Question

Please solve q 60 ?

Answer 1

(2)

Explanation:

As given in the figure, `M` is the midpoint of line segment `AB`. Let `X and Y` be the centers of the two smaller semicircles on `AM` and `MB`.
Let `C` is the point where perpendicular bisector of `AB` meets the greater semicircle which has its center at `M`.

Let `r` be radius of circle and center `O`, which touches all three semi-circles.

From figure line segments `XM, MO and OX` form a right triangle. Also these are `R/2, R-r, and R/2+r` respectively. Using Pythagoras theorem we get

`(R/2)^2 + (R-r)^2 = (R/2+r)^2`
`=>R^2/4 + R^2 - 2rR + r^2 = R^2/4 + rR + r^2`
`=>R^2 - 3rR = 0`
`=>R(R-3r) = 0`

Two solutions are `R=0, (R-3r) = 0`
As `R` can not be zero, we have

`r = R/3`

As radius of the greater semicircle `R = (AB)/2`

`=> r = (AB)/6`