Question

please answer with perfect expanation Q.41

Answer 1

Let r=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, A=36cm (given)
Then, PE=FQ= 1/4 x AB = 1/4 = x 36 = 9cm
PR = r+AB/4 = r +36/4 = r+9 = RQ
PRQ is an Isosceles triangle.
Since, M is the midpoint of PQ, RM⊥PQ.
Now, MR=CM-CR = 1/2(36) - r = 18-r
In ΔPMR,
By pythagoras Theorem,
PM + MR = PR
9 + (18-r) = (9+r)
405 - 36r = 81 - 405
-54r = -324
r = 6
Shaded area = Area of semicircle ABC - Area of semicircle AME - Area of semicircle MBD - Area of circle CED
= πr²/2 - πr²/2 - πr²/2 - πr²
= π(18)²/2 - π(9)²/2 - π(9)²/2 - π(6)²
= π/2 [(324 - 81 - 81] - 36π
= π/2 [162] - 36π
= 81π - 36π
= 45π cm²
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