A semicircular piece of paper has center O, as shown above. Its diamet

Question

https://gmatclub.com/forum/download/file.php?id=26515 A semicircular piece of paper has center O, as shown above. Its diameter A’A is coated with adhesive. If the adhesive is used to fuse radii OA’ and OA along their entire lengths (so that points A and A’ coincide, points P and P’ coincide, and so on), a cone is formed as shown above. If point B divides the original semicircle into two identical arcs, what is the measure of angle AOB in the folded cone? A. 45º B. Between 45º and 60º C. 60º D. Between 60º and 90º E. 90º 20130909_1.gif

Bunuel · Accepted Answer

MANHATTAN GMAT OFFICIAL SOLUTION:

If you’re having trouble visualizing the relationship between the flat piece of paper and the 3D cone, get out your scissors and make a cone yourself. (You can’t do this on the real test, of course, but this is a challenge problem—you probably wouldn’t see a problem this hard on the real test!)

There are two key relationships between the semicircle (the flat piece of paper) and the cone (the 3D shape). First, the radius of the semicircle becomes the slant height of the cone. The slant height is the length measured along the outer surface of the cone, from A’ to O in the diagram). Second, the “half-circumference” of the semicircle (i.e., the length just of the rounded part, not including the diameter) becomes the circumference of the cone’s base (the circle on the bottom of the cone).

Let R stand for the radius of the original semicircle. If there were a full circle the circumference would be 2πR, so the “half-circumference” (through points A, B, and A’) is half of that, or just πR. This length becomes the full circumference of the cone’s circular base.

Find the radius, r, of the cone’s circular base:

2πr = πR
r = R/2

The diameter of the cone’s base is thus 2(R/2) = R.

As observed above, the cone’s slant height is also R, since it’s the same as the radius OA or OA’ of the original semicircle.

Now consider the triangle AOB. The two sides OA and OB are both equal to the cone’s slant height, so each is R units. The bottom side AB is a diameter of the cone’s base, so its length is also R units.

The triangle AOB is equilateral! The measure of each angle must be 60°, so the measure of angle AOB is 60°.

The correct answer is (C).

maxville23 · Answer

find perimeter of cone: AA' = 2*BA = 2*r*pi -> solve for diameter -> AA'/pi = 2*r <=> 2*BA/pi = 2*r

find length BA: BA = 2*pi*r*(90/360) = 1/2 *pi*r = 1/2 *pi*OA -> simplify -> (2*BA) / pi = OA

combine both statements: 2*BA / pi = 2*r and 2*BA / pi = OA -> OA = 2r

By definition OA and OB are the same. Also, Angles OBA and OAB are the same. As OA = 2r the three sites of the cone are equilateral. 
Therefore, BOA has to be 60°.

C

RAJINI123 · Answer

Is the question asking us to fins the 3 dimensional
 Angle or the 2dimensional ?? 
If it's the 2 dimensional then the angle is 60. Answer choice C.
Solution.

Let the radius OA be 10 units. Length of arc ABA' IS 10pi units.
Length of arc ABA' is the circumference of base of cone. Hence radius of cone is 5 units slant height is 10 units ( radius OA) so height of cone is 5√3 units using Pythagoras theorem. So its a 30, 60, 90 ∆. So angle AOB is 60. Hope I'm right.

maxville23 · Answer

Bunuel

Could you please provide an official answer?

askhere · Answer

See attached figure.

Lets suppose:
Radius of the original semicircle = r
 The arc length ABA'   =  (pi)*r  ( Circumference of semicircle )

After folding the semicircle into a cone observe the following:
1. The base of the cone is a circle whose circumference (ABA') is circumference of semicircle (pi)*r
 Lets say  radius of base of cone is r1 , So 2*(pi)*r1 = (pi)*r; So r1 = r/2. So  diameter of base of cone = r 
2. Notice that OAB forms a triangle. OA and OB are radius of the original semicircle i.e.  r . 
3. AB, the third side of triangle is nothing but the diameter of base of cone which is r. 
4. OAB is thus an equilateral triangle. So angle AOB will be equal to 60.

Answer :  C

Shobhit7 · Answer

Let the radius of large circle be R and of small circle be r.
 
Circumference of large semicircle is pi*R
Circumference of small full circle is 2*pi*r

Now, both of these circumferences are same in length because large semicircle is stretched to form small full circle. 
Therefore, 2*pi*r = pi*R
2r = R
d = R, where d is the diameter of small circle.
As diameter is AB
Therefore AB = R

From large circle, we know that OA=OB=R (Radius of large circle)

So, in triangle AOB in small circle, each length equals R, thus forming an equilateral triangle.
Hence, Angle AOB= 60

Ans C

bumpbot · Answer

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A semicircular piece of paper has center O, as shown above. Its diamet

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