Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 12.1, 3 Given figure depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions. Diameter of gold circle = 21 cm So, Radius of gold = π·πππππ‘ππ/2 = 21/2 cm = 10.5 cm Now, Area of gold = πrGOLD2 = 22/7 Γ 10.5 Γ 10.5 = 346.5 cm2 Now , Radius of red = Radius of gold + width of band = 10.5 + 10.5 = 21 cm Area of (red + gold ) = πrRED2 = 22/7Γ21Γ21 = 1386 cm2 Area of red = Area of (Red + gold ) β area of gold = 1386 β 346.5 = 1039.5 Radius of blue = radius of red + width of band = 21 + 10.5 = 31.5 cm Area of (Gold + red + blue) = Οr2 = 22/7Γ(31.5)2 = 22/7Γ31.5Γ31.5 = 22 Γ 4.5 Γ31.5 = 3118.5 cm2 Area of blue = Area of (red + gold + blue) β Area of (red + gold) = 3118.5 β 1386 = 1732.5 cm2 Radius of black = radius of blue + width of band = 31.5 + 10.5 = 42 cm Area of (Gold + red + blue + black) = Οr2 = 22/7 Γ (42)2 = 22/7 Γ 42 Γ 42 = 5544 Area of black portion = (Area of red + gold + blue + black) β (Area of red + gold + blue) = 5544 β 3118 . 5 = 2425.5 cm2 Radius of white = radius of black + width of band = 42 + 10.5 = 52.5 cm Area of (Gold + red + blue + black + white) = Οr2 = 22/7 Γ (52.5)2 = 22/7 Γ 52.5 Γ 52.5 = 8662.5 Area of white portion = (Area of gold + red + blue + black + white ) β (Area of red + gold + blue + black) = 8662.5 β 5544 = 3118 .5 cm2

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.