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9 years, 4 months ago
9 years, 4 months ago
Since the decrease is uniform, the distance illuminated after 2 hours = 5 meters
Area initially illuminated = $$\pi * 9^2$$
Area illuminated after 2 hours = $$\pi * 5^2$$
Required area = $$81 \pi - 25 \pi = 56*22/7 = 176 m^2$$
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