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The outline of a sign for an ice-cream store is made by placing 3/4

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Bunuel
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The outline of a sign for an ice-cream store is made by placing 3/4 [#permalink] New post   21 Jan 2021, 07:00
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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) \(3\pi + 3\sqrt{3}\)

(B) \(3\pi + 6\sqrt{3}\)

(C) \(3\pi + 2\sqrt{33}\)

(D) \(4\pi + 3\sqrt{3}\)

(E) \(4\pi + 6\sqrt{3}\)

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Bunuel
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Re: The outline of a sign for an ice-cream store is made by placing 3/4 [#permalink] New post   03 Feb 2021, 09:30
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Bunuel 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) \(3\pi + 3\sqrt{3}\)

(B) \(3\pi + 6\sqrt{3}\)

(C) \(3\pi + 2\sqrt{33}\)

(D) \(4\pi + 3\sqrt{3}\)

(E) \(4\pi + 6\sqrt{3}\)

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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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The outline of a sign for an ice-cream store is made by placing 3/4 [#permalink] New post   Updated on: 25 Jan 2021, 05:31
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Bunuel 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) \(3\pi + 3\sqrt{3}\)

(B) \(3\pi + 6\sqrt{3}\)

(C) \(3\pi + 2\sqrt{33}\)

(D) \(4\pi + 3\sqrt{3}\)

(E) \(4\pi + 6\sqrt{3}\)

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The length of the arc for the perimeter = (3/4) * 2 \(\pi\) * 2 = \(3\pi\)

The central angle for the chord = 1/4 * 360 = 90 degrees

The Triangle formed is an isosceles right triangle with sides = 2. The hypotenuse, which is the base of the triangle. By Pythagoras's theorem, this is \(\sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}\)

half the base = \(\sqrt{2}\) and the height is 5. Again by PT, the side = \(\sqrt{(\sqrt{2})^2 + 5^2} = \sqrt{2 + 25} = \sqrt{27} = 3\sqrt{3}\)

The perimeter = \(3\pi + 2*3\sqrt{3} = 3\pi + 6\sqrt{3} \)


Option B

Arun Kumar
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Originally posted by CrackverbalGMAT on 22 Jan 2021, 03:48.
Last edited by CrackverbalGMAT on 25 Jan 2021, 05:31, edited 1 time in total.
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Re: The outline of a sign for an ice-cream store is made by placing 3/4 [#permalink] New post   23 Jan 2021, 06:10
Shouldn't it be B ? in the solution above the second side of the triangle was not added.

We need to calculate the complete perimeter, so the Circle part will have 3pi and 3 root 3 will be have to be multiplied by 2 which gives = 3pi + 6 root 3
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Re: The outline of a sign for an ice-cream store is made by placing 3/4 [#permalink] New post   26 Sep 2022, 05:11
I have a question from the solutions provided. Do we not need to assume that AC=BC here, for the triangle to be isosceles? Can we simply assume without it being mentioned in the question stem?

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Re: The outline of a sign for an ice-cream store is made by placing 3/4 [#permalink] New post   18 Oct 2023, 10:13
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Re: The outline of a sign for an ice-cream store is made by placing 3/4 [#permalink]
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