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A sphere is inscribed in a right cone with base radius 12 cm and height 24 cm, as shown. The radius of the sphere can be expressed as \(a\sqrt{c} - a\) cm. What is the value of \(a+c\)?

 Jan 2, 2019
 #1
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Let the point where the top of the circle meets the base of the cone =    A

Let the radius of the cone = AB = 12

Let the center of the sphere = C

Let the radius of the sphere meet the side of the cone at D

Let the bottom "tip" of the cone = E

 

So......CD  meets the side of the cone at a right angle

And  BA  =  BD  = 12   [  tangents  drawn from the same point ]

 

So....Triangle CDE is a right triangle

 

The slant height of the cone, BE =   √ [ 24^2 + 12^2]  = √720 =  √ [ 144 * 5]  = 12√5

 

So DE  = BE - BD =    12√5 - 12

And  CE =  24 - r

And CD = r

 

So.....by the Pythagorean Theorem, we have that

 

CE^2 = CD^2 + DE^2

 

(24 - r)^2  = r^2 + (12√5 - 12)^2       simplify

 

r^2 - 48r + 576  = r^2 + 144(√5 - 1)^2

 

-48r + 576   =  144 (5 - 2√5 + 1)

 

-48r + 576   =  144(6 - 2√5)

 

-48r + 576  = 864 - 288√5

 

48r  =  288√5 - 288            

 

48r =  288 (√5 - 1)

 

r = 6 (√5 - 1)   =  6√5 - 6

 

So     a  +  c   =    6 + 5  =   11

 

 

 

cool cool cool

 Jan 3, 2019
edited by CPhill  Jan 3, 2019

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