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6.5: Area, Surface Area and Volume Formulas

  • Page ID
    51016
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    Area formulas

    Let \(b\) = base

    Let \(h\) = height

    Let \(s\) = side

    Let \(r\) = radius

    Table 6.5.1: Area formulas

    Shape Name

    Shape

    Area Formula

    Rectangle

    clipboard_efe15c30b1007547b342fc746a7efcf20.png

    \(A=bh\)

    Square

    clipboard_ec24caea3d5296d300e6f1f98eccae4d4.png

    \(\begin{array}{l}
    A=b h \\
    A=s^{2}
    \end{array}\)

    Parallelogram

    clipboard_eb485ee8fb60488e5bde456d41c01adeb.png

    \(A=bh\)

    Triangle

    clipboard_e2212c051d12fc4634a8ee3f1eda28490.png

    \(A=\dfrac{1}{2} b h\)

    Circle

    clipboard_e062a8ba2f01f7dbdd5050dc75b4afcfb.png

    \(A=\pi r^{2}\)

    Trapezoid

    clipboard_e4f0ce691e022af3e89a66a7c366a6375.png

    \(A=\dfrac{1}{2} h\left(b_{1}+b_{2}\right)\)

    Surface Area Formulas

    Variables:

    \(SA\) = Surface Area

    \(B\) = area of the base of the figure

    \(P\) = perimeter of the base of the figure

    \(h\) = height

    \(s\) = slant height

    \(r\) = radius

    Table 6.5.2: Surface Area formulas

    Geometric Figure 

    Surface Area Formula 

    Surface Area Meaning 

    clipboard_e81f9b4881086fecb9335c925c494c4e6.png 

    \(S A=2 B+P h\) 

    Find the area of each face. Add up all areas. 

    clipboard_e917a9764b931cea8719511a37f526a89.png 

    \(S A=B+\dfrac{1}{2} s P\) 

    Find the area of each face. Add up all areas. 

    clipboard_eaf05e34afb977462650d27c4706fb637.png 

    \(S A=2 B+2 \pi r h\) 

    Find the area of the base, times 2, then add the areas to the areas of the rectangle, which is the circumference times the height.  

    clipboard_ea5d578ab0347abea37383a22a2492701.png

    \(S A=4 \pi r^{2}\)

    Find the area of the great circle and multiply it by 4. 

    clipboard_e9c547851612b12562b5185fe0605b0ad.png

    \(S A=B+\pi r S\) 

    Find the area of the base and add the product of the radius times the slant height times PI.  

    Volume Formulas

    Variables:

    \(SA\) = Surface Area

    \(B\) = area of the base of the figure

    \(P\) = perimeter of the base of the figure

    \(h\) = height

    \(s\) = slant height

    \(r\) = radius

    Table 6.5.3: Volume formulas

    Geometric Figure 

    Volume Formula 

    Volume Meaning 

    clipboard_e9a0de49a1375afc8775962a50a30a81b.png

    \(V=B h\) 

    Find the area of the base and multiply it by the height 

    clipboard_e3a949ff5ec42fae21bd6c9a6773a69dc.png

    \(V=\dfrac{1}{3} B h\) 

    Find the area of the base and multiply it by 1/3 of the height. 

    clipboard_e88f8e5771fdb263d07c69b2ddf933536.png 

    \(V=B h\) 

    Find the area of the base and multiply it by the height. 

    clipboard_e24e7962afe42ee50f4fc0758451cd560.png 

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

    Find the area of the great circle and multiply it by the radius and then multiply it by 4/3. 

    clipboard_ed295e65c6d76699d3776472f1c4b2c17.png 

    \( V=\dfrac{1}{3} B h\)

    Find the area of the base and multiply it by 1/3 of the height. 

    Example \(\PageIndex{1}\)

    Find the area of a circle with diameter of 14 feet.

    clipboard_e41b579d2e55c37d8145b71fc46929023.png
    Figure 6.5.1

    Solution

    \[\begin{aligned}A&=\pi r^{2}\\&=\pi(7)^{2}\\&=49 \pi \text {feet}^{2}\\&=153.86 \text {feet}^{2} \end{aligned} \nonumber \]

    Example \(\PageIndex{2}\)

    Find the area of a trapezoid with a height of 12 inches, and bases of 24 and 10 inches.

    clipboard_eaa167458276c034c1497fbee20c48d82.png
    Figure 6.5.2

    Solution

    \[\begin{aligned} A&=\dfrac{1}{2} h\left(b_{1}+b_{2}\right)\\ &=\dfrac{1}{2}(12)(24+10)\\ &=6(34)\\ &=204 \text { inches}^2 \end{aligned}\nonumber \]

    Example \(\PageIndex{3}\)

    Find the surface area of a cone with a slant height of 8 cm and a radius of 3 cm.

