Geometry and Measurement - Ninth Grade Math
1. Area and Perimeter: Word Problems
Area: The amount of space inside a 2D shape (measured in square units: $\text{cm}^2$, $\text{m}^2$, $\text{ft}^2$)
Perimeter: The total distance around the outside of a 2D shape (measured in linear units: cm, m, ft)
Perimeter: The total distance around the outside of a 2D shape (measured in linear units: cm, m, ft)
Area Formulas for 2D Shapes
| Shape | Area Formula | Variables |
|---|---|---|
| Square | $A = s^2$ | $s$ = side length |
| Rectangle | $A = l \times w$ | $l$ = length, $w$ = width |
| Triangle | $A = \frac{1}{2}bh$ | $b$ = base, $h$ = height |
| Circle | $A = \pi r^2$ | $r$ = radius, $\pi \approx 3.14$ |
| Parallelogram | $A = bh$ | $b$ = base, $h$ = height |
| Trapezoid | $A = \frac{1}{2}(b_1 + b_2)h$ | $b_1, b_2$ = bases, $h$ = height |
| Rhombus | $A = \frac{1}{2}d_1 d_2$ | $d_1, d_2$ = diagonals |
Perimeter Formulas for 2D Shapes
| Shape | Perimeter Formula | Variables |
|---|---|---|
| Square | $P = 4s$ | $s$ = side length |
| Rectangle | $P = 2l + 2w$ or $P = 2(l + w)$ | $l$ = length, $w$ = width |
| Triangle | $P = a + b + c$ | $a, b, c$ = side lengths |
| Circle (Circumference) | $C = 2\pi r$ or $C = \pi d$ | $r$ = radius, $d$ = diameter |
| Regular Polygon | $P = n \times s$ | $n$ = number of sides, $s$ = side length |
Example 1: A rectangular garden has a length of 15 meters and width of 8 meters. Find the area and perimeter.
Area: $A = l \times w = 15 \times 8 = 120 \text{ m}^2$
Perimeter: $P = 2(l + w) = 2(15 + 8) = 2(23) = 46 \text{ m}$
Area: $A = l \times w = 15 \times 8 = 120 \text{ m}^2$
Perimeter: $P = 2(l + w) = 2(15 + 8) = 2(23) = 46 \text{ m}$
Example 2: A circular pizza has a radius of 10 inches. Find the area and circumference. (Use $\pi = 3.14$)
Area: $A = \pi r^2 = 3.14 \times 10^2 = 3.14 \times 100 = 314 \text{ in}^2$
Circumference: $C = 2\pi r = 2 \times 3.14 \times 10 = 62.8 \text{ in}$
Area: $A = \pi r^2 = 3.14 \times 10^2 = 3.14 \times 100 = 314 \text{ in}^2$
Circumference: $C = 2\pi r = 2 \times 3.14 \times 10 = 62.8 \text{ in}$
Example 3: A triangle has a base of 12 cm and height of 8 cm. Find the area.
Area: $A = \frac{1}{2}bh = \frac{1}{2} \times 12 \times 8 = \frac{96}{2} = 48 \text{ cm}^2$
Area: $A = \frac{1}{2}bh = \frac{1}{2} \times 12 \times 8 = \frac{96}{2} = 48 \text{ cm}^2$
Word Problem Tips:
• Identify the shape involved
• Write down given information
• Choose the correct formula
• Substitute values and solve
• Include proper units in your answer
• Identify the shape involved
• Write down given information
• Choose the correct formula
• Substitute values and solve
• Include proper units in your answer
2. Volume
Volume: The amount of space occupied by a 3D object (measured in cubic units: $\text{cm}^3$, $\text{m}^3$, $\text{ft}^3$)
Volume Formulas for 3D Shapes
| 3D Shape | Volume Formula | Variables |
|---|---|---|
| Cube | $V = s^3$ | $s$ = side length |
| Rectangular Prism (Cuboid) | $V = l \times w \times h$ | $l$ = length, $w$ = width, $h$ = height |
| Cylinder | $V = \pi r^2 h$ | $r$ = radius, $h$ = height |
| Sphere | $V = \frac{4}{3}\pi r^3$ | $r$ = radius |
| Cone | $V = \frac{1}{3}\pi r^2 h$ | $r$ = radius, $h$ = height |
| Pyramid | $V = \frac{1}{3}Bh$ | $B$ = base area, $h$ = height |
| Prism | $V = Bh$ | $B$ = base area, $h$ = height |
General Volume Principle:
• Prisms & Cylinders: $V = (\text{Base Area}) \times \text{Height}$
• Pyramids & Cones: $V = \frac{1}{3} \times (\text{Base Area}) \times \text{Height}$
• Sphere: $V = \frac{4}{3}\pi r^3$
• Prisms & Cylinders: $V = (\text{Base Area}) \times \text{Height}$
• Pyramids & Cones: $V = \frac{1}{3} \times (\text{Base Area}) \times \text{Height}$
• Sphere: $V = \frac{4}{3}\pi r^3$
Example 1: Find the volume of a rectangular box with length 10 cm, width 6 cm, and height 4 cm.
