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. Show that : AB = 6 × r

Answer 1

Let O, P and Q be the centers of the circle and semicircle.

Join OP and OQ.

OR = OS = r

And `AP = PM = MQ = QB = (AB)/4`

Now, `OP = OR + RP = r + (AB)/4`  ...(Since PM = RP = radii of same circle)

Similarly, `OQ = OS + SQ = r + (AB)/4`

`OM = LM - OL = (AB)/2 - r`

Now in right ΔOPM,

OP² = PM² + OM²

`=> (r + (AB)/4)^2 = ((AB)/4)^2 + ((AB)/2 - r)^2`

`=> r^2 + (AB^2)/16 + (rAB)/2 = (AB^2)/16 + (AB^2)/4 + r^2 - rAB`

`=> (rAB)/2 = (AB^2)/4 - rAB`

`=> (AB^2)/4 = (rAB)/2 + rAB`

`=> (AB^2)/4 = (3rAB)/2`

`=> (AB)/4 = 3/2 r`

`=> AB = 3/2 rxx 4 = 6r`

Hence AB = 6 × r