Geometry Problem

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__Kit
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Geometry Problem

Postby __Kit » Thu Jun 07, 2007 7:06 am UTC

-NOT HOMEWORK-

In the following diagram, each blue line is 24 centimetres long, and the red line is 16 centimetres long. Assume the shape in the centre of the circle is a perfect square, and it is exactly centred in the circle. Also assume that all three coloured lines are parallel to each other.

Image

What is the radius of the circle, in centimetres?
=]

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Cosmologicon
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Postby Cosmologicon » Thu Jun 07, 2007 7:33 am UTC

Working in centimeters, call x the distance from the center of the square to any of its four corners. Then the radius of the circle is 16 + x. Draw the radius that goes from the center to the point where a blue line touches the circle. This is the hypotenuse of a right triangle whose leg lengths are 24 and x. So x^2 + 24^2 = (16 + x)^2, so x = 10. Thus the radius is 26 centimeters. How does that sound?

The_Spectre
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Postby The_Spectre » Thu Jun 07, 2007 2:29 pm UTC

Cosmologicon wrote:Working in centimeters, call x the distance from the center of the square to any of its four corners. Then the radius of the circle is 16 + x. Draw the radius that goes from the center to the point where a blue line touches the circle. This is the hypotenuse of a right triangle whose leg lengths are 24 and x. So x^2 + 24^2 = (16 + x)^2, so x = 10. Thus the radius is 26 centimeters. How does that sound?

You should probably also invoke some sort of symmetry argument to show that the red line lies along a radius of the circle, since this isn't stated in the original problem.

The red line must be an extension of a diagonal of the square, or the blue lines which are parallel to it wouldn't be equally long.

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Marbas
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Postby Marbas » Fri Jun 08, 2007 1:47 am UTC

I got 28...
Where did I go wrong?


First I thought that because the line intersected the corner at that point that it had also intersected the diameter and was therefore halfway across the circle.

Image


So the angle between the square and the line is 45. <--(I'm pretty sure this is where it all goes belly up, but if Cosmo's solution works this should work too, else Cosmo's triangle is not a right triangle.)

Image

So if we draw another line from the edge of the circle we get a 45-45-90.

Image


So, 2(x^2)=24^2

2(x^2)=576
x^2=288
x=12(2^.5) is one side of the square.

So now we have to find half of the diagonal. So using Pythagoreans theorem:

2(x^2) = 24

24/2 = 12

12+16 = 28 = r

Halp?

EDIT:
I think found my mistake.
My triangle doesn't have to be a right triangle.
If that's not it can someone please tell me?

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Postby shill » Fri Jun 08, 2007 3:34 am UTC

Marbas, I don't understand your third diagram. It looks like you're assuming that the point where the blue line touches the edge of the circle is a continuation of the side of the square.

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Postby Marbas » Fri Jun 08, 2007 4:01 am UTC

shill wrote:Marbas, I don't understand your third diagram. It looks like you're assuming that the point where the blue line touches the edge of the circle is a continuation of the side of the square.


Yeah... I noticed that. Didn't you read the 'EDIT' part?


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