If the curved surface area of a sphere increased by 44%, then by what percentage would its volume increase?
72.8%
Let the initial radius of the sphere be r, the initial curved surface area be A, and the initial volume be V.
The formula for the curved surface area of a sphere is $A = 4 \pi r^2$.
The formula for the volume of a sphere is $V = \frac{4}{3} \pi r^3$.
The curved surface area increases by 44%. The new area, A', is 1.44 times the original area.
$A' = A \times (1 + 0.44) = 1.44A$
Let the new radius be r'.
$4 \pi r'^2 = 1.44 \times (4 \pi r^2)$
$r'^2 = 1.44 r^2$
Taking the square root of both sides:
$r' = \sqrt{1.44 r^2} = 1.2 r$
This shows the radius increases by 20% (since $1.2r = r \times (1 + 0.20)$).
Now, let's find the new volume, V', using the new radius r'.
$V' = \frac{4}{3} \pi r'^3$
Substitute $r' = 1.2 r$:
$V' = \frac{4}{3} \pi (1.2 r)^3$
$V' = \frac{4}{3} \pi (1.728 r^3)$
$V' = 1.728 \times (\frac{4}{3} \pi r^3)$
Since $V = \frac{4}{3} \pi r^3$, we have:
$V' = 1.728 V$
The percentage increase in volume is calculated as:
Percentage Increase = $ \frac{V' - V}{V} \times 100\% $
Substitute $V' = 1.728V$:
Percentage Increase = $ \frac{1.728 V - V}{V} \times 100\% $
Percentage Increase = $ \frac{0.728 V}{V} \times 100\% $
Percentage Increase = $ 0.728 \times 100\% = 72.8\% $
Therefore, the volume would increase by 72.8%.
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