BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Ice Cream

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 172
Joined: Mon Jun 23, 2008 12:00 pm
Thanked: 3 times
Followed by:1 members

Ice Cream

by smclean23 » Fri Jul 18, 2008 5:24 pm
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
Top
Source: — Problem Solving |
Junior | Next Rank: 30 Posts
Posts: 16
Joined: Thu Jul 03, 2008 5:58 pm
Location: Indonesia
Thanked: 2 times

Re: Ice Cream

by cornell » Fri Jul 18, 2008 5:53 pm
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
smclean23,
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
Life is about choices
Top
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

Re: Ice Cream

by Stuart@KaplanGMAT » Fri Jul 18, 2008 8:50 pm
cornell wrote:
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
smclean23,
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
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.

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).
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3380
Joined: Mon Mar 03, 2008 1:20 am
Thanked: 2256 times
Followed by:1535 members
GMAT Score:800

by lunarpower » Wed Apr 15, 2009 10:15 pm
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).

--
Ron has been teaching various standardized tests for 20 years.

--

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi

--

Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.

Yves Saint-Laurent

--

Learn more about ron
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3380
Joined: Mon Mar 03, 2008 1:20 am
Thanked: 2256 times
Followed by:1535 members
GMAT Score:800

by lunarpower » Wed Apr 15, 2009 10:15 pm
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.
Ron has been teaching various standardized tests for 20 years.

--

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi

--

Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.

Yves Saint-Laurent

--

Learn more about ron
Top
Master | Next Rank: 500 Posts
Posts: 151
Joined: Wed Feb 11, 2009 2:45 am
Thanked: 3 times

by deepoe » Thu Apr 16, 2009 4:15 am
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 :$
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3380
Joined: Mon Mar 03, 2008 1:20 am
Thanked: 2256 times
Followed by:1535 members
GMAT Score:800

by lunarpower » Fri Apr 17, 2009 12:32 am
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 :$
nope, sorry, 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]
Ron has been teaching various standardized tests for 20 years.

--

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi

--

Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.

Yves Saint-Laurent

--

Learn more about ron
Top
Master | Next Rank: 500 Posts
Posts: 151
Joined: Wed Feb 11, 2009 2:45 am
Thanked: 3 times

by deepoe » Fri Apr 17, 2009 3:37 am
Ok thanks!


But you chose √2 because radius were 2?
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3380
Joined: Mon Mar 03, 2008 1:20 am
Thanked: 2256 times
Followed by:1535 members
GMAT Score:800

by lunarpower » Fri Apr 17, 2009 2:47 pm
deepoe wrote:Ok thanks!


But you chose √2 because radius were 2?
that's the standard ratio in the 45-45-90 triangle: the hypotenuse is exactly √2 times as long as each leg.

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.

--

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi

--

Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.

Yves Saint-Laurent

--

Learn more about ron
Top
Post new topic Post Reply
• Page 1 of 1