Please solve q 60 ?

Answer 1

(2)

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#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.