The diameter of a circle is 16. By what number must the radius be decreased in order to decrease the area of the circle by $48\pi$?
The area of a circle is given by the formula $A = \pi r^2$, where $r$ is the radius and $\pi$ is a constant. The diameter of a circle is twice the radius, so if the diameter is 16, then the radius is 8. The area of the circle with radius 8 is $A = \pi (8)^2 = 64\pi$. I want to find a new radius $r'$ such that the area of the circle with radius $r'$ is $64\pi - 48\pi = 16\pi$. Using the formula for the area of a circle, I can write $16\pi = \pi (r')^2$. Dividing both sides by $\pi$, I get $16 = (r')^2$. Taking the square root of both sides, I get $\pm 4 = r'$. Since the radius of a circle must be positive, I ignore the negative root and take $r' = 4$. This means that the radius must be decreased by $8 - 4 = 4$ units to decrease the area of the circle by $48\pi$. 