The Central Idea
Different spatial questions require different kinds of measurement.
- Circumference measures a circular boundary.
- Area measures a two-dimensional region.
- Surface area measures the outer surface of a three-dimensional object.
- Volume measures three-dimensional interior space.
Their units reveal their dimensions:
\[ \boxed{ \text{length}\sim L,\qquad \text{area}\sim L^2,\qquad \text{volume}\sim L^3 }. \]From Perimeter to Circumference
The previous lesson defined perimeter as the total length of a closed boundary. A circle also has a closed boundary, but its boundary is curved.
The length of that boundary is called the circumference.
Circumference therefore plays the same measurement role for a circle that perimeter plays for a polygon.
Radius and Diameter
A circle is determined by a center and a radius.
The radius \(r\) is the distance from the center to any point on the circle.
The diameter \(d\) passes through the center and connects two points on the circle.
\[ \boxed{d=2r}. \]The Number \(\pi\)
For every Euclidean circle, the ratio of circumference to diameter is the same constant:
\[ \frac{C}{d}=\pi. \]Therefore,
\[ \pi\approx3.14159265\ldots \]The decimal expansion continues indefinitely without repeating.
The Circumference Formula
Since
\[ \frac{C}{d}=\pi, \]multiplying by \(d\) gives
\[ C=\pi d. \]Because \(d=2r\),
\[ \boxed{C=2\pi r}. \]Both formulas describe the same boundary length.
Example: Circumference from Radius
A circle has radius
\[ r=5\text{ cm}. \]Then
\[ C=2\pi(5)=10\pi\text{ cm}. \]The exact circumference is
\[ \boxed{10\pi\text{ cm}}. \]Numerically,
\[ C\approx31.42\text{ cm}. \]Example: Circumference from Diameter
A circular table has diameter
\[ d=1.4\text{ m}. \]Use
\[ C=\pi d. \]Then
\[ C=1.4\pi\text{ m}\approx4.40\text{ m}. \] \[ \boxed{C\approx4.40\text{ m}}. \]Circumference Uses Linear Units
Circumference is a boundary length, so it is measured in ordinary linear units:
\[ \text{cm},\qquad \text{m},\qquad \text{ft},\qquad \text{mi}. \]Even though the circle occupies a two-dimensional region, its circumference measures only the one-dimensional boundary.
What Is Area?
Area measures the size of a two-dimensional region.
Informally, area asks how many unit squares are needed to cover a region without gaps or overlaps.
A square whose side length is one unit has area
\[ 1\text{ square unit}. \]Why Area Uses Square Units
Area combines two independent length dimensions.
If a rectangle is \(4\text{ cm}\) long and \(3\text{ cm}\) wide, then it can be covered by
\[ 4\cdot3=12 \]unit squares.
Its area is therefore
\[ \boxed{12\text{ cm}^2}. \]Area of a Rectangle
If a rectangle has length \(l\) and width \(w\), then
\[ \boxed{A=lw}. \]This formula expresses the number of unit squares that fit across the two dimensions.
Area of a Square
A square is a rectangle whose side lengths are equal.
If each side has length \(s\), then
\[ \boxed{A=s^2}. \]The exponent \(2\) reflects the two-dimensional nature of area.
Area of a Parallelogram
A parallelogram can be cut and rearranged into a rectangle with the same base \(b\) and perpendicular height \(h\).
Therefore,
\[ \boxed{A=bh}. \]The height must be perpendicular to the chosen base.
Area of a Triangle
Two congruent copies of a triangle can be arranged to form a parallelogram.
Therefore, a triangle occupies half the area of a parallelogram with the same base and height:
\[ \boxed{A=\frac12 bh}. \]Example: Triangle Area
A triangle has base
\[ b=10\text{ m} \]and perpendicular height
\[ h=7\text{ m}. \]Then
\[ A=\frac12(10)(7)=35. \] \[ \boxed{A=35\text{ m}^2}. \]Area of a Trapezoid
Let the parallel bases of a trapezoid have lengths \(b_1\) and \(b_2\), with perpendicular height \(h\).
