ps set

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ps set

by thumpin_termis » Fri Jul 06, 2007 5:29 pm
See attached file.

Getting the first part of the equation is no problem, but how do you get the perimeter of the cone part with the given info?


OA is B.
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by givemeanid » Fri Jul 06, 2007 6:53 pm
We have AD = 5
Radius = 2

The cone is 3/4 circumference of the circle = 3/4 * 2 * pi * 2 = 3*pi

Now, consider semi-circle PCBQ.
The arcs PC + BQ = 1/4 of the entire circle (since cone is 3/4 and the semi-cirlce on top is 1/2. So that leaves 1/4). There are 360 degrees in a circle. So 1/4 = 90degrees. Measure of arc CB = 90 degress. Since arc CB is subtended by angle COB, angle COB = 90.

In triangle COB, CO = OB = radius = 2 and angle COB = 90
This is a 45-45-90 triangle. So, CB = 2 * sqrt2 and CD = DB = 1/2 (2 * sqrt2) = sqrt2

In triangle ADB, AD^2 + DB^2 = BA^2
5^2 + (sqrt2)^2 = BA^2
BA = sqrt27 = 3sqrt3 = AC

Perimeter = 3/4 circumference of circle + BA + AC = 3*pi + 3sqrt3 + 3sqrt3 = 3*pi + 6sqrt3
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So It Goes

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by thumpin_termis » Fri Jul 06, 2007 7:33 pm
Whoo! That's a doozy. Thanks a bunch.

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by moneyman » Sat Jul 07, 2007 7:52 am
Can u explain this part pls..

The arcs PC + BQ = 1/4 of the entire circle (since cone is 3/4 and the semi-cirlce on top is 1/2. So that leaves 1/4). There are 360 degrees in a circle. So 1/4 = 90degrees. Measure of arc CB = 90 degress. Since arc CB is subtended by angle COB, angle COB = 90.
Maxx

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by thumpin_termis » Sat Jul 07, 2007 10:48 am
From given info, the part of the ice cream that sits on top of the cone is 1/4 of the circumference. Since the entire circumference of the circle has 360 degrees when measured from the center, the angle from the center measuring 1/4th of this would have (360/4)=90 degrees.

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by wizardofwashington » Mon Sep 17, 2007 7:34 am
Brilliant ..You guys are amazing..I got part way through the solution and then had to make an intelligent guess..GivemeanID, you are truly my inspiration..
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