Wrap a semicircle into a cone with height 10 units. What's the cone's radius and slant height?
March 11, 2010 1:14 PM   Subscribe

There is a cone whose lateral surface is a semicircle (when flattened). The height of the cone is 10 inches. What formula can I use to find the radius of the cone, and the slant height of the cone?

No, you are not doing my homework. I am answer checking some geometry problems, I can't access their text online, and google is failing me. I don't just want to know the answer, I'd like to know how to actually solve a problem like this. Help.
posted by 23skidoo to Education (11 answers total) 1 user marked this as a favorite
 
start with what's known:

C = 2*PI*R, therefore the circumference of the cone is C/2 = PI*R
that makes the the radius of the cone's base (r) = (PI*R)/(2*PI) = R/2

Pythagorus tells us that the height of the cone is:
sqrt(R^2 - r^2) = sqrt(R^2 - R^2/4) = (R/2)*sqrt(3)

I leave the rest as an exercise.
posted by mce at 1:40 PM on March 11, 2010


On preview, you can supplement mce's math with some of the equations from wikipedia, specifically "slant height", which is equal to the radius of your hemisphere.
posted by misterbrandt at 1:43 PM on March 11, 2010


This will help you.
posted by yoyoceramic at 1:47 PM on March 11, 2010


I love you.
posted by 23skidoo at 1:48 PM on March 11, 2010


So


[SPOILER ALERT]


the radius is (5 times the square root of 3)/3 and the slant height is (10 times the square root of 3)/3, right?
posted by Madamina at 2:05 PM on March 11, 2010


The trick to solving problems like this without looking up equations that you will forget is visualization. I'm typing as I think the problem through. If I get to where I think I have an answer at the end, I'll post it.

Picture in your mind rolling and unrolling the cone. You know it's a semicircle when unrolled. When rolled, the two edges come together as you form the point and you know that those edges used to be radii on the semicircle. Realize that their length doesn't change as you do this.

Now think of what happens to the round part of the semicircle. As you roll it up into a cone, that edge forms a circle that makes up the base of the cone. Realize that when you do this, the original half-circumference length doesn't change, it just turns into a circle with the same full circumference.

Now realize that with the cone sitting on the table, the radius of the new circular base is perpendicular to the cone's height. This makes a nice right triangle, which is one thing you always want to look for in problems like this. One leg is the line shooting up from the center of the base to the tip, the other leg is the radius of the base, the hypotenuse is those two edges that came together as you formed the point of the cone.

Now that I'm this far, I know a lot of relationships but I don't see a clear way to find the radius of the original semicircle ("slant height") or the cone radius. But I know that these are both parts of my right triangle and if I get one, combined with the original information of the height being 10, I can get the other.

Thinking about cones a little bit, I picture starting with various chunks of circle. If I have less than a semicircle, my mind's eye tells me the cone will be taller. If I have more, the cone will be shorter. Heck if I have a full circle, my cone is zero height.

If I picture semicircles of different radii and making them all into cones, I notice they all curl up to form the same angles, but they have different heights. I can picture making a stack of them, like those nesting dolls, and that all their sides are parallel. I'm feeling pretty confident that cones made out of semicircles have a certain proportion between their semicircle radius and height. But I don't know what it is.

So what I think I'll do now is work backward and give my original semicircle a known radius (say 1). This gives me a nice number for the half-circumference (pi). If I roll that up into a cone, I know my cone's base's circumference (still pi) and the cone's slant height (still 1). So I can get the cone's base's radius using simple knowledge about circles, and then the cone's height using simple knowledge about right triangles.

Now I know the proportions of a cone made out of a semicircle! Using simple knowledge about proportions, I can solve the original problem. If I was clear enough about the thought process, I bet you can too! Please let me know how it goes.
posted by fritley at 2:16 PM on March 11, 2010


I don't think you have enough information to determine anything specific about the cone other than it's height.

There are some constraints, though.

If your semicircle came from a circle whose circumference is C, say that the length of the curved part of the semicircle is pC. (p for proportion, multiplied by C).

The cone's base is a circle with circumference = pC.
The radius of this circle is pC/(2*pi)
The slant height is the radius of the circle that the semicircle came from.
So the slant height is C/(2*pi)

The height of the cone -- 10 in your case -- is a function of the radius of the original circle and the proportion of the circle that you use to construct the cone.

The relationship can be expressed as, 10^2 + (pC/2pi)^2 = (C/2pi)^2
So, C = sqrt[((20*pi)^2)/(1-p^2)] and p = sqrt[1 - (20*pi/C)^2 ].

This means that with a fixed p or a fixed C, you can determine the other, and then find the slant height and radius of the cone, with the identities I specified above.

On preview: Oh. You meant a semicircle where p = 1/2.
So, you get that C = ~72, so slant height is ~11.5 and radius is ~5.8.
posted by sentient at 2:36 PM on March 11, 2010


semicircle does explicitly mean half of a circle.
posted by fritley at 3:14 PM on March 11, 2010


I made the same mistake, sentient. Then I just gave up.
posted by muddgirl at 3:46 PM on March 11, 2010


You can see any cone sitting on a circular base as the surface generated by the rotation around the apex of a particular isosceles triangle sitting on its base (the isosceles triangle which is the 'shadow' of the cone cast on any wall perpendicular to the base by light rays parallel to the base and each other and perpendicular to the wall).

This cone happens to be generated by the very special case of an equilateral triangle.
posted by jamjam at 4:24 PM on March 11, 2010


semicircle does explicitly mean half of a circle.

Ah, good to know. :)
posted by sentient at 4:41 PM on March 11, 2010


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