Question

AB is a line segment & M is its mid- point. Semi- circles are drawn with AM, MB & AB as diameters on the same side of the line AB. A circle C (o, r) is drawn so that it touches all the three semicircles. Prove that r = 1/6 AB

Answer 1

Let AB = x cm , Then AM = $\frac{x}{2}$ and PM = PN = $\frac{x}{4}$

Using Pythagoras Theorem,

$\Rightarrow$ $O{P}^2=P{M}^2+O{M}^2$

$\Rightarrow$ $(x/4+r{)}^2=(x/4{)}^2+(x/2-r{)}^2$

$\Rightarrow$ $(x/4{)}^2+r^2+(xr)/2=(x/4{)}^2+(x/2{)}^2+r^2–xr$

$\Rightarrow$ $r=\frac{1}{6}x$

$\Rightarrow$ $r=\frac{1}{6}AB$