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 = /2 = 21/2 cm = 10.5 cm
Now,
Area of gold = rGOLD2
= 22/7 10.5 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 ) = rRED2
= 22/7 21 21
= 1386 cm2
Area of red = Area of (Red + gold ) area of gold
= 1386 346.5
= 1039.5
Radius of blue = radius of red + width of band
= 21 + 10.5
= 31.5 cm
Area of (Gold + red + blue) = r2
= 22/7 (31.5)2
= 22/7 31.5 31.5
= 22 4.5 31.5
= 3118.5 cm2
Area of blue = Area of (red + gold + blue) Area of (red + gold)
= 3118.5 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) = r2
= 22/7 (42)2
= 22/7 42 42
= 5544
Area of black portion
= (Area of red + gold + blue + black) (Area of red + gold + blue)
= 5544 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) = r2
= 22/7 (52.5)2
= 22/7 52.5 52.5
= 8662.5
Area of white portion
= (Area of gold + red + blue + black + white )
(Area of red + gold + blue + black)
= 8662.5 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 12 years. He provides courses for Maths and Science at Teachoo.