4(2 * pi * 2) = 3*pi
Cc = 2AB (AB is the side's length of the triangle)

AB^2 = (1/2BC)^2 + AH^2 = (1/2 * 2sqrt2) + 5^2 = 27
=> AB = 3sqrt3

=> C = 3*pi + 6sqrt3
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Re: ice cream cone [#permalink] New post 16 May 2007, 21:08
kirakira wrote:
Sergey_is_cool wrote:
try this


(B) it is

C = Cc + Ct

Cc = 3/4(2 * pi * 2) = 3*pi
Cc = 2AB (AB is the side's length of the triangle)

AB^2 = (1/2BC)^2 + AH^2 = (1/2 * 2sqrt2) + 5^2 = 27
=> AB = 3sqrt3

=> C = 3*pi + 6sqrt3


I do not understand this. Could you explain this in itsy bitsy bits pls? :?
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Re: ice cream cone [#permalink] New post 17 May 2007, 06:19
kirakira wrote:
Sergey_is_cool wrote:
try this


(B) it is

C = Cc + Ct

Cc = 3/4(2 * pi * 2) = 3*pi
Cc = 2AB (AB is the side's length of the triangle)

AB^2 = (1/2BC)^2 + AH^2 = (1/2 * 2sqrt2) + 5^2 = 27
=> AB = 3sqrt3

=> C = 3*pi + 6sqrt3


hey, i'm kinda week in geometry. Can you explain how you got 2sqrt2 ?
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Re: ice cream cone [#permalink] New post 17 May 2007, 07:33
Sergey_is_cool wrote:
kirakira wrote:
Sergey_is_cool wrote:
try this


(B) it is

C = Cc + Ct

Cc = 3/4(2 * pi * 2) = 3*pi
Cc = 2AB (AB is the side's length of the triangle)

AB^2 = (1/2BC)^2 + AH^2 = (1/2 * 2sqrt2) + 5^2 = 27
=> AB = 3sqrt3

=> C = 3*pi + 6sqrt3


hey, i'm kinda week in geometry. Can you explain how you got 2sqrt2 ?


Triangle ABC with heigth AH, the circle with the center O.

AB^2 = (1/2BC)^2 + AH^2

The triangle BCO is a right triangle (because chord BC has the length of 1/4 circumference of the circle)

=> BC^2 = BO^2 + CO^2 = 8
=> BC = 2sqrt2
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 [#permalink] New post 19 May 2007, 02:22
the key for solving the above problem is knowing how to find the perimeter of a square inscribed inside a circle!

part one

finding the circal circumference

circal r = 2

circumference whole circle 2*pi*r = 4*pi

finding 3/4 of circumference = 3/4*4*pi = 3*pi

part two

draw a square inscribed inside a circle (look at the attachment)

the diagonal of the square is r=2*2=4 (diameter of circle).

the square forms two 45-45-90 triangles.

sqrt(2)*x=4

x=4/sqrt(2)

so the ratio is 1:1:sqrt(2) = 4/sqrt(2):4/sqrt(2):4

meaning the line in the attachment marked (1) equal 4/sqrt(2) and half the line is [4/sqrt(2)]/2

now:

from the triangle with height 5 we can make two smaller right triangles with base [4/sqrt(2)]/2.

using the Pythagorean theorem will yield [[4/sqrt(2)]/2]^2+5^2=x^2

2+25 = x^2

x = sqrt(27)

so we can conclude that the sign circumference is pi*3+2*sqrt(27) or in other words pi*3+3*2sqrt(3) = pi*3+6sqrt(3)

and as was said before the answer is then (B)

:-D
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 [#permalink] New post 14 Sep 2007, 21:40
where did you get the idea to inscribe the square within the 3/4 circle? this is partly a rhetorical question.
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Re: ice cream cone [#permalink] New post 15 Sep 2007, 07:07
Sergey_is_cool wrote:
try this


The main hint here is that the circumference is 3/4 the rest 1/4 should be considered as 90 degree angle of the triangle. This will easen work a lot.

Ans: B
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 [#permalink] New post 16 Sep 2007, 22:44
Anyone trying POE here...

1st part is easy... 3pi. I eliminated 4 and 5.

The second part needs to be greater than 10(2*height). eliminate A

So, u r left with B and C. I chose B before working it out the right way.
  [#permalink] 16 Sep 2007, 22:44
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