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tunnel entrance

by uptowngirl92 » Sun Jun 21, 2009 2:58 am
The figure above shows the shape of a tunnel entrance. If the curved portion is of a circle and the base of the entrance is 12 feet across, what is the perimeter, in feet, of the curved portion of the entrance'?



(A) 9π

(B) 12π

(C) 9 pie root2
(D) 18π

(E)9 pie/root2
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by rah_pandey » Sun Jun 21, 2009 3:20 am
I think one more variable needs to be given to evaluate the question. It can be height of the tunnel or angle that the flat portion makes at the center. Can you please check

What is the source of the problem
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Re: tunnel entrance

by dtweah » Sun Jun 21, 2009 4:45 am
uptowngirl92 wrote:The figure above shows the shape of a tunnel entrance. If the curved portion is of a circle and the base of the entrance is 12 feet across, what is the perimeter, in feet, of the curved portion of the entrance'?



(A) 9π

(B) 12π

(C) 9 pie root2
(D) 18π

(E)9 pie/root2
1. Complete the circle so that you have a full circle.

2. Let segment AB be the base of lenght 12.

3. Make O the center of the circle

4. Draw segments AC and BC. You now have a triangle in the circle.

5. Draw radius to all vertices of the triangle. Let r be length of radius. OA=OB=OC. Draw Diameter CE which bisects AB. Let D be the midpoint of AB. You now have a median CD. Any line from one vertex of a triangle passing through the centroid that bisects aan opposite side is a Median. There is a relationship between CO and OD. 2/3 of the length of each median is between the vertex of a triangle and the centriod of the triangle and 1/3 of the length is b/w the centroid and the midpoint at the opposide side. This is OD. O is centroid because you can draw all the medians and they pass thru it. We are only interested in median CD. This allows us to find the Redius of the circle. Using right triangle ODC, OD=r/3 and hypotenuse OC=r, and DC=6, we can find r=9root2/2. So we know the radius. The Key question is is this an Equalateral triangle? If it is then we we will take 2/3 of the circumference and that will give the answer. If not we are doomed.

6. How do we show that this is an equilateral triangle? Note that in your drawing you should see 3 isoceles triangles: AOC, COB, and BOA. Label the base angles in each as x, y z. You derive
2x +2y +2z=180 or
x + y +z =90

Now take right triangle ADC. summing its angles you have
2x +z =90, since angle ADC is 90. Equate the two 90 degrees we have
2x +z =x + y +z
X=Y.

Doing same analysis using the other right triangle shows shows that X=Y=Z which means we we have an equalateral triangle.

From here on you can use two approaches to find perimeter in question.

2/3 (2 pi r) = 2/3 x 2 pi 9 root2/2= 6pi root2 is the preffered approach.

Or

Arc length = Central angle ( in radians) x raduis x 2. You multiply by 2 b/c there are two arcs involved arc AC and arc CB.

Arc ACB= 120pi/180 x 9 root2/2 x 2 ( Central angle is 120 b/c x=y=z=30)

This gives 6pi root 2.

The answer is not among the choices but I think the 9 in C may have been mistakenly written for 6.
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by abhinav85 » Tue Jun 23, 2009 10:09 am
IMO C.

In the given figure just complete the circle by drawing the fourth arc.

after doing that just make a square inside the circle by joining the 3 sides
with the given one side that is 12 feet.

Now with the help of the diognal we can calculate the diameter i.e 12root2.

The radius will be 12 root 2 /2. i.e 6root2.

now use the formula to calculate the arc i.e theta/360 x 2pier.
by this we will get the length of the arc that is 90/360 2 pie 6root2.i.e 3root2pie.


To get the perimetre of the curved area multiply 3 with 3root2pie i.e 9root2pie.

Hence choice C.
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by dtweah » Tue Jun 23, 2009 10:18 am
abhinav85 wrote:IMO C.

In the given figure just complete the circle by drawing the fourth arc.

after doing that just make a square inside the circle by joining the 3 sides
with the given one side that is 12 feet.

Now with the help of the diognal we can calculate the diameter i.e 12root2.

The radius will be 12 root 2 /2. i.e 6root2.

now use the formula to calculate the arc i.e theta/360 x 2pier.
by this we will get the length of the arc that is 90/360 2 pie 6root2.i.e 3root2pie.


