Question

AB is a line segment and M is its mid-point. Semi-circles are drawn with AM, MB and AB as diameters on the same side of AB. A circle is drawn to touch all the three semi-circles. Its radius is:

• A. $\frac{AB}{3}$
• B. $\frac{2}{3}AB$
• C. $\frac{AB}{6}$
• D. $\frac{3}{4}AB$

Answer 1

The correct option is

C

$\frac{AB}{6}$

Let

AB=x & OD =r

then, MD=$\frac{x}{2}$

By pythagorus theorem in triangle OPM

$(OP{)}^2=(PM{)}^2+(OM{)}^2$.....(1)

So,$OM=MD-OD=\frac{x}{2}-r$ [From the diagram]

And $MP=AM-AP=$$\frac{x}{2}-\frac{x}{4}=\frac{x}{4}$

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

$={(\frac{x}{2}-r)}^2+\frac{x^2}{16}$

$={(\frac{x}{4}+r)}^2$

Hence,$r=\frac{x}{6}=\frac{AB}{6}$

Hence, the correct answer is option $\left(C\right)$