Question

(Note: In this problem, a "cylinder" means the can-shaped figure that we used to recognize as a cylinder, not the cylinder defined in the sense as in Section \(12.6 .\).

Problem: Three circular cylinders, each with radius \(\sqrt{3}\), are standing tangent to one another on the plane \(\Omega\), as shown in the figure above. Let \(P, Q\), and \(R\) denote the centers of the upper circular bases of the shortest cylinder, the next, shortest cylinder, and the tallest cylinder, respectively. Suppose the following:

\(* \triangle Q P R\) is an isoscoles triangle.

* The measure of the anglo between plans \(Q P R\) and plane \(\Omega\) is \(60^{\circ}\).

* The heights of the three cylinders are \(8, a\), and \(b\), with \(8<a<b\).

Find the value of \(a+b\).

Answer 1

Difference between the heights of smallest and largest cylinder = b–8

Distance between the axis of two cylinders = 2√3

We know that line PR is inclined at 60°

Therefore, tan 60° = (b–8)/2√3

=> 6 = b–8 => b = 14

Now we get :

For triangle to be isosceles PQ=QR

So a–8 = b–a => a = 11