To get the perimetre of the curved area multiply 3 with 3root2pie i.e 9root2pie.

Hence choice C.
How do you determine that the other sides are 12ft? Because one segment is 12 ft in a circle dsnt mean if you connect segments to it each will have 12 ft.

Also even if it is a square how do you know the square will be entirely in the circle? I drew a square that is outside the circle so logic doesn't follow.
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by abhinav85 » Tue Jun 23, 2009 10:23 am
How do you determine that the other sides are 12ft? Because one segment is 12 ft in a circle dsnt mean if you connect segments to it each will have 12 ft.

Also even if it is a square how do you know the square will be entirely in the circle? I drew a square that is outside the circle so logic doesn't follow.
Dude i don't know how u made a square outside a circle???
And if u see the answer choices and the problem this is the only logic.

BTW what is the OA???
Last edited by abhinav85 on Tue Jun 23, 2009 10:27 am, edited 1 time in total.
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by ssmiles08 » Tue Jun 23, 2009 10:27 am
IMO there is something missing in this question. WHat is n? are we supposed to assume its the radius? diameter?

What is the source of this question?
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by abhinav85 » Tue Jun 23, 2009 10:29 am
Hey ssmiles08

Its from Gmat prep test.

that is not n in answer choices its pie.
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by truplayer256 » Tue Jun 23, 2009 10:32 am
The correct answer should be C.) 9*pi*sqrt(2). Here's why:

If the curved portion is 3/4 of the circle, then the remaining portion is 1/4 of the circle or 90 degrees. So now we have a triangle with a vertex of 90 degrees. If we bisect the triangle, we get two 45-45-90 triangles with a base that's half of 12 or 6. This means that the radius of the circle is the hypotenuse of one of the 45-45-90 triangles, which is 6*sqrt(2). Diameter equals 2*radius, so 2*6*sqrt(2)=12*sqrt(2). Now to find the perimeter of the curved portion, we take the circumference of the circle and multiply it by 3/4, so:

pi*12*sqrt(2)*3/4=9*pi*sqrt(2). Btw, there's a diagram of this question on 1000 PS Section 23 #20.
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by ssmiles08 » Tue Jun 23, 2009 10:36 am
OOH thanks abhinav85, my eyes were decieving me :)

anyway I think OP posted the question wrong.

I searched around and the question actually says:

The figure above shows the shape of a tunnel entrance. If the curved portion is 3/4 of a circle and the base of the entrance is 12 feet across, what is the perimeter, in feet, of the curved portion of the entrance'?

3/4 of a circle is left out.
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by abhinav85 » Tue Jun 23, 2009 11:14 am
Hey trueplayer

[/quote]The correct answer should be C.) 9*pi*sqrt(2). Here's why:

If the curved portion is 3/4 of the circle, then the remaining portion is 1/4 of the circle or 90 degrees. So now we have a triangle with a vertex of 90 degrees. If we bisect the triangle, we get two 45-45-90 triangles with a base that's half of 12 or 6. This means that the radius of the circle is the hypotenuse of one of the 45-45-90 triangles, which is 6*sqrt(2). Diameter equals 2*radius, so 2*6*sqrt(2)=12*sqrt(2). Now to find the perimeter of the curved portion, we take the circumference of the circle and multiply it by 3/4, so:

pi*12*sqrt(2)*3/4=9*pi*sqrt(2). Btw, there's a diagram of this question on 1000 PS Section 23 #20.


Can u send me the 1000ps ques???
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by dtweah » Tue Jun 23, 2009 4:01 pm
ssmiles08 wrote:OOH thanks abhinav85, my eyes were decieving me :)

anyway I think OP posted the question wrong.

I searched around and the question actually says:

The figure above shows the shape of a tunnel entrance. If the curved portion is 3/4 of a circle and the base of the entrance is 12 feet across, what is the perimeter, in feet, of the curved portion of the entrance'?

3/4 of a circle is left out.
If this is part of the problem then it is trivial.
you will have a centrial angle that is 90, with OA and OB =r and hypotenuse =12
r^2 +r^2==144
r^2=72
r=6 root2

3/4 2 pi 6root 2

9pi root 2.
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Re: tunnel entrance

by dumb.doofus » Tue Jun 23, 2009 6:08 pm
dtweah wrote:
uptowngirl92 wrote:The figure above shows the shape of a tunnel entrance. If the curved portion is of a circle and the base of the entrance is 12 feet across, what is the perimeter, in feet, of the curved portion of the entrance'?



