GEOM 1 | Lesson 3 | Practice Solution

 1).  Volume  = \(6• 10• 10\) or \(600\) cu in

\(SA = ph + 2B\), where \(p\) is the perimeter of the base and \(B\) is the area of the base. Therefore:  \(SA=40•6 + 2(100)\) or 440 sq in.

2).  The tennis balls have a diameter of 2 inches.

Volume of the container:  \(V = π(1)^2(6)\) or \(V ≈18.85\) cu in.

Volume of the ball:  \(V = \frac{4π}{3}• (1)^3\) or \(V ≈4.19\) cu in

Volume of air:  Volume of the container- Volume of 3 balls:  \(18.85 – 3(4.19)\) or 6.28 cu in.

3).  The area of the base is \(3600\)sq ft.  To find the altitude :  \(\sqrt{50^2 -30^2}\) or \(40’\).

Volume: \(\Large\frac{1}{3}\normalsize 3600•40\)  or \(48,000\) cu ft.

The surface area:   \(\Large\frac{1}{2}\normalsize 240•40\) or \(4800\) sq ft

 

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