Its area is
\[ \boxed{ A=\frac12(b_1+b_2)h }. \]The formula may be interpreted as average base length times height.
Area of a Circle
A circle of radius \(r\) has area
\[ \boxed{A=\pi r^2}. \]The factor \(r^2\) reflects two-dimensional scaling, while \(\pi\) accounts for the circular geometry.
Example: Area of a Circle
A circle has radius
\[ r=6\text{ cm}. \]Its area is
\[ A=\pi(6)^2=36\pi\text{ cm}^2. \]Therefore,
\[ \boxed{A=36\pi\text{ cm}^2} \]or approximately
\[ A\approx113.10\text{ cm}^2. \]Circumference and Area Measure Different Things
For a circle of radius \(r\),
\[ C=2\pi r \]measures the boundary, while
\[ A=\pi r^2 \]measures the interior region.
Circumference has linear units. Area has square units.
Composite Areas
Many regions are built from simpler shapes.
A composite area can often be found by:
- adding the areas of nonoverlapping pieces;
- subtracting removed regions;
- rearranging pieces into familiar forms.
The essential rule is to account for every part of the target region exactly once.
Example: Area by Decomposition
Suppose an L-shaped region can be split into rectangles with dimensions
\[ 8\text{ m}\times3\text{ m} \]and
\[ 4\text{ m}\times2\text{ m}. \]Then
\[ A=8(3)+4(2)=24+8=32. \] \[ \boxed{A=32\text{ m}^2}. \]Area by Subtraction
Sometimes it is easier to begin with a larger familiar region and subtract a missing piece.
If a \(10\text{ cm}\times8\text{ cm}\) rectangle has a \(3\text{ cm}\times2\text{ cm}\) rectangular corner removed, then
\[ A=10(8)-3(2). \] \[ A=80-6=74. \] \[ \boxed{A=74\text{ cm}^2}. \]What Is Surface Area?
Surface area measures the total area of the boundary surface of a three-dimensional object.
For a polyhedron, surface area is the sum of the areas of its faces.
Curved solids require curved-surface formulas, but the underlying idea is the same: measure the two-dimensional boundary of a three-dimensional object.
Surface Area Still Uses Square Units
A three-dimensional object has surfaces that are themselves two-dimensional.
Surface area is therefore measured in square units:
\[ \text{cm}^2,\qquad \text{m}^2,\qquad \text{ft}^2. \]The object is three-dimensional, but the quantity being accumulated is area.
Surface Area of a Rectangular Prism
A rectangular prism with length \(l\), width \(w\), and height \(h\) has three pairs of congruent rectangular faces.
Their areas are
\[ lw,\qquad lh,\qquad wh. \]Therefore,
\[ \boxed{ SA=2lw+2lh+2wh }. \]Example: Rectangular-Prism Surface Area
Let
\[ l=8\text{ cm},\qquad w=5\text{ cm},\qquad h=3\text{ cm}. \]Then
\[ SA=2(8)(5)+2(8)(3)+2(5)(3). \] \[ SA=80+48+30=158. \] \[ \boxed{SA=158\text{ cm}^2}. \]Surface Area of a Cube
A cube has six congruent square faces.
If each edge has length \(s\), each face has area \(s^2\), so
\[ \boxed{SA=6s^2}. \]Nets and Surface Area
A net unfolds the surfaces of a three-dimensional object into a two-dimensional arrangement.
Nets make surface area visible as an ordinary area-summing problem.
For polyhedra, a reliable strategy is:
- identify every face;
- find each face area;
- add the face areas exactly once.
Surface Area of a Right Circular Cylinder
A closed right circular cylinder consists of two circular bases and one curved lateral surface.
The two bases contribute
\[ 2\pi r^2. \]When the curved surface is unwrapped, it becomes a rectangle whose width is the circumference \(2\pi r\) and whose height is \(h\).
Thus the lateral area is
\[ 2\pi rh. \]Total surface area is
\[ \boxed{ SA=2\pi r^2+2\pi rh }. \]Lateral Surface Area
Sometimes a problem asks only for the side surface and excludes the bases.
This is called lateral surface area.
