A bowling ball cannot weigh more than 16 pounds and must have a diameter of \(8 \frac{1}{2}\) inches. How many square inches are in the surface area of a bowling ball before the finger holes are drilled? Express your answer as a common fraction in terms of \(\pi\).

Logic May 29, 2019

#1**+3 **

The maximum weight of a bowling ball doesn't matter for this question...maybe that is just a red herring.

We can find the surface area of a sphere only knowing its diameter.

surface area of sphere = 4 π ( radius )^{2}

radius = ( \(\frac12\) )( diameter ) = ( \(\frac12\) )( 8 + \(\frac12\) ) = ( \(\frac12\) )( \(\frac{17}{2}\) ) = \(\frac{17}{4}\)

surface area of sphere = 4 π ( \(\frac{17}{4}\) )^{2}

surface area of sphere = 4 π ( \(\frac{289}{16}\) )

surface area of sphere = \(\frac{289\pi}{4}\) (square inches)

hectictar May 29, 2019

#1**+3 **

Best Answer

The maximum weight of a bowling ball doesn't matter for this question...maybe that is just a red herring.

We can find the surface area of a sphere only knowing its diameter.

surface area of sphere = 4 π ( radius )^{2}

radius = ( \(\frac12\) )( diameter ) = ( \(\frac12\) )( 8 + \(\frac12\) ) = ( \(\frac12\) )( \(\frac{17}{2}\) ) = \(\frac{17}{4}\)

surface area of sphere = 4 π ( \(\frac{17}{4}\) )^{2}

surface area of sphere = 4 π ( \(\frac{289}{16}\) )

surface area of sphere = \(\frac{289\pi}{4}\) (square inches)

hectictar May 29, 2019