(A) 9π

(B) 12π

(C) 9 pie root2
(D) 18π

(E)9 pie/root2
1. Complete the circle so that you have a full circle.

2. Let segment AB be the base of lenght 12.

3. Make O the center of the circle

4. Draw segments AC and BC. You now have a triangle in the circle.

5. Draw radius to all vertices of the triangle. Let r be length of radius. OA=OB=OC. Draw Diameter CE which bisects AB. Let D be the midpoint of AB. You now have a median CD. Any line from one vertex of a triangle passing through the centroid that bisects aan opposite side is a Median. There is a relationship between CO and OD. 2/3 of the length of each median is between the vertex of a triangle and the centriod of the triangle and 1/3 of the length is b/w the centroid and the midpoint at the opposide side. This is OD. O is centroid because you can draw all the medians and they pass thru it. We are only interested in median CD. This allows us to find the Redius of the circle. Using right triangle ODC, OD=r/3 and hypotenuse OC=r, and DC=6, we can find r=9root2/2. So we know the radius. The Key question is is this an Equalateral triangle? If it is then we we will take 2/3 of the circumference and that will give the answer. If not we are doomed.

6. How do we show that this is an equilateral triangle? Note that in your drawing you should see 3 isoceles triangles: AOC, COB, and BOA. Label the base angles in each as x, y z. You derive
2x +2y +2z=180 or
x + y +z =90

Now take right triangle ADC. summing its angles you have
2x +z =90, since angle ADC is 90. Equate the two 90 degrees we have
2x +z =x + y +z
X=Y.

Doing same analysis using the other right triangle shows shows that X=Y=Z which means we we have an equalateral triangle.

From here on you can use two approaches to find perimeter in question.

2/3 (2 pi r) = 2/3 x 2 pi 9 root2/2= 6pi root2 is the preffered approach.

Or

Arc length = Central angle ( in radians) x raduis x 2. You multiply by 2 b/c there are two arcs involved arc AC and arc CB.

Arc ACB= 120pi/180 x 9 root2/2 x 2 ( Central angle is 120 b/c x=y=z=30)

This gives 6pi root 2.

The answer is not among the choices but I think the 9 in C may have been mistakenly written for 6.
My two cents..

Dude. I really dont think your approach is correct. All your assumptions and logics at the start of the solution are based upon the assumption that the circumcenter coincides with the centroid..this happens only in an equilateral triangle. After assuming that, you are trying to prove that it is in fact an equilateral triangle.. obviously, you'll get it as equilateral only, since to start with you assumed it..

Basically what's wrong in your thought process is that you implied: O is the centroid of the triangle.. It is possible ONLY when the triangle is an equilateral triangle.. There are two concepts here..
1. Circumcenter: all the perpendicular bisectors of the sides of a triangle pass through a point called the circumcenter.. now in your case, circumcenter will be same as centroid ONLY IF the triangle is equilateral.
2. Centroid: As we all know and you mentioned too is a point where the lines from vertx to the mid point of the other sides of a triangle meet.. we know that such lines are called medians..

Problem is that you are confusing between these two concepts.. if my triangle is not equilateral triangle, then the circumcenter and centroid cannot coincide..
So I dont think that your logic is correct and also that this question cannot be solved without that missing information about the major arc being 3/4th of the circle's perimeter.
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Re: tunnel entrance

by dtweah » Wed Jun 24, 2009 12:53 am
dumb.doofus wrote:
dtweah wrote:
uptowngirl92 wrote:The figure above shows the shape of a tunnel entrance. If the curved portion is of a circle and the base of the entrance is 12 feet across, what is the perimeter, in feet, of the curved portion of the entrance'?



(A) 9π

(B) 12π

(C) 9 pie root2
(D) 18π

(E)9 pie/root2
1. Complete the circle so that you have a full circle.

