SOLUTION: A 6 inch and a 18 inch diameter poles are placed tangent to each other and bound together with wire. Find the length of the shortest wire that will go around them.

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Question 167262: A 6 inch and a 18 inch diameter poles are placed tangent to each other and bound together with wire. Find the length of the shortest wire that will go around them.
Answer by Mathtut(3670) (Show Source):
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|AB| = |BF| = R(large circle radius), |CD| = r(small circle radius) The belt section AD and FG are tangent to the large pulley at A and F respectively, so angle DAB as well as GFB are right angles. The line DE is constructed parallel to CB so DE=BC.
we know that EA=6 and ED=12(hypothenuse)
Since triangle AED is a right triangle
we have
AD=FG therefore FG=10.39 inches

Now since ED and BC are parallel, angle AED and angle ABC are equal. Angle AED is the inverse tangent of |AD|/|AE| (opposite/adjacent). that would be equal to arctan of 1.7333 which equals 63.88degrees. The measure of angle ABF is twice the measure of angle ABC and hence you can now find the length of the longer of the two arcs joining A and F. since the angle we are looking for is ABF which = 2(63.88)=127.76. given the radius of 9 and angle of 127.76 we arrive at the small arc distance FA=20.07 inches since the entire circumference equals =56.52. we subtract small arc FA from the entire circumference to get large arc FA. 56.52-20.07 equals
Angle BCD is 180 degrees minus angle ABC and hence you can also find the length of the appropriate arc on the smaller pulley. 180-63.88= 116.12degrees double that and subtract from 360 degrees to get small circle arc DG(smaller portion)
360-232.24= 127.76....how about that!!!. Now given arc angle of 127.76 and a radius of 3 we arrive at arc length of
thusly, adding up all the lengths we have =