For a right circular cylinder,
\[ \boxed{LA=2\pi rh}. \]Always determine whether a problem asks for lateral area or total surface area.
What Is Volume?
Volume measures the amount of three-dimensional space occupied by a solid.
Informally, volume asks how many unit cubes are needed to fill a solid without gaps or overlaps.
A cube with edge length one unit has volume
\[ 1\text{ cubic unit}. \]Why Volume Uses Cubic Units
Volume combines three independent length dimensions.
A rectangular prism measuring
\[ 4\text{ cm}\times3\text{ cm}\times2\text{ cm} \]contains
\[ 4\cdot3\cdot2=24 \]unit cubes.
Therefore,
\[ \boxed{V=24\text{ cm}^3}. \]Volume of a Rectangular Prism
A rectangular prism with length \(l\), width \(w\), and height \(h\) has volume
\[ \boxed{V=lwh}. \]Since \(lw\) is the area of the base, this may also be written
\[ \boxed{V=Bh}, \]where \(B\) denotes base area.
Volume of a Cube
A cube has equal length, width, and height.
If each edge has length \(s\), then
\[ \boxed{V=s^3}. \]The exponent \(3\) reflects the three-dimensional nature of volume.
Volume of a Prism
More generally, a prism with base area \(B\) and perpendicular height \(h\) has volume
\[ \boxed{V=Bh}. \]The shape of the base may vary. What matters is that congruent cross-sections extend through the prism along its height.
Volume of a Cylinder
A cylinder follows the same base-area-times-height principle.
Its circular base has area
\[ B=\pi r^2. \]Therefore,
\[ \boxed{V=\pi r^2h}. \]Example: Cylinder Volume
A cylinder has radius \(4\text{ cm}\) and height \(9\text{ cm}\).
Then
\[ V=\pi(4)^2(9) =144\pi\text{ cm}^3. \] \[ \boxed{V=144\pi\text{ cm}^3}. \]Approximately,
\[ V\approx452.39\text{ cm}^3. \]Volume of a Pyramid
A pyramid with base area \(B\) and perpendicular height \(h\) has one-third the volume of a prism with the same base and height.
\[ \boxed{ V=\frac13 Bh }. \]Volume of a Cone
A cone is the circular-base analogue of a pyramid.
Since its base area is \(\pi r^2\),
\[ \boxed{ V=\frac13\pi r^2h }. \]Volume of a Sphere
A sphere of radius \(r\) has volume
\[ \boxed{ V=\frac43\pi r^3 }. \]The cubic dependence on \(r\) is consistent with the three-dimensional nature of volume.
Surface Area of a Sphere
A sphere of radius \(r\) has surface area
\[ \boxed{ SA=4\pi r^2 }. \]Notice that surface area depends on \(r^2\), while volume depends on \(r^3\). This difference becomes important when objects are scaled.
Boundary versus Interior
Surface area and volume answer different questions.
For a box:
- surface area measures the cardboard needed to cover the outside;
- volume measures how much space is available inside.
The first measures a two-dimensional boundary. The second measures a three-dimensional interior.
Area and Volume Are Additive
If a region is decomposed into nonoverlapping pieces, its total area is the sum of the areas of the pieces.
\[ A=A_1+A_2+\cdots+A_n. \]Similarly, if a solid is decomposed into nonoverlapping pieces,
\[ V=V_1+V_2+\cdots+V_n. \]This principle supports composite-figure methods and later ideas of accumulation in calculus.
Composite Volumes
Complex solids can often be decomposed into familiar prisms, cylinders, pyramids, cones, or other standard solids.
Add the volumes of pieces that are present and subtract the volumes of cavities or removed pieces.
The method is structurally the same as composite-area reasoning, but the units are cubic.
Example: Volume by Subtraction
A rectangular block has dimensions
\[ 10\text{ cm}\times8\text{ cm}\times6\text{ cm}. \]A rectangular cavity of dimensions
\[ 4\text{ cm}\times3\text{ cm}\times2\text{ cm} \]is removed.
