Two solid spherical balls with centres P and Q touch each other. The balls lie inside and in contact with a hemispherical bowl of center R. Given that PQ=13 cm, QR=16cm and PR=19cm, calculate the radii of the bowl and the two spherical balls

Sorry - I misread the conditions. Suppose the radius of ball P is r.

Then the radius of ball R is 13-r.
Two radii RP and RQ both touch circle R.
So, R = 19+r = 16+(13-r)
r = 5
so R = 19+5 = 16+8 = 24

Extra credit.

If Q is at (0,0) and R is at (0,16), where is P?

To solve this problem, we can use the concept of similarity in triangles. Let's denote the radii of the two spherical balls as r1 and r2, and the radius of the hemispherical bowl as R.

First, let's consider the right triangle PQR. Notice that PR is the hypotenuse, and PQ and QR are the two legs of the triangle. According to the Pythagorean theorem, we can write:

PQ^2 + QR^2 = PR^2
13^2 + 16^2 = 19^2
169 + 256 = 361
425 = 361

This equation is not true, which means there is an error in the given data. Please double-check the measurements provided or confirm if there is any other information missing, so I can assist you further.

Sorry. If PR=19 and circle P touches the bowl, then the radius of the bowl must be 19+19 = 38

But, by the same token, it must also be 16+16 if QR=16