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

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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?

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.

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

thanks

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

Attachment:

untitled.PNG [ 6.49 KiB | Viewed 14008 times ]

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: \(P=3\pi+3\sqrt{3}+3\sqrt{3}=3\pi+6\sqrt{3}\).

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

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

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20 Jun 2013, 23:50

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I Will give it a try. I supposed everyone got 3pi part. Next its isosceles triangle. Meaning both sides if added should be even. So among options of 6,3 and 2 u can blindly choose 6 since 2 is absurd here. Hope it helps.

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

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09 Oct 2013, 05:26

Hi ,

I found the answer like this the cirumfrence length will be 3pi (3/4*2pi*2) now the total perimeter will 3pi+2*equal side of the isoleces triangle so our answer will be in the form of 3pi+2*equal side of the isoleces which is only B (3pi+2*(3sqrt3))

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

thanks

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

Attachment:

untitled.PNG

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.[/quot

hey bunuel, as per your attached figure CANT we take the height inside the triangle to be a perpendicular? And as per 30 - 60 -90 theorem we see bc equals 10/sq rt 3, which we rationalize and get side bc as 10*sq rt 3. since this is an equilateral triangle we can get perimiter of 2 sides= (20*sq rt 3)/3=approx 6*sq rt 3..

also please explain how you took BD as half of AB.

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

thanks

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

Attachment:

untitled.PNG

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.[/quot

hey bunuel, as per your attached figure CANT we take the height inside the triangle to be a perpendicular? And as per 30 - 60 -90 theorem we see bc equals 10/sq rt 3, which we rationalize and get side bc as 10*sq rt 3. since this is an equilateral triangle we can get perimiter of 2 sides= (20*sq rt 3)/3=approx 6*sq rt 3..

also please explain how you took BD as half of AB.

Triangle ABC is isosceles, thus its height to the base and the median to the base are the same, thus D is the midpoint of AB.

I cannot understand the rest of your post.
_________________

I tried extending the sides of the triangle to make a larger one with the base being the diameter of the triangle. After that, the given triangle and the new one are similar, and the height is 5 and 7, the base is x and 4, and solved from there using the Pythagorean theory (height (5)^2 + x/2^2) = side of triangle. but I got the wrong answer. Can someone look it over and say if I have a mistake in calculation or a wrong hypothesis?

I tried extending the sides of the triangle to make a larger one with the base being the diameter of the triangle. After that, the given triangle and the new one are similar, and the height is 5 and 7, the base is x and 4, and solved from there using the Pythagorean theory (height (5)^2 + x/2^2) = side of triangle. but I got the wrong answer. Can someone look it over and say if I have a mistake in calculation or a wrong hypothesis?

Not clear what you've done there: "the diameter of the triangle" does not make sense, not clear which is "given triangle" and "new one", why they are similar, etc.

Please attach a diagram and write your reasoning so that it's clear.
_________________

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

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20 Dec 2014, 03:27

Bunuel wrote:

hirendhanak wrote:

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

thanks

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

Attachment:

untitled.PNG

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.

HI Bunuel,

I have one doubt here. why are we not adding this length also in calculating parameter AB=hypotenuse=2\sqrt{2}.

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

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21 Dec 2014, 12:09

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?

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

thanks

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

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 [b]big arc AB + AC + BC: \(P=3\pi+3\sqrt{3}+3\sqrt{3}=3\pi+6\sqrt{3}\).

Answer: B.

HI Bunuel,

I have one doubt here. why are we not adding this length also in calculating parameter AB=hypotenuse=2\sqrt{2}.

Could you please provide u r comments on this.

Thanks

Because AB is not a part of the perimeter of the sign. Perimeter is the distance around a figure, the length of the boundary.
_________________

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

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23 Dec 2014, 10:44

Hi, I used another method to find the result but it gives me an approximative result that abviously leads to the B answer:

1- for the circle, the result is: 3pi. 2- for the perimeter of the triangle: the triangle is an isosceles one, the heigh divides it into two 30° 60° 90° triangles. So the length of the sides of each one is: x (short side), x\sqrt{3} ; 2x (hypotenuse).

so when I calculate everything I find: 3pi+(20\sqrt{3}/3). 20/3 yields to 6 (approximately).

Why I don't find the exact result when I use this method?! any answers please Thanks!

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

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25 Dec 2014, 10:13

Bunuel wrote:

hirendhanak wrote:

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

thanks

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

Attachment:

untitled.PNG

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.

Hi,

I was wondering, did we need to use the angle values etc? Couldn't we just draw the radius from the (assumed) center of the circle to the base of the triangle and then use the pythagorean to find the lengths of the sides?

Tha basis of the trianle is anyway in the circle, and the ends of it are touching the circumference of the circle. We can then draw the 2 radii, to the ends of the base of the trianegle and just use the pythagorean to calculate the base. And from there, move on to the rest of the sides. Right...?

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

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04 Mar 2015, 03:51

Bunuel wrote:

hirendhanak wrote:

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

thanks

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

Attachment:

untitled.PNG

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}=

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