A truncated cone has horizontal bases with radii 18 and 2. A sphere is tangent to the top, bottom, and lateral surface of the truncated cone. What is the radius of the sphere?
+773 Consider a trapezoidal (label it ABCD as follows) cross-section of the truncate cone along a diameter of the bases:
image here.
Above, E, F, and G are points of tangency. By the Two Tangent Theorem, BF = BE = 18 and CF = CG = 2, so BC = 20.
We draw H such that it is the foot of the altitude Segment HD to Segment AB:
image here.
By the Pythagorean Theorem, \(r = \dfrac{DH}{2} = \dfrac{\sqrt{(20)^2 - (16)^2}}{2} = \boxed{6}\).
Hope this helps,
- PM
+773 If the first solution confuses you, here is a second solution.
Create a trapezoid with inscribed circle O exactly like in the solution above, and extend lines AD and BC from the solution above and label the point at where they meet H. Because \(\frac{\overline{GC}}{\overline{BE}} = \frac{1}{9}\), \(\frac{\overline{HG}}{\overline{HE}} = \frac{1}{9}\). Let \(\overline{HG} = x\) and \(\overline{GE} = 8x\).
Because these are radii, \(\overline{GO} = \overline{OE} = \overline{OF} = 4x\). \(\overline{OF} \perp \overline{BH}\) , so \(\overline{OF}^2 + \overline{FH}^2 = \overline{OH}^2\). Plugging in, we get \(4x^2 + \overline{FH}^2 = 5x^2 \), so \(\overline{FH} = 3x\).Triangles OFH and BEH are similar, so \(\frac{\overline{OF}}{\overline{BE}} = \frac{\overline{FH}}{\overline{EH}}\), which gives us \(\frac{4x}{18} = \frac{3x}{9x}\). Solving for x, we get \(x = 1.5\) and \(4x = \boxed{6}\).
Hope this helps,
- PM
You plagiarized both of your answers.
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_19
Nothing new here, you’ve plagiarized many times before.
Is this the only way you can correctly answer the questions?
You’ve implied others have plagiarized, but you blatantly do it yourself, so you are not only a plagiarizer you are a hypocrite!
Why are you here?
+773 I gave links to both of my answers...
in the response below, you didn't even spell bs correctly...
