# 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