If the radius of a sphere is decreased by 48%, then by what percent does its surface area decrease?

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Accepted Answer

**Solution:**

1. Let the original radius of the sphere be *r*.

2. The original surface area of the sphere is given by:
   
   $$4\pi r^2$$

3. The radius is decreased by 48%, so the new radius is:
   
   $$r_{new} = r - 0.48r = 0.52r$$

4. The new surface area becomes:
   
   $$4\pi (0.52r)^2 = 4\pi r^2 (0.52)^2 = 4\pi r^2 (0.2704)$$

5. The decrease in surface area is:
   
   $$4\pi r^2 - 4\pi r^2 (0.2704) = 4\pi r^2 (1 - 0.2704) = 4\pi r^2 (0.7296)$$

6. The percentage decrease in surface area is:
   
   $$\frac{4\pi r^2 - 4\pi r^2 (0.2704)}{4\pi r^2} 	imes 100 = 0.7296 	imes 100 = 72.96\%$$

**Answer:** 72.96\%

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