    clipboard_e7f0e0d855883d455b9c17bd32f5fbc15.png
    Figure 6.5.3

    Solution

    \[\begin{aligned}
    SA&= B+\pi rS\\ &=\left(\pi r^{2}\right)+\pi rs\\ &=\left(\pi\left(3^{2}\right)\right)+\pi(3)(8) \\
    &=9 \pi+24 \pi\\ &=33 \pi \text {cm}^{2}\\ &=103.62 \text {cm}^{2}
    \end{aligned} \nonumber \]

    Example \(\PageIndex{4}\)

    Find the surface area of a rectangular pyramid with a slant height of 10 yards, a base width (b) of 8 yards and a base length (h) of 12 yards.

    clipboard_e1966ee3a811a2998d43c8e1a01f65351.png
    Figure 6.5.4

    Solution

    \[\begin{aligned}
    SA&=B+\dfrac{1}{2} s P\\ 
    &=(b h)+\dfrac{1}{2} s(2 b+2 h) \\
    &=(8)(12)+\dfrac{1}{2}(10)(2(8)+2(12)) \\
    &=96+\dfrac{1}{2}(10)(16+24) \\
    &=96+5(40) \\
    &=296 \text { yards}^{2}
    \end{aligned} \nonumber \]

    Example \(\PageIndex{5}\)

    Find the volume of a sphere with a diameter of 6 meters.

    clipboard_e8f88daf0b42e79373a2f60aef45758dc.png
    Figure 6.5.5

    Solution

    \[\begin{aligned} V&=\dfrac{4}{3} \pi r^{3}\\ &=\dfrac{4}{3} \pi(3)^{3}\\ &=\dfrac{4}{3}(27 \pi)\\ &=36 \pi \text { meters }^{3}\\ &=113.04 \text { meters }^{3} \end{aligned} \nonumber \]

    Partner Activity 1

    1. Find the area of a triangle with a base of 40 inches and a height of 60 inches.
    2. Find the area of a square with a side of 15 feet.
    3. Find the surface area of Earth, which has a diameter of 7917.5 miles. Use 3.14 for PI.
    4. Find the volume of a can a soup, which has a radius of 2 inches and a height of 3 inches. Use 3.14 for PI.

    Practice Problems

    (Problems 1 – 4) Find the area of each circle with the given parameters. Use 3.14 for PI. Round your answer to the nearest tenth.

    1. Radius = 9 cm
    2. Diameter = 6 miles
    3. Radius = 8.6 cm
    4. Diameter = 14 meters

    (Problems 5 – 8) Find the area of each polygon. Round answers to the nearest tenth.

    1. clipboard_e7c11d2a06b604c10f855daeef8996806.png
    2.  clipboard_e71f2d376b673e2eeedce58fde84d357b.png
    3. clipboard_e2f13d1085ad717ca038c1450b452d3d1.png
    4. clipboard_e1abd666b733c9eb5f7c9c6b081cdd56a.png

    (Problems 9 – 12) Name each figure.

    1.  clipboard_e9e2cc2664937ad831fbb674ad9999f14.png
    2. clipboard_e0434f0d23c6a5e226b248f81e884b497.png
    3. clipboard_e73a8e4539aa29805f2121f8d54380633.png
    4. clipboard_e1b38ca5b795760abf9ea3f4ed234a5f3.png

    (Problems 13 – 17) Find the surface area of each figure. Leave your answers in terms of PI, if the answer contains PI. Round all other answers to the nearest hundredth.

    1. clipboard_e2209ea68bc005aa5e461f595277d41d9.png
    2. clipboard_e6cc4dcb798c0d02022f26a2f8f7598c0.png
    3. clipboard_e3a982ab4b0487b87494a4fc61271e4e0.png
    4. clipboard_e4b6f66b4963ff0e17e58b75a7b6405ba.png
    5. clipboard_e584c2ff25992a84eaa7e40796c297b23.png

    (Problems 18 – 25) Find the volume of each figure. Leave your answers in terms of PI, for answers that contain PI. Round all other answers to the nearest hundredth.

    1. clipboard_e0446db1e9293874d2e55a0a205226560.png
    2. clipboard_eac0f896eb03b29a9f16544a127879fbd.png
    3. clipboard_ee5baa989cb3bd5eb303b8cf115df882b.png
    4. clipboard_e5b24dffcb97d63b1510c5f204e131818.png
    5. clipboard_ebc8278c77a6141f1298d394b314e229e.png
    6. clipboard_e6599552f2317bca580dc5b815ff3baf1.png
    7. clipboard_ef7f0cda8e3c1436d52ef454dbff99d00.png
    8. clipboard_ef61bb9b9104ce69d8becd339515a44e1.png

    Extension: Methods of Teaching Mathematics

    Part 1

    Assessments:

    1. What is the Difference between Formative and Summative Assessments? Which One is More Important?
    2. Formative Assessment Examples and When to Use Them
    3. Summative Assessment Examples and When to Use Them

    Part 2

    Write a Formative and Summative Assessment for Your Lesson Plan

    Part 3

    Make sure you are working on Khan Academy throughout the semester.


    This page titled 6.5: Area, Surface Area and Volume Formulas is shared under a not declared license and was authored, remixed, and/or curated by Amy Lagusker.

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