2. Let segment AB be the base of lenght 12.

3. Make O the center of the circle

4. Draw segments AC and BC. You now have a triangle in the circle.

5. Draw radius to all vertices of the triangle. Let r be length of radius. OA=OB=OC. Draw Diameter CE which bisects AB. Let D be the midpoint of AB. You now have a median CD. Any line from one vertex of a triangle passing through the centroid that bisects aan opposite side is a Median. There is a relationship between CO and OD. 2/3 of the length of each median is between the vertex of a triangle and the centriod of the triangle and 1/3 of the length is b/w the centroid and the midpoint at the opposide side. This is OD. O is centroid because you can draw all the medians and they pass thru it. We are only interested in median CD. This allows us to find the Redius of the circle. Using right triangle ODC, OD=r/3 and hypotenuse OC=r, and DC=6, we can find r=9root2/2. So we know the radius. The Key question is is this an Equalateral triangle? If it is then we we will take 2/3 of the circumference and that will give the answer. If not we are doomed.

6. How do we show that this is an equilateral triangle? Note that in your drawing you should see 3 isoceles triangles: AOC, COB, and BOA. Label the base angles in each as x, y z. You derive
2x +2y +2z=180 or
x + y +z =90

Now take right triangle ADC. summing its angles you have
2x +z =90, since angle ADC is 90. Equate the two 90 degrees we have
2x +z =x + y +z
X=Y.

Doing same analysis using the other right triangle shows shows that X=Y=Z which means we we have an equalateral triangle.

From here on you can use two approaches to find perimeter in question.

2/3 (2 pi r) = 2/3 x 2 pi 9 root2/2= 6pi root2 is the preffered approach.

Or

Arc length = Central angle ( in radians) x raduis x 2. You multiply by 2 b/c there are two arcs involved arc AC and arc CB.

Arc ACB= 120pi/180 x 9 root2/2 x 2 ( Central angle is 120 b/c x=y=z=30)

This gives 6pi root 2.

The answer is not among the choices but I think the 9 in C may have been mistakenly written for 6.
My two cents..

Dude. I really dont think your approach is correct. All your assumptions and logics at the start of the solution are based upon the assumption that the circumcenter coincides with the centroid..this happens only in an equilateral triangle. After assuming that, you are trying to prove that it is in fact an equilateral triangle.. obviously, you'll get it as equilateral only, since to start with you assumed it..

Basically what's wrong in your thought process is that you implied: O is the centroid of the triangle.. It is possible ONLY when the triangle is an equilateral triangle.. There are two concepts here..
1. Circumcenter: all the perpendicular bisectors of the sides of a triangle pass through a point called the circumcenter.. now in your case, circumcenter will be same as centroid ONLY IF the triangle is equilateral.
2. Centroid: As we all know and you mentioned too is a point where the lines from vertx to the mid point of the other sides of a triangle meet.. we know that such lines are called medians..

Problem is that you are confusing between these two concepts.. if my triangle is not equilateral triangle, then the circumcenter and centroid cannot coincide..
So I dont think that your logic is correct and also that this question cannot be solved without that missing information about the major arc being 3/4th of the circle's perimeter.
After going thru my thought process you reach the wrong conclusions. Because I proved equalateral triangle last doesnot mean equalater triangle cannot be proved first. The centroid of a circle and center of the an equilateral triangle coincide. The proof of the equalateral triangle is given. Reverse the order as you want. After proving, then follow the rest of the analysis using the centroid. All conclusions follow.
So I dont' see how you can reach a conclusion that the problem cannot be solved. You will have a problem if the proof of equalateral triangle is invalid. The proof of the traingle is based on the Center of the circle not on the centrod. You don't say so yet you reach a strange conclusion based on the positioning of the proof of the triangle. . Positioning can easily be reversed. I stand absolutely by the given solution in the absence of 3/4 info. You have to show that reasoning used in the proof is wrong and not just the assumption. I can agree on the assumption but reversing resolves that problem and does not affect the solution at all.
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by rah_pandey » Wed Jun 24, 2009 1:19 am
Dtweah,
Does your proof say that all inscribed triangles in a circle are equilateral.I Hope NOT. If answer is NO then what specific info you have used to prove that the triangle that is drawn is equilateral. If i understand your proof right then you have assumed that the circumcenter and centroid coincide(this is true for equilateral triangles). Also how can you say that triangle ADC is right angled. I know any bisector of the chord passing through the center will be perpendicular to the chord but any line from vertex C on the circle passing through the mid point of the chord may not pass through the center of the circle.

I think your proof is wrong and as dumdoofus said you assumed one thing and later on proved the same.

Regards,
Rahul
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