Ex 11.1, 9 - Chapter 11 Class 10 Areas related to Circles

Transcript

Ex 11.1, 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 Brooch is made with silver wire in 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 = 2𝜋𝑟 = 2×22/7×35/2 = 22 × 5 = 110 mm Wire used in 5 diameters Silver wire used in 5 diameters = 5 × Diameter of circle = 5 × 35 = 175 mm Now, Total wire is used = Length of wire in circle + Wire used in 5 diameters = 110 + 175 = 285 mm Ex 11.1, 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 = 𝟏/𝟏𝟎× Area of all sectors of the brooch Now, Area of all sectors of the brooch = Area of circle = 𝝅𝒓𝟐 = 22/7×(35/2)^2 = 22/7×35/2×35/2 = (𝟏𝟏 × 𝟓 × 𝟑𝟓)/𝟐 Hence, Area of each sector of the brooch = 𝟏/𝟏𝟎× Area of all sectors of the brooch = 1/10×(11 × 5 × 35)/2 = 𝟑𝟖𝟓/𝟒 mm2 ∴ Area of each sector = 385/4 mm2