The outline of a sign for an ice-cream store is made by placing 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 sq rt 3
(B) 3 pi + 6 sq rt 3
(C) 3 pi + 2 sq rt 33
(D) 4 pi + 3 sq rt 3
(E) 3 pi + 6 sq rt 3
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cornell
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smclean23,smclean23 wrote:The outline of a sign for an ice-cream store is made by placing 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 sq rt 3
(B) 3 pi + 6 sq rt 3
(C) 3 pi + 2 sq rt 33
(D) 4 pi + 3 sq rt 3
(E) 3 pi + 6 sq rt 3
I can't find the picture above. It helps if we know the shape of the sign, so we can calculate the perimeter of isosceles triangle
But i assume the circle perimeter as 2(pi)r =2(pi)2 = 4 (pi)
The only answer with 4(pi) is D
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Stuart@KaplanGMAT
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That's a great strategy, but I'm guessing that it's not the right answer; chances are that only part of the circumference makes up the perimiter of the sign.cornell wrote:smclean23,smclean23 wrote:The outline of a sign for an ice-cream store is made by placing 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 sq rt 3
(B) 3 pi + 6 sq rt 3
(C) 3 pi + 2 sq rt 33
(D) 4 pi + 3 sq rt 3
(E) 3 pi + 6 sq rt 3
I can't find the picture above. It helps if we know the shape of the sign, so we can calculate the perimeter of isosceles triangle
But i assume the circle perimeter as 2(pi)r =2(pi)2 = 4 (pi)
The only answer with 4(pi) is D
However, there's no way to answer the question without knowing exactly where the circle is in reference to the triangle (i.e. how much overlap there is between the two shapes - we really need to know the angle measure of the triangle at the centre of the circle).
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lunarpower
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here's the image:
https://s3.supload.com/free/0415092036.jpg/view/
* the part including pi is the easy part: it's 3/4 of the circumference. (they don't even hide this with any sort of tricky rephrase.)
this part is thus (3/4)(2*pi*2), or 3*pi.
therefore, it's either (a), (b), or (c).
at this point, by far the easiest thing to do is estimate.
3√3 is approximately 3*1.7, or 5.1. (alternately, it's √9 * √3, or √27, which is a lil more than 5)
6√3 is approximately 3*1.7, or 10.2. (alternately, it's √36 * √3, or √108, which is a lil more than 10)
2√33 is approximately 2*(about 5.6 or 5.7), or 11.2-11.4. (alternately, it's √4 * √33, or √132, which is right smack between 11 and 12)
the two sides of the triangle included in the perimeter obviously don't add to just 5.1, so (a) is out. then just guess between (b) and (c).
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https://s3.supload.com/free/0415092036.jpg/view/
* the part including pi is the easy part: it's 3/4 of the circumference. (they don't even hide this with any sort of tricky rephrase.)
this part is thus (3/4)(2*pi*2), or 3*pi.
therefore, it's either (a), (b), or (c).
at this point, by far the easiest thing to do is estimate.
3√3 is approximately 3*1.7, or 5.1. (alternately, it's √9 * √3, or √27, which is a lil more than 5)
6√3 is approximately 3*1.7, or 10.2. (alternately, it's √36 * √3, or √108, which is a lil more than 10)
2√33 is approximately 2*(about 5.6 or 5.7), or 11.2-11.4. (alternately, it's √4 * √33, or √132, which is right smack between 11 and 12)
the two sides of the triangle included in the perimeter obviously don't add to just 5.1, so (a) is out. then just guess between (b) and (c).
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lunarpower
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THE "TEXTBOOK SOLUTION"
* draw two radii, one to each of the endpoints of the horizontal segment.
* because of the "3/4 of the perimeter" thing, you've just created a 45-45-90 triangle.
* therefore, the length of the horizontal segment is 2√2.
* therefore, the legs of each of the 2 right triangles (created by splitting the isosceles triangle in half) are √2 and 5.
* using the pythagorean theorem, each hypotenuse is found by
hypotenuse^2 = (√2)^2 + 5^2
hypotenuse^2 = 27
hypotenuse = √27 = 3√3
therefore ans = (b).
by the way, (c) is what you get if you forget that you have to use half of the quantity 2√2 in the pythagorean theorem.
* draw two radii, one to each of the endpoints of the horizontal segment.
* because of the "3/4 of the perimeter" thing, you've just created a 45-45-90 triangle.
* therefore, the length of the horizontal segment is 2√2.
* therefore, the legs of each of the 2 right triangles (created by splitting the isosceles triangle in half) are √2 and 5.
* using the pythagorean theorem, each hypotenuse is found by
hypotenuse^2 = (√2)^2 + 5^2
hypotenuse^2 = 27
hypotenuse = √27 = 3√3
therefore ans = (b).
by the way, (c) is what you get if you forget that you have to use half of the quantity 2√2 in the pythagorean theorem.
Ron has been teaching various standardized tests for 20 years.
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So the circumference of the circle is 3/4 x 2 PI 2 = 4 pi x 3/4 = 3 pi
H = 5 so, 5 = s sqr root 3 / 2
10 = s sqr root 3
10 / sqr root 3 ( is I think estimated 1.5 ) = ( estimated ) on 6
So B???
Please correct me if my calculation is wrong :$
H = 5 so, 5 = s sqr root 3 / 2
10 = s sqr root 3
10 / sqr root 3 ( is I think estimated 1.5 ) = ( estimated ) on 6
So B???
Please correct me if my calculation is wrong :$
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lunarpower
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nope, sorry, wrong.deepoe wrote:So the circumference of the circle is 3/4 x 2 PI 2 = 4 pi x 3/4 = 3 pi
H = 5 so, 5 = s sqr root 3 / 2
10 = s sqr root 3
10 / sqr root 3 ( is I think estimated 1.5 ) = ( estimated ) on 6
So B???
Please correct me if my calculation is wrong :$
the above calculations, involving √3, are only valid for 30°-60°-90° triangles. this isn't a 30°-60°-90° triangle, so you can't use the 1-√3-2.
remember not to get the "30°-60°-90° disease". in other words, DO NOT assume that a triangle is 30°-60°-90° just because it doesn't look isosceles.
most right triangles are NOT any sort of "special" triangle - i.e., not 30°-60°-90°, not 45°-45°-90°, not 3-4-5, nor anything else special. they're just random triangles for which you have to use the pythagorean theorem - like the triangle in this problem.
on the other hand, the triangle that you create by drawing the two aforementioned radii is special (it's a 45°-45°-90° triangle).[/list]
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lunarpower
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that's the standard ratio in the 45-45-90 triangle: the hypotenuse is exactly √2 times as long as each leg.deepoe wrote:Ok thanks!
But you chose √2 because radius were 2?
since both of the radii are the same, you could also figure this out using the pythagorean theorem: 2^2 + 2^2 = (horizontal segment)^2. you'll get the same result.
Ron has been teaching various standardized tests for 20 years.
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Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
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Voit esittää kysymyksiä Ron:lle myös suomeksi
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Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
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