Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 12.2, 9 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 divide the circle into 10 equal sectors as shown in figure. Find : the total length of the silver wire required Here, Brooch is made with silver wire in the form of a circle. Diameter of the brooch = 35 mm Radius of brooch = 35/2 mm Since wire is used in making 5 diameter & circle. So, Total wire is used = length of wire in circle + Wire used in 5 diameters Length of wire in circle Length of wire in circle = circumference of circle = 2๐๐ = 2ร22/7ร35/2 = 22 ร 5 = 110 mm Silver wire used in 5 diameters = 5 ร Diameter of circle = 5 ร 35 = 175 mm Now, Total wire is used = length of wire in circle + Wire used in 5 diameters = 110 + 175 = 285 mm Ex 12.2, 9 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 divide the circle into 10 equal sectors as shown in figure. Find : (ii) the area of each sector of the brooch. Since the wire divides the circle into 10 equal sectors Area of each sector of the brooch = 1/10ร Area of all sectors of the brooch Area of all sectors of the brooch = area of circle = ๐๐2 = 22/7ร(35/2)^2 = 22/7ร35/2ร35/2 = (11 ร 5 ร 35)/2 Area of each sector of the brooch = 1/10ร Area of all sectors of the brooch = 1/10ร(11 ร 5 ร 35)/2 = 385/4 mm2 โด Area of each sector = 385/4 mm2

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.