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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 \(\pi\).

 May 29, 2019

Best Answer 

 #1
avatar+8829 
+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)

 May 29, 2019
 #1
avatar+8829 
+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

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