n the given figure, diameter AB is 12 cm long. AB is trisected at points P and Q. Find the area of the shaded region.

AP+PQ+QB=12cm
3AP=12cm (trisected)
AP=12/3= 4 cm
similarly, AP=PQ=QB=4cm

now,
AQ= 2AP= 8cm
and
PB=2QB=8cm
R=AQ/2=4cm
r= AP/2= 2cm
therefore,
area of shaded region= 2x (1/2 pie R^2 - 1/2 pie r^2)
or, 2x 132/7 = 37.7 cm^2 (approx.)

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