Question

answer..... ......... ..........

Answer 1

Solution:
Let D be the centre of one of the semicircles.
Draw line DC & MC connecting the centres C,D & M.
Let R be the radius of the bigger semicircle, such that R/2 is the radius of the smaller semicircle & let r be the radius of the circle.

Here R = AB/2 = 36/2 = 18 cm

In ∆DCM,
DM = R/2 = 18/2 = 9cm
DC = R/2 + r = 18/2 + r = 9+r
& MC = R - r = 18-r

we know that ∆DCM is right angle triangle, so the Pythagorean Theorem is,
DM^2 + MC^2 = DC^2
or 9^2 + (18-r)^2 = (9+r)^2
or 81 + 324 - 36r +r^2 = 81 + 18r + r^2
or 324 = 18r + 36r
or r = 324/54
or r = 6 cm

Therefore,
Area of shaded region = area of bigger semicircle - area of 2 smaller semicircles - area of the circle
= πR^2 - 2×π×(R/2)^2 - πr^2
= π×18^2 - 2×π×(18/2)^2 - π×6^2
= 1017.87 - 508.94 - 113.09
= 395.84 cm^2

Hence, the area of shaded region is 395.84 cm^2