The remaining volume is
\[ V=10(8)(6)-4(3)(2). \] \[ V=480-24=456. \] \[ \boxed{V=456\text{ cm}^3}. \]Unit Consistency
All measurements in a geometric formula must use compatible units before they are combined.
For example, if a rectangle is \(2\text{ m}\) long and \(50\text{ cm}\) wide, convert one measurement first:
\[ 50\text{ cm}=0.5\text{ m}. \]Then
\[ A=2(0.5)=1\text{ m}^2. \]Converting Area Units
Area conversion factors must be squared.
Since
\[ 1\text{ m}=100\text{ cm}, \]it follows that
\[ 1\text{ m}^2 =(100\text{ cm})^2 =10{,}000\text{ cm}^2. \]A common mistake is to multiply by \(100\) rather than by \(100^2\).
Converting Volume Units
Volume conversion factors must be cubed.
Since
\[ 1\text{ m}=100\text{ cm}, \]then
\[ 1\text{ m}^3 =(100\text{ cm})^3 =1{,}000{,}000\text{ cm}^3. \]Dimensional exponents apply to the conversion factor as well as to the unit.
Linear, Square, and Cubic Units
A useful dimensional comparison is
\[ \boxed{ \begin{aligned} \text{length}&:\quad \text{unit},\\ \text{area}&:\quad \text{unit}^2,\\ \text{volume}&:\quad \text{unit}^3. \end{aligned} } \]The exponent on the unit records how many independent length dimensions are being measured.
Checking Formulas by Dimensions
Units provide a powerful error check.
Consider the rectangle-area formula
\[ A=lw. \]Since both \(l\) and \(w\) have dimension \(L\),
\[ L\cdot L=L^2, \]which is appropriate for area.
Dimensional Check for Volume
For a rectangular prism,
\[ V=lwh. \]Dimensionally,
\[ L\cdot L\cdot L=L^3. \]This confirms that the formula produces a volume rather than a length or area.
Exact Values with \(\pi\)
When possible, keep \(\pi\) in the answer until a decimal approximation is actually needed.
For a circle of radius \(7\),
\[ A=49\pi \]is exact.
Replacing \(\pi\) too early with \(3.14\) reduces precision and can accumulate rounding error.
Measured Dimensions and Approximate Results
If dimensions come from physical measurement, computed areas and volumes are also based on approximate data.
For example, if a measured side is reported as
\[ 4.2\text{ cm}, \]the value represents the precision of the measuring process, not an infinitely exact mathematical length.
Bounds and measurement error will be developed later in this section.
Perimeter and Area Can Change Differently
Two figures can have the same perimeter but different areas.
For example, a \(6\times2\) rectangle and a \(4\times4\) square both have perimeter
\[ 16, \]but their areas are
\[ 12 \]and
\[ 16, \]respectively.
Boundary length alone does not determine enclosed area.
Surface Area and Volume Can Change Differently
A solid's surface area and volume respond differently to changes in scale.
If every length is multiplied by a factor \(k\), then:
\[ \text{surface area is multiplied by }k^2, \]while
\[ \text{volume is multiplied by }k^3. \]A later lesson will develop these scaling laws systematically.
Example: Doubling a Cube
A cube with edge length \(s\) has
\[ SA=6s^2 \]and
\[ V=s^3. \]If the edge becomes \(2s\), then
\[ SA'=6(2s)^2=4(6s^2)=4SA, \]while
\[ V'=(2s)^3=8s^3=8V. \]Doubling length quadruples surface area and multiplies volume by eight.
Common Error: Using Perimeter When Area Is Needed
A floor measuring \(8\text{ m}\) by \(5\text{ m}\) requires an area calculation:
\[ A=8(5)=40\text{ m}^2. \]The perimeter
\[ 2(8)+2(5)=26\text{ m} \]answers a different question about the boundary.
Common Error: Using Diameter as Radius
Suppose a circle has diameter \(10\text{ cm}\).
Then the radius is
\[ r=5\text{ cm}. \]The area is
\[ A=\pi(5)^2=25\pi\text{ cm}^2, \]not
\[ 100\pi\text{ cm}^2. \]Common Error: Forgetting the Second Base of a Cylinder
A closed cylinder has two circular bases.
