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

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

05 Apr 2008, 22:17

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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) 3pi + 3 sqrt 3
(B) 3pi + 6 sqrt 3
(C) 3pi + 3 sqrt 33
(D) 4pi + 3 sqrt 3
(E) 4pi + 6 sqrt 3

[Reveal] Spoiler: OA

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Jimmy Low, Frankfurt, Germany
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Re: 1000PS: Section 7 Question 18 [#permalink]

05 Apr 2008, 23:11

jimmylow wrote:

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) 3pi + 3 sqrt 3
(B) 3pi + 6 sqrt 3
(C) 3pi + 3 sqrt 33
(D) 4pi + 3 sqrt 3
(E) 4pi + 6 sqrt 3

The circumference of the ice cream part is 0.75 * 2 * pi *r = 3pi.

Now the triangle portion. While there might be a way to geometrically calculate this, we can easily arrive at the answer by backsolving. We know that the triangle is an isosceles triangle. This means that, as its drawn, the two sides that would be enclosed by the perimeter are equal.

We also know that the height is 5ft and the base is less than 4 ft. We also have only 3 choices, A B and C. We can eliminate A because the sum of the 2 sides are definetely > 10ft. C cannot be because its way too much. Leaving B as the only answer.

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Re: 1000PS: Section 7 Question 18 [#permalink]

23 Nov 2010, 00:54

can someone explain this problem in a better way... Bunuel it would be great if you can take the lead.

thanks
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Re: 1000PS: Section 7 Question 18 [#permalink]

23 Nov 2010, 03:52

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hirendhanak wrote:

can someone explain this problem in a better way... Bunuel it would be great if you can take the lead.

thanks

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As 3/4 of the circumference of a circle is placed on top of a triangle then the perimeter of big arc AB will be 3/4 of a circumference so \frac{3}{4}*2\pi{r}=3\pi, so the answer is either A, B, or C.

Next, as big arc is 3/4 of a circumference then small arc is 1/4 of a circumference or 1/4*360=90 degrees, so \angle{AOB}=90. Now, as AO=BO=r=2 then AB=hypotenuse=2\sqrt{2} --> BD=\frac{AB}{2}=\sqrt{2} --> AC=BC=\sqrt{BD^2+DC^2}=\sqrt{2+25}=3\sqrt{3}.

The whole perimeter equals to big arc AB + AC + BC: P=3\pi+3\sqrt{3}+3\sqrt{3}=3\pi+6\sqrt{3}.

Answer: B.
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Re: 1000PS: Section 7 Question 18 [#permalink]

23 Nov 2010, 05:19

Nice explaination Bunuel...

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Re: geometry [#permalink]

08 Jul 2011, 22:56

ssarkar 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) 3pi +3 sq root 3
(B) 3pi +6 sq root 3
(C) 3pi +2 sq root 33
(D) 4pi +3 sq root 3
(E) 4pi +6 sq root 3

Its B .

3/4 * 2pi*2 = 3pi .........................1

now is the tricky part.
let the centre of circle be O. draw two line OA and OB.
the bigger arc is 3/4( 3/4*360) , hence the smaller will 1/4 = 90
hence traingle OAB is a right angled triangle,
with OA = OB = 2= radius

hence AB= 2*sqrt 2
now we know that D is the mid point of A and B
AC = BC
use pytho. theorem
sqrt( BD^2 + DC^2) = sqrt( 2+25) = 3 sqrt 3

now perimeter of triangle = 2*3sqrt 3 = 6sqrt 3............................2
total perimeter
1+2

3pi + 6 sqrt 3

hence B

PS: Apologies for the awkward diagram .. at office

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

08 Feb 2012, 06:37

Merging similar topics.
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Re: The outline of a sign for an ice-cream store is made by [#permalink] 08 Feb 2012, 06:37

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