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.

From figure,

OR = OS = r.

AP = PM = MQ = QB = $\dfrac{AB}{4}$

Also,

As, RP = SQ = PM [Radius of semi-circle with center P]

⇒ OP = OR + RP = r + $\dfrac{AB}{4}$,

⇒ OQ = OS + SQ = r + $\dfrac{AB}{4}$,

⇒ OM = LM - OL = $\dfrac{AB}{2} - r$

In right angle △OPM,

OP² = OM² + PM²

$\Rightarrow \Big(r + \dfrac{AB}{4}\Big)^2 = \Big(\dfrac{AB}{4}\Big)^2 + \Big(\dfrac{AB}{2} - r\Big)^2 \\[1em] \Rightarrow r^2 + \dfrac{AB^2}{16} + \dfrac{AB.r}{2} = \dfrac{AB^2}{16} + \dfrac{AB^2}{4} + r^2 - AB.r \\[1em] \Rightarrow r^2 - r^2 + \dfrac{AB.r}{2} + AB.r = \dfrac{AB^2}{16} - \dfrac{AB^2}{16} + \dfrac{AB^2}{4} \\[1em] \Rightarrow \dfrac{3AB.r}{2} = \dfrac{AB^2}{4} \\[1em] \Rightarrow \dfrac{AB^2}{AB} = \dfrac{3 \times 4r}{2} \\[1em] \Rightarrow AB = 6 \times r.$

Hence, proved that AB = 6 × r.