Question

In Fig. 15.95, 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 cirlces. Find the area of the shaded region,

Answer 1

Let r = the radius of the cirlce = 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 = RQ =
$\frac{1}{4}\times AB=\frac{1}{4}\times 36=9cm$
$\Rightarrow PR=$
$r+\frac{AB}{4}=r+\frac{36}{4}=r+9=RQ$
$\Rightarrow \Delta PRQ$ is in isosceles thraingle.
Since, M is the mid - point of $PQ,RM\perp PQ$
Now, MR = CM - CR =
$\frac{1}{2}\left(36\right)-r=18-r$
In $\Delta PMR,$
By pytagoras theorem,
$(PM{)}^2+(MR{)}^2=(PR{)}^2\Rightarrow (9{)}^2+(18-r{)}^2=(9+r{)}^2\Rightarrow 81+324+r^2-36r=81+r^2+18r\Rightarrow 405-36r=81+18r\Rightarrow -36r-18r=81+18r\Rightarrow -54r=-324\Rightarrow r=6$
Shaded area = Area of semicircle ABC - Area of semicirlce AME - Area of semicirlce MBD - Area of circle CED
$=\pi \frac{\pi r^2}{2}-\frac{\pi r^2}{2}-\frac{\pi r^2}{2}-\pi r^2=\frac{\pi (18{)}^2}{2}-\frac{\pi (9{)}^2}{2}-\frac{\pi (9{)}^2}{2}-\pi (6{)}^2=\frac{\pi }{2}[(18{)}^2-(9{)}^2]-36\pi =\frac{\pi }{2}[324-81-81]-36\pi =\frac{\pi }{2}[162]-36\pi =81\pi -36\pi =45\pi$