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# help

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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

#1
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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
+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