. Fig. 12.3 depicts an archery target marked with its five scoring regions 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.Solution:The radius of 1st\xa0circle, r1\xa0= 21/2 cm (as diameter D is given as 21 cm)So, area of gold region = π r12\xa0= π(10.5)2\xa0= 346.5 cm2Now, it is given that each of the other bands is 10.5 cm wide,So, the radius of 2nd\xa0circle, r2\xa0= 10.5cm+10.5cm = 21 cmThus,∴ Area of red region = Area of 2nd\xa0circle − Area of gold region = (πr22−346.5) cm2= (π(21)2\xa0− 346.5) cm2= 1386 − 346.5= 1039.5 cm2Similarly,The radius of 3rd\xa0circle, r3\xa0= 21 cm+10.5 cm = 31.5 cmThe radius of 4th\xa0circle, r4\xa0= 31.5 cm+10.5 cm = 42 cmThe Radius of 5th\xa0circle, r5\xa0= 42 cm+10.5 cm = 52.5 cmFor the area of nth\xa0region,A = Area of circle n – Area of circle (n-1)∴ Area of blue region (n=3) = Area of third circle – Area of second circle= π(31.5)2\xa0– 1386 cm2= 3118.5 – 1386 cm2= 1732.5 cm2∴ Area of black region (n=4) = Area of fourth circle – Area of third circle= π(42)2\xa0– 1386 cm2= 5544 – 3118.5 cm2= 2425.5 cm2∴ Area of white region (n=5) = Area of fifth circle – Area of fourth circle= π(52.5)2\xa0– 5544 cm2= 8662.5 – 5544 cm2= 3118.5 cm2
Ok