If the radius of a circle is decreased by 50%, its area is reduced by:

Question

Examoogle · Accepted Answer

The correct answer is Option C: 75%. Use the successvie percentage change formula: $a + b + \frac{ab}{100}$. Since radius decreases by 50% twice in area (r²), change = $-50 - 50 + \frac{(-50 	imes -50)}{100} = -100 + 25 = -75\%$.

Step-by-Step Solution & Concepts:
Shortcut Trick: Use the successvie percentage change formula: $a + b + \frac{ab}{100}$. Since radius decreases by 50% twice in area (r²), change = $-50 - 50 + \frac{(-50 	imes -50)}{100} = -100 + 25 = -75\%$.

Formula Used: $$A = \pi r^2$$

Given: Initial radius of circle r_1 = r. The radius is decreased by 50%, so the new radius r_2 = 0.5r.

Steps: Define Initial Area: Let the original radius be $r$. The initial area is $A_1 = \pi r^2$.. Define New Radius: Since the radius is decreased by 50%, the new radius $r_2$ is $r - 0.5r = 0.5r$.. Calculate New Area: The new area $A_2 = \pi (0.5r)^2 = \pi (0.25 r^2) = 0.25 \pi r^2$.. Calculate Percentage Decrease: The reduction in area is $A_1 - A_2 = A_1 - 0.25 A_1 = 0.75 A_1$. Expressed as a percentage, this is $0.75 	imes 100\% = 75\%$.

Answer: C is correct because a 50% decrease in radius results in the new area being 25% of the original, which corresponds to a 75% reduction.

Exam Tip: This logic applies to any square-based relationship; for example, if the side of a square is halved, its area also decreases by 75%.

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