Question 83e60

Sep 6, 2017

$\text{a)}$ Eight tennis balls

$\text{b)}$ $\text{1.}$ $268.08$ ${\text{cm}}^{3}$

$\text{ " "2.}$ $3267.26$ ${\text{cm}}^{3}$

$\text{c)}$ 34.4%

$\text{d)}$ $1734.16$ ${\text{cm}}^{2}$

Explanation:

$\text{a)}$ The radius of both the cylindrical container and each tennis ball is $4$ $\text{cm}$.

So, one ball can be filled at a time.

Also, as the radius is $4$ $\text{cm}$, the diameter of each ball must be $8$ $\text{cm}$.

If we had eight balls, they would occupy $8 \times 8$ $\text{cm}$ $= 64$ $\text{cm}$ in length.

The length of the container is $65$ $\text{cm}$, so we can exactly fit in it a maximum of eight tennis balls.

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$\text{b)}$

$\text{1.}$ The volume of a sphere is given as $V = \frac{4}{3} \pi {r}^{3}$:

$R i g h t a r r o w \text{Volume of each tennis ball} = \frac{4}{3} \times \pi \times {\left(4\right)}^{3}$ ${\text{cm}}^{3}$

$R i g h t a r r o w \text{Volume of each tennis ball} = \frac{4}{3} \times \pi \times 64$ ${\text{cm}}^{3}$

$R i g h t a r r o w \text{Volume of each tennis ball} = \frac{256}{3} \times \pi$ ${\text{cm}}^{3}$

$R i g h t a r r o w \text{Volume of each tennis ball} = 268.08257311$ ${\text{cm}}^{3}$

$\therefore \text{Volume of each tennis ball} = 268.08$ ${\text{cm}}^{3}$

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$\text{2.}$ The volume of a cylinder is given as $V = \pi {r}^{2} h$; where $h$ is the height, or length, of the cylinder:

$R i g h t a r r o w \text{Volume of the cylindrical container} = \pi \times {\left(4\right)}^{2} \times 65$ ${\text{cm}}^{3}$

$R i g h t a r r o w \text{Volume of the cylindrical container} = \pi \times 16 \times 65$ ${\text{cm}}^{3}$

$R i g h t a r r o w \text{Volume of the cylindrical container} = \pi \times 1040$ ${\text{cm}}^{3}$

$R i g h t a r r o w \text{Volume of the cylindrical container} = 3267.2563597$ ${\text{cm}}^{3}$

$\therefore \text{Volume of the cylindrical container} = 3267.26$ ${\text{cm}}^{3}$

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$\text{c)}$ From part $\text{b.}$ $\text{1.}$, the volume of each ball is around $268.08$ ${\text{cm}}^{3}$.

So the volume of all eight balls that can fit in the container is $8 \times 268.08$ ${\text{cm}}^{3} = 2144.64$ ${\text{cm}}^{3}$.

$\frac{2144.64 {\text{ cm"^(3))(3267.26 " cm}}^{3}}{=} 0.656$, or roughly 65.6% of the container is occupied by the eight balls.

Therefore, 100% - 65.6% = 34.4% of the container is not occupied by the balls.

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$\text{d)}$ The surface area of a cylinder is given as $A = 2 \pi r h + 2 \pi {r}^{2}$:

$R i g h t a r r o w \text{Surface area of the container} = 2 \times \pi \times 4 \times 65 + 2 \times \pi \times {4}^{2}$ ${\text{cm}}^{2}$

$R i g h t a r r o w \text{Surface area of the container} = 520 \times \pi + 32 \times \pi$ ${\text{cm}}^{2}$

$R i g h t a r r o w \text{Surface area of the container} = 552 \times \pi$ ${\text{cm}}^{2}$

$R i g h t a r r o w \text{Surface area of the container} = 1734.1591448$ ${\text{cm}}^{2}$

$\therefore \text{Surface area of the container} = 1734.16$ ${\text{cm}}^{2}$