If the curved surface area of a sphere increased by 44%, then by what percentage would its volume increase?
72.8%
Sphere Surface Area and Volume Relationship
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 $V = \frac{4}{3} \pi r^3$.
Calculating New Radius from Area Increase
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 $1.2r = r \times (1 + 0.20)$).
Calculating Volume Increase
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:
$V' = 1.728 V$
Determining Percentage Volume Increase
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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