Its total surface area is
\[ 2\pi r^2+2\pi rh, \]not
\[ \pi r^2+2\pi rh. \]A diagram or net can help verify that every surface has been counted.
Common Error: Confusing Surface Area with Volume
For a rectangular prism,
\[ 2lw+2lh+2wh \]measures the outer surface, while
\[ lwh \]measures the interior volume.
The units provide an immediate check: square units for surface area and cubic units for volume.
Common Error: Using Slant Height as Perpendicular Height
Volume formulas such as
\[ V=\frac13Bh \]use the perpendicular height \(h\), not a slanted edge or slant height.
The required height measures perpendicular separation between the base and the opposite vertex or parallel plane.
Common Error: Converting Units Linearly
Since
\[ 1\text{ m}=100\text{ cm}, \]it does not follow that
\[ 1\text{ m}^2=100\text{ cm}^2. \]The correct conversion is
\[ \boxed{ 1\text{ m}^2=10{,}000\text{ cm}^2 }. \]For volume, the conversion factor must be cubed.
Common Error: Rounding Too Early
When formulas contain \(\pi\), radicals, or measured decimal values, keep enough precision through intermediate steps.
Prefer
\[ V=72\pi \]until the final stage rather than repeatedly replacing \(\pi\) with a short decimal.
A Reliable Procedure for Circumference
- Identify whether the given measure is radius or diameter.
- Convert between them if necessary using \(d=2r\).
- Use \(C=2\pi r\) or \(C=\pi d\).
- Keep \(\pi\) exact until a decimal is needed.
- Report the result in linear units.
A Reliable Procedure for Area
- Identify the two-dimensional region being measured.
- Choose a formula appropriate to the shape.
- Identify the required perpendicular heights when relevant.
- Decompose or subtract regions if the figure is composite.
- Convert all lengths to compatible units.
- Compute and simplify.
- Report the result in square units.
A Reliable Procedure for Surface Area
- Identify every exposed surface.
- Use a net when it helps reveal the faces.
- Find the area of each face or curved surface.
- Determine whether bases are included.
- Add each required surface exactly once.
- Report the result in square units.
A Reliable Procedure for Volume
- Identify the three-dimensional solid or its component solids.
- Choose the correct volume formula.
- Use perpendicular height where the formula requires it.
- Convert dimensions to compatible units.
- Add or subtract component volumes when necessary.
- Keep exact constants such as \(\pi\) until the end.
- Report the result in cubic units.
Example: Full Circle Workflow
A circle has diameter
\[ d=18\text{ cm}. \]The radius is
\[ r=9\text{ cm}. \]Circumference:
\[ C=\pi d=18\pi\text{ cm}. \]Area:
\[ A=\pi r^2=81\pi\text{ cm}^2. \]Thus,
\[ \boxed{C=18\pi\text{ cm}} \]and
\[ \boxed{A=81\pi\text{ cm}^2}. \]Example: Full Rectangular-Prism Workflow
A rectangular prism has dimensions
\[ 10\text{ cm}\times6\text{ cm}\times4\text{ cm}. \]Surface area:
\[ SA=2(10)(6)+2(10)(4)+2(6)(4). \] \[ SA=120+80+48=248\text{ cm}^2. \]Volume:
\[ V=10(6)(4)=240\text{ cm}^3. \]Therefore,
\[ \boxed{SA=248\text{ cm}^2} \]and
\[ \boxed{V=240\text{ cm}^3}. \]Example: Full Cylinder Workflow
A closed cylinder has radius \(3\text{ m}\) and height \(8\text{ m}\).
Circumference of each base:
\[ C=2\pi(3)=6\pi\text{ m}. \]Area of each base:
\[ A=\pi(3)^2=9\pi\text{ m}^2. \]Total surface area:
\[ SA=2\pi(3)^2+2\pi(3)(8) =18\pi+48\pi =66\pi\text{ m}^2. \]Volume:
\[ V=\pi(3)^2(8)=72\pi\text{ m}^3. \]Therefore,
\[ \boxed{SA=66\pi\text{ m}^2} \]and
\[ \boxed{V=72\pi\text{ m}^3}. \]Practice 1: Circumference
Find the exact circumference of a circle with radius \(7\text{ cm}\).
