6.5: Area, Surface Area and Volume Formulas
- Page ID
- 51016
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Area formulas
Let \(b\) = base
Let \(h\) = height
Let \(s\) = side
Let \(r\) = radius
Shape Name |
Shape |
Area Formula |
---|---|---|
Rectangle |
\(A=bh\) |
|
Square |
\(\begin{array}{l} |
|
Parallelogram |
\(A=bh\) |
|
Triangle |
\(A=\dfrac{1}{2} b h\) |
|
Circle |
\(A=\pi r^{2}\) |
|
Trapezoid |
\(A=\dfrac{1}{2} h\left(b_{1}+b_{2}\right)\) |
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
Geometric Figure |
Surface Area Formula |
Surface Area Meaning |
---|---|---|
|
\(S A=2 B+P h\) |
Find the area of each face. Add up all areas. |
|
\(S A=B+\dfrac{1}{2} s P\) |
Find the area of each face. Add up all areas. |
|
\(S A=2 B+2 \pi r h\) |
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. |
\(S A=4 \pi r^{2}\) |
Find the area of the great circle and multiply it by 4. |
|
\(S A=B+\pi r S\) |
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
Geometric Figure |
Volume Formula |
Volume Meaning |
---|---|---|
\(V=B h\) |
Find the area of the base and multiply it by the height |
|
\(V=\dfrac{1}{3} B h\) |
Find the area of the base and multiply it by 1/3 of the height. |
|
|
\(V=B h\) |
Find the area of the base and multiply it by the height. |
|
\(V=\dfrac{4}{3} \pi r^{3}\) |
Find the area of the great circle and multiply it by the radius and then multiply it by 4/3. |
|
\( V=\dfrac{1}{3} B h\) |
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.
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.
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.
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.
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.
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
- Find the area of a triangle with a base of 40 inches and a height of 60 inches.
- Find the area of a square with a side of 15 feet.
- Find the surface area of Earth, which has a diameter of 7917.5 miles. Use 3.14 for PI.
- 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.
- Radius = 9 cm
- Diameter = 6 miles
- Radius = 8.6 cm
- Diameter = 14 meters
(Problems 5 – 8) Find the area of each polygon. Round answers to the nearest tenth.
(Problems 9 – 12) Name each figure.
(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.
(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.
Extension: Methods of Teaching Mathematics
Part 1
Assessments:
- What is the Difference between Formative and Summative Assessments? Which One is More Important?
- Formative Assessment Examples and When to Use Them
- 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.