show that AB = 6 x r

Question

AB is a line segment and M is its mid-point. three semi-circles are drawn with AM, MB and AB as diameters on the same side of the line AB. a circle with radius r unit is drawn so that it touches all the three semi-circles

Answer 1

Pranav,

 

The circle you are looking for is the one that fits in the space above M and

just touches each of the semicircles.

 

Label the midpoints of the semicircles AM and MB, P and Q respectively.

Label the centre of your small contiguous circle R.

Draw the two missing sides of triangle PRQ and extend MR to touch circle AB

 

Let the length AB=x

 

Then

PM=x/4

PR=x/4+r

RM=x/2-r

 

Applying Pythagoras' Theorem to triangle PRM: PM^2+RM^2=PR^2

 

So: (x^2)/16+(x^2)/4+r^2-xr=(x^2)/16 +r^2+xr/2

 

Dividing by x and collecting terms:  x/4-r=r/2 so x=6r