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The outline of a sign for an ice-cream store is made by

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Director
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The outline of a sign for an ice-cream store is made by  [#permalink]

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New post 13 Aug 2008, 05:18
The outline of a sign for an ice-cream store is made by placing 3/4 of the circumference of a circle with radius 2 feet on top of an isosceles triangle with height 5 feet, as shown above. What is the perimeter, in feet, of the sign?
(A) 3(pie) + 3(sqrt3)
(B) 3(pie) + 6(sqrt3)
(C) 3(pie) + 2(sqrt33)
(D) 4(pie) + 3(sqrt3)
(E) 4(pie) + 6(sqrt3)

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New post 13 Aug 2008, 05:36
1
B

Heres how

perimeter of the circular part
3/4* circumference
3/4 * 2 *pi * 2 = 3 * pi

calculate the base of the isosceles triangle, connect the base of the triangle to centre of circle to form triangle.

the arc = 1/4th of circumference
so the central angle is 90 => right angle triangle
two sides of the triangle = radius of circle = 2 each

so base of isosceles triangle sqrt(8)

height of external isoceles triangle= 5

length of the side of external traingle a: a2 = 5^2 + [sqrt(8)/2]^2 = 25 + 2 = 27 i.e length of side = 3* sqrt(3)

perimeter = 3* pi + 2* 3 * sqrt(3)
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New post 13 Aug 2008, 06:02
Exactly the same way I got to the answer, with only 1 difference.

Instead of using \(\sqrt{8}\) I used \(2\sqrt{2}\) and then half of that was \(\sqrt{2}\) to use for finding the length of the hypotnuse.
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New post 13 Aug 2008, 06:06
Thanks .. never thought about the 45/45/90 traingle inside the circle. Thanks guys!
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New post 13 Aug 2008, 06:12
Almost all geometry questions on GMAT involving triangles will involve some form of special triangle if you have to calculate the lengths of the sides or hypotnuse.

rao_1857 wrote:
Thanks .. never thought about the 45/45/90 traingle inside the circle. Thanks guys!

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New post 13 Aug 2008, 06:24
alpha_plus_gamma wrote:
B

Heres how

perimeter of the circular part
3/4* circumference
3/4 * 2 *pi * 2 = 3 * pi

calculate the base of the isosceles triangle, connect the base of the triangle to centre of circle to form triangle.

the arc = 1/4th of circumference
so the central angle is 90 => right angle triangle
two sides of the triangle = radius of circle = 2 each

so base of isosceles triangle sqrt(8)

height of external isoceles triangle= 5

length of the side of external traingle a: a2 = 5^2 + [sqrt(8)/2]^2 = 25 + 2 = 27 i.e length of side = 3* sqrt(3)

perimeter = 3* pi + 2* 3 * sqrt(3)



How did you get the base of isosceles triangle sqrt(8) ?
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New post 13 Aug 2008, 06:48
1
We're told that "3/4 of a circle is placed on a triangle". This means that 270 degrees of the 360 is represented by the ice cream part of the sign. 90 degrees (1/4th of the 360) represents the measurement of the interior angle that gives us the base of the traingle where it connects to the circle. Becuase this is 90 degrees and we know the distance between the end points of the base to the center of the circule is the radius and the length is equal, it creates a 45:45:90 triangle, or a \(1:1:\sqrt{2}\)

If the radius is 2 and we have to two of them, it's 2^2 + 2^2 = base of triange ^2

4 + 4 = base^2 so base is \(\sqrt{8}\)
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Re: tough one geometery  [#permalink]

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New post 13 Aug 2008, 06:59
1
sarzan wrote:
alpha_plus_gamma wrote:
B

Heres how

perimeter of the circular part
3/4* circumference
3/4 * 2 *pi * 2 = 3 * pi

calculate the base of the isosceles triangle, connect the base of the triangle to centre of circle to form triangle.

the arc = 1/4th of circumference
so the central angle is 90 => right angle triangle
two sides of the triangle = radius of circle = 2 each

so base of isosceles triangle sqrt(8)

height of external isoceles triangle= 5

length of the side of external traingle a: a2 = 5^2 + [sqrt(8)/2]^2 = 25 + 2 = 27 i.e length of side = 3* sqrt(3)

perimeter = 3* pi + 2* 3 * sqrt(3)



How did you get the base of isosceles triangle sqrt(8) ?
Attachment:
geometry.JPG
geometry.JPG [ 11.12 KiB | Viewed 916 times ]


I am bad at drawing, but the attached image may help you.

Arc AB = 1/4 ( since remaining part is 3/4 of circumference)

therefore the central angle is 1/4 of 360 = 90 degrees. making OAB right angle triangle.
sides of the triangle are radii of circle = 2
and the hypotenuse = \(sqrt (2^2 + 2^2)\)
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Re: tough one geometery  [#permalink]

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New post 13 Aug 2008, 08:57
rao_1857 wrote:
The outline of a sign for an ice-cream store is made by placing 3/4 of the circumference of a circle with radius 2 feet on top of an isosceles triangle with height 5 feet, as shown above. What is the perimeter, in feet, of the sign?
(A) 3(pie) + 3(sqrt3)
(B) 3(pie) + 6(sqrt3)
(C) 3(pie) + 2(sqrt33)
(D) 4(pie) + 3(sqrt3)
(E) 4(pie) + 6(sqrt3)



good question
agree with B.
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Re: tough one geometery  [#permalink]

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New post 13 Aug 2008, 10:02
alpha_plus_gamma wrote:
sarzan wrote:
alpha_plus_gamma wrote:
B

I am bad at drawing, but the attached image may help you.

Arc AB = 1/4 ( since remaining part is 3/4 of circumference)

therefore the central angle is 1/4 of 360 = 90 degrees. making OAB right angle triangle.
sides of the triangle are radii of circle = 2
and the hypotenuse = \(sqrt (2^2 + 2^2)\)



Thanks for the drawing. Really made things clear. KUDOS !

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Re: tough one geometery &nbs [#permalink] 13 Aug 2008, 10:02
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