AB=36cm.M is the midpoint of AB .There semicircles are drawn on AB AM and BM as diameters . Find the area of shaded region.
Answers
Answer:
Step-by-step explanation:
hey buddy!!!
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 = RQ =1\4×ab=1/4×36=9cm
⇒PR =r + ab \4 = r + 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=1 \2(36) - r = 18 - r
In Δ PMR,
By pythagoras theorem,
Shaded area = Area of semicircle ABC-Area of semicircle AME-Area of semicircle MBD-Area of circle CEDAR
hope it helps!!!
Answer:
Step-by-step explanation: diameter = 36/2 =16 cm
area of circle = pi r² = 3.14 * 8 * 8 = 200.96 cm²
since there are 2 semi circles area of shaded portion is 200.96 cm²