Today i am solving ncert class 10 chapter Areas Related To Circles it is very hard to solve exercise 12.1 question no. 3 please help me to solve this .Give me the best and easiest solution of this chapter. 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

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# 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.Q.3

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The radius of 1

^{st}Â circle, r_{1}Â = 21/2 cm (as diameter D is given as 21 cm)So, area of gold region = Ï€ r

_{1}^{2Â }= Ï€(10.5)^{2Â }= 346.5 cm^{2}Now, it is given that each of the other bands is 10.5 cm wide,

So, the radius of 2

^{nd}Â circle, r_{2}Â = 10.5cm+10.5cm = 21 cmThus,

âˆ´ Area of red region = Area of 2

^{nd}Â circle âˆ’ Area of gold region = (Ï€r_{2}^{2}âˆ’346.5) cm^{2}= (Ï€(21)

^{2}Â âˆ’ 346.5) cm^{2}= 1386 âˆ’ 346.5

= 1039.5 cm

^{2}Similarly,

The radius of 3

^{rd}Â circle, r_{3}Â = 21 cm+10.5 cm = 31.5 cmThe radius of 4

^{th}Â circle, r_{4}Â = 31.5 cm+10.5 cm = 42 cmThe Radius of 5

^{th}Â circle, r_{5}Â = 42 cm+10.5 cm = 52.5 cmFor the area of n

^{thÂ }region,A = Area of circle n â€“ Area of circle (n-1)

âˆ´ Area of blue region (n=3) = Area of third circle â€“ Area of second circle

= Ï€(31.5)

^{2}Â â€“ 1386 cm^{2}= 3118.5 â€“ 1386 cm

^{2}= 1732.5 cm

^{2}âˆ´ Area of black region (n=4) = Area of fourth circle â€“ Area of third circle

= Ï€(42)

^{2}Â â€“ 1386 cm^{2}= 5544 â€“ 3118.5 cm

^{2}= 2425.5 cm

^{2}âˆ´ Area of white region (n=5) = Area of fifth circle â€“ Area of fourth circle

= Ï€(52.5)

^{2}Â â€“ 5544 cm^{2}= 8662.5 â€“ 5544 cm

^{2}= 3118.5 cm

^{2}