12.1 ka question no 3

. 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 circle, r1 = 21/2 cm (as diameter D is given as 21 cm) So, area of gold region = π r12 = π(10.5)2 = 346.5 cm2 Now, it is given that each of the other bands is 10.5 cm wide, So, the radius of 2nd circle, r2 = 10.5cm+10.5cm = 21 cm Thus, ∴ Area of red region = Area of 2nd circle − Area of gold region = (πr22−346.5) cm2 = (π(21)2 − 346.5) cm2 = 1386 − 346.5 = 1039.5 cm2 Similarly, The radius of 3rd circle, r3 = 21 cm+10.5 cm = 31.5 cm The radius of 4th circle, r4 = 31.5 cm+10.5 cm = 42 cm The Radius of 5th circle, r5 = 42 cm+10.5 cm = 52.5 cm For the area of nth region, 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 – 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 – 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 – 5544 cm2 = 8662.5 – 5544 cm2 = 3118.5 cm2

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