Let O be the centre of the circle and let the points where the arcs meet
be C and D respectively. ABCD is a square since its sides are all equal
to the radius of the arc CD and ÐACB=90° (angle in a
semicircle).
In triangle OCB, CB2 = OC2 + OB2; hence CB=Ö2 cm. The area of
the segment bounded by arc CD and diameter CD is equal to the area of
section BCD - the area of the triangle BCD, i.e.
æ ç
è
14
p(Ö2)2-
12
×Ö2× Ö2
ö ÷
ø
cm2,
i.e.
(
12
p- 1)
cm2.
The unshaded area in the original figure is, therefore, (p-2) cm2.
Now the area of the circle is p cm2, and hence the shaded area is
2 cm2.