Practice 2: Circle Area
Find the exact area of a circle with diameter \(12\text{ m}\).
Practice 3: Triangle Area
Find the area of a triangle with base \(15\text{ cm}\) and perpendicular height \(8\text{ cm}\).
Practice 4: Trapezoid Area
A trapezoid has parallel bases \(9\text{ m}\) and \(15\text{ m}\), with height \(6\text{ m}\). Find its area.
Practice 5: Cube Surface Area
Find the surface area of a cube with edge length \(5\text{ cm}\).
Practice 6: Rectangular-Prism Volume
Find the volume of a rectangular prism measuring \(9\text{ m}\times4\text{ m}\times3\text{ m}\).
Practice 7: Cylinder Volume
Find the exact volume of a cylinder with radius \(5\text{ cm}\) and height \(10\text{ cm}\).
Practice 8: Pyramid Volume
A pyramid has base area \(48\text{ m}^2\) and perpendicular height \(9\text{ m}\). Find its volume.
Practice 9: Unit Conversion
Convert
\[ 2.5\text{ m}^2 \]to square centimeters.
Practice 10: Dimensional Reasoning
Every length of a solid is multiplied by \(3\). By what factor do its surface area and volume change?
Practice Answers
Practice 1
\[ C=2\pi(7)=14\pi. \] \[ \boxed{C=14\pi\text{ cm}}. \]Practice 2
The radius is \(6\text{ m}\), so
\[ A=\pi(6)^2=36\pi. \] \[ \boxed{A=36\pi\text{ m}^2}. \]Practice 3
\[ A=\frac12(15)(8)=60. \] \[ \boxed{A=60\text{ cm}^2}. \]Practice 4
\[ A=\frac12(9+15)(6) =\frac12(24)(6) =72. \] \[ \boxed{A=72\text{ m}^2}. \]Practice 5
\[ SA=6(5)^2=150. \] \[ \boxed{SA=150\text{ cm}^2}. \]Practice 6
\[ V=9(4)(3)=108. \] \[ \boxed{V=108\text{ m}^3}. \]Practice 7
\[ V=\pi(5)^2(10)=250\pi. \] \[ \boxed{V=250\pi\text{ cm}^3}. \]Practice 8
\[ V=\frac13(48)(9)=144. \] \[ \boxed{V=144\text{ m}^3}. \]Practice 9
Since
\[ 1\text{ m}^2=10{,}000\text{ cm}^2, \]then
\[ 2.5\text{ m}^2 =2.5(10{,}000) =25{,}000\text{ cm}^2. \] \[ \boxed{25{,}000\text{ cm}^2}. \]Practice 10
Surface area scales with the square of the length factor:
\[ 3^2=9. \]Volume scales with the cube of the length factor:
\[ 3^3=27. \]Therefore,
\[ \boxed{\text{surface area factor}=9} \]and
\[ \boxed{\text{volume factor}=27}. \]What to Remember
- Circumference is the length of a circle's boundary.
- \(C=2\pi r=\pi d\).
- Area measures a two-dimensional region and uses square units.
- Surface area measures the two-dimensional boundary of a three-dimensional object.
- Volume measures three-dimensional interior space and uses cubic units.
- Composite measurements can be found by decomposition and accumulation.
- Unit conversions must respect dimensional exponents.
- Surface area and volume are different quantities even for the same solid.
- Exact forms involving \(\pi\) should usually be retained until approximation is needed.
- Dimensional reasoning provides a useful check on formulas and units.
The central structural distinction is simple: boundaries, regions, surfaces, and interiors have different dimensions and therefore require different measurements.
Where We Go Next
Length, circumference, area, surface area, and volume measure spatial extent. The next lesson, Angular Measure in Degrees and Radians, turns from size to rotation and direction.
Angles will provide a second major measurement system for spatial relationships and will connect geometry to circles, trigonometry, periodicity, and later analysis.