Measurement and Geometry Unit

3.7 Volume and Surface Area

  1. An oil barrel is approximately 15 feet tall and has a diameter of 10 feet. How many cubic meters of oil is needed to fill the barrel so it is 75% full?

    2. Volume of a Cylinder

    \( \begin{equation} V= \pi r^{2} h \end{equation} \)

    \( \begin{equation} V= \pi * 5^{2} * 15 \end{equation} \)

    \( \begin{equation} V = 1178 \text{ ft}^{3} \text { when full} \end{equation} \)

    75% full: \( V = 1178 \text{ ft}^{3}\times 0.75 = 883.6 \text{ ft}^{3} \)

    \(\frac{883.6\;ft^3}{tub}\times\frac{1^3\;yd^3}{3^3\;ft^3}\times\frac{1^3\;m^3}{1.09^3yd^3}=25.27\;m^3\)

  2. You plan to wrap the 55” TV you bought for your partner for Christmas. The box it came in is 56” long, 8” deep and 36” tall. How many square feet of wrapping paper will you need to wrap it?

    3. Surface Area of a Rectangular Prism

    SA = 2LW + 2LH + 2WH

    SA = 2(56)(8) + 2(56)(36) + 2(8)(36)

    \( SA = 5504 \text{ in}^{2} \)

    \(\frac{5504\;in^2}{box}\times\frac{1^2\;ft^2}{12^2\;in^2}=38.2\;ft^2 \)

  3. The circumference of a standard basketball is approximately 75 centimeters. The diameter of a basketball goal rim is 18 inches.
    1. How much bigger is the diameter of the rim than the basketball in inches?

      2. Circumference of a Circle: \(C=\pi d\)

      The circumference of the basketball is 75 cm.

      \(75=\pi d\)

      \( d=\frac{75}{\pi} =23.87 cm\)

      \( \frac{23.87\;cm}{ball}\times\frac{1\;in}{2.54\;cm}=9.4\;in \)

      The diameter of a standard basketball is 9.4 inches.

      The diameter of the goal is 18 inches.

      18 - 9.4 = 8.6

      The rim is 8.6 inches larger in circumference than the basketball.

    2. How many square centimeters of rubber does it take to cover the outer surface of a standard basketball?

      3. Surface Area of a Sphere

      \( SA = 4 \pi r^{2} \)

      The diameter of the sphere (from part a) is 23.87 cm

      The radius is \(\frac12\) the diameter.

      \( r = \frac{23.87}{2}=11.935 \;cm \)

      \( SA = 4 \pi * 11.935^{2} =1790.01 \text{ cm}^{2} \)

      It will take \( 1790.01 \text{ cm}^{2} \) to cover the outer surface of a standard basketball.

    3. How many cubic inches of air does it take to fill a standard basketball?

      4. Volume of a Sphere

      \( V = \frac{4}{3} \pi r^{3} \)

      The diameter of the basketball (from part a) is 9.4 inches.

      The radius is \(\frac12\) the diameter.

      \( r = \frac{9.4}{2}=4.7 \;in \)

      \( V = \frac{4}{3} \pi * 4.7^{3} =434.89 \text{ in}^{3}\)

      The volume of the basketball is \( 434.89 \text{ in}^{3} \) .