An area reduction of 90% corresponds to what percent diameter reduction?

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Multiple Choice

An area reduction of 90% corresponds to what percent diameter reduction?

Explanation:
Diameter changes follow the square-root rule with area. For a circle, area is proportional to the square of the diameter, so if the area is reduced to 10% of its original value (a 90% drop), then the new diameter must satisfy (D_new / D_old)^2 = 0.10. Taking the square root gives D_new / D_old ≈ 0.316, meaning the diameter is about 31.6% of its original size. That’s a reduction of about 68.4% in diameter. Among the given choices, the closest percentage is 75%, since rounding to the nearest whole number is common in these contexts.

Diameter changes follow the square-root rule with area. For a circle, area is proportional to the square of the diameter, so if the area is reduced to 10% of its original value (a 90% drop), then the new diameter must satisfy (D_new / D_old)^2 = 0.10. Taking the square root gives D_new / D_old ≈ 0.316, meaning the diameter is about 31.6% of its original size. That’s a reduction of about 68.4% in diameter. Among the given choices, the closest percentage is 75%, since rounding to the nearest whole number is common in these contexts.

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