Here, we have
r = 15 cm, ө = 60°
Let OACBO be the given sector and OAB is a triangle. Then
(i) Area of the sector (OACBO)
(ii) Area of the triangle (AOB)
Now,
Area of Minor segment (ACBA)
= Area of sector (OACBO) - Area of triangle (AOB)
= (117.75 - 97.313) cm2
= 20.437 cm2
and Area of Major segment (ABDA)
= Area of circle-Area of Minor segment
= (πr3 - 20.437) cm2
= (3.14 × 15 × 15 - 20.437) cm2
= (706.5 - 20.437) cm2
= 686.063 cm2.
A horse tied to a corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see Fig. 12.11). Find
(i) The area of that part of the field in which the horse can graze.
(ii) The increase in the grazing area if the rope were 10 m long instead of 5m.
A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divides the circle into 10 equal sectors as shown in figure. Find
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.