Question

surface area formulas
$sa_{cylinder}=2a_{circle}+a_{rectangle}$
$sa_{cylinder}=2\pi r^{2}+2\pi rh$
$sa_{cone}=\pi r^{2}+\pi rs$
$s = \sqrt{r^{2}+h^{2}}$
$sa_{sphere}=4\pi r^{2}$
answer the following questions, showing all your work for full marks.

1. find the surface area of a sphere with a diameter of 5 cm.
2. find the diameter of the basketball to the nearest tenth of an inch given the surface area is 271.6 square inches.

Explanation:

Step1: Find radius for first question

Given diameter $d = 5$ cm, radius $r=\frac{d}{2}=\frac{5}{2}=2.5$ cm.

Step2: Calculate surface - area of sphere for first question

Use the formula $SA_{sphere}=4\pi r^{2}$. Substitute $r = 2.5$ cm. So $SA_{sphere}=4\pi(2.5)^{2}=4\pi\times6.25 = 25\pi\approx 25\times 3.14 = 78.5$ $cm^{2}$.

Step3: Set up equation for second question

Given $SA_{sphere}=271.6$ square inches and $SA_{sphere}=4\pi r^{2}$. So $4\pi r^{2}=271.6$.

Step4: Solve for $r$ in second question

First, divide both sides of the equation $4\pi r^{2}=271.6$ by $4\pi$. $r^{2}=\frac{271.6}{4\pi}$. Since $\pi\approx3.14$, $r^{2}=\frac{271.6}{4\times3.14}=\frac{271.6}{12.56}=21.6242$. Then $r=\sqrt{21.6242}\approx4.65$ inches.

Step5: Find diameter for second question

Since $d = 2r$, $d=2\times4.65 = 9.3$ inches.

Answer:

1. The surface - area of the sphere is approximately $78.5$ $cm^{2}$.
2. The diameter of the basketball is approximately $9.3$ inches.