GEOM 1 | Lesson 3 | Practice (Application Problems) Solutions

  1. Volume  = \(6\cdot10\cdot10\) or \(600\) cu in

\(SA = ph+2B\), where \(p\) is the perimeter of the base and \(B\) is the area of the base. Therefore:  \(SA=40\cdot6+2(100)\) or 440 sq in.

  1. 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:  \(V = \dfrac{4π}{3}• (1)^3\) 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.

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

Volume: \(\dfrac{1}{3}3600\cdot40\)  or \(48,000\) cu ft.

The surface area:   \(\dfrac{1}{2}50\cdot60\cdot4\) or \(6000\) sq ft

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Go to Lesson 4: Application of Area vs Volume

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