In Fig. AB = 36 cm and M is mid-point of AB. Semi-circles are drawn on AB, AM and MB as diameters. A circle with centre C touches all the three circles. Find the area of the shaded region.

Let r = the radius of the circle=CR.

Consider AMB is a straight line such that AM=MB.
Semicircles are drawn with AB,AM and MB as diameters.

A circle is drawn with centre C such that CM is perpendicular to AB, and such that the circle is tangent to all

three semicircles.

As, AB=36 cm (given)

Then, PE = $\frac{1}{4}\times AB=\frac{1}{4}\times 36=9cm$

=> $PR=r+\frac{AB}{4}=r+\frac{36}{4}=r+9=RQ$

⇒Δ PRQ is n isosceles triangle.
Since, M is the mid-point of PQ, RM ⊥PQ.

Now, MR = CM-CR= $\frac{1}{2}\times 36-r=18-r$

In Δ PMR,

By pythagoras theorem,

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

$=>9^2+(18-r{)}^2=(9+r{)}^2$

$=>81+324+r^2-36r=81+r^2+18r$

$=>405-36r=81+18r$

$=>-36r-18r=81-405$

$=>-54r=-324$

$=>r=6$

Shaded area = Area of semicircle ABC-Area of semicircle AME-Area of semicircle MBD-Area of circle CED

$=\frac{\pi r^2}{2}-\frac{\pi r^2}{2}-\frac{\pi r^2}{2}-\pi r^2$

$=\frac{\pi \times {18}^2}{2}-\frac{\pi \times 9^2}{2}-\frac{\pi \times 9^2}{2}-\pi \times 6^2$

$=\frac{\pi ({18}^2-9^2-9^2)}{2}-36\pi$

$=\frac{\pi (324-81-81)}{2}-36\pi$

$=\frac{\pi \left(162\right)}{2}-36\pi$

$=81\pi -36\pi =45\pi$