$V = l \times w \times h = 10 \times 6 \times 4 = 240 \text{ cm}^3$
$V = l \times w \times h = 10 \times 6 \times 4 = 240 \text{ cm}^3$
Example 2: A cylinder has a radius of 5 m and height of 12 m. Find its volume. (Use $\pi = 3.14$)
$V = \pi r^2 h = 3.14 \times 5^2 \times 12 = 3.14 \times 25 \times 12 = 942 \text{ m}^3$
$V = \pi r^2 h = 3.14 \times 5^2 \times 12 = 3.14 \times 25 \times 12 = 942 \text{ m}^3$
Example 3: A sphere has a radius of 6 inches. Find its volume. (Use $\pi = 3.14$)
$V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times 3.14 \times 6^3 = \frac{4}{3} \times 3.14 \times 216 = 904.32 \text{ in}^3$
$V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times 3.14 \times 6^3 = \frac{4}{3} \times 3.14 \times 216 = 904.32 \text{ in}^3$
Example 4: A cone has a radius of 4 cm and height of 9 cm. Find its volume.
$V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times 3.14 \times 4^2 \times 9 = \frac{1}{3} \times 3.14 \times 16 \times 9 = 150.72 \text{ cm}^3$
$V = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times 3.14 \times 4^2 \times 9 = \frac{1}{3} \times 3.14 \times 16 \times 9 = 150.72 \text{ cm}^3$
3. Surface Area
Surface Area: The total area of all surfaces (faces) of a 3D object (measured in square units: $\text{cm}^2$, $\text{m}^2$)
Lateral Surface Area (LSA): Area of the sides only (excluding bases)
Total Surface Area (TSA): Area of all faces including bases
Lateral Surface Area (LSA): Area of the sides only (excluding bases)
Total Surface Area (TSA): Area of all faces including bases
Surface Area Formulas for 3D Shapes
| 3D Shape | Total Surface Area Formula | Variables |
|---|---|---|
| Cube | $SA = 6s^2$ | $s$ = side length |
| Rectangular Prism | $SA = 2(lw + lh + wh)$ | $l$ = length, $w$ = width, $h$ = height |
| Cylinder | $SA = 2\pi r^2 + 2\pi rh$ or $SA = 2\pi r(r + h)$ | $r$ = radius, $h$ = height |
| Sphere | $SA = 4\pi r^2$ | $r$ = radius |
| Cone | $SA = \pi r^2 + \pi r l$ or $SA = \pi r(r + l)$ | $r$ = radius, $l$ = slant height |
| Square Pyramid | $SA = s^2 + 2sl$ | $s$ = base side, $l$ = slant height |
Lateral Surface Area (LSA) Formulas:
• Cylinder: $LSA = 2\pi rh$ (curved surface only)
• Cone: $LSA = \pi rl$ (curved surface only)
• Rectangular Prism: $LSA = 2h(l + w)$ (4 sides, no top/bottom)
• Cylinder: $LSA = 2\pi rh$ (curved surface only)
• Cone: $LSA = \pi rl$ (curved surface only)
• Rectangular Prism: $LSA = 2h(l + w)$ (4 sides, no top/bottom)
Example 1: Find the surface area of a cube with side length 5 cm.
$SA = 6s^2 = 6 \times 5^2 = 6 \times 25 = 150 \text{ cm}^2$
$SA = 6s^2 = 6 \times 5^2 = 6 \times 25 = 150 \text{ cm}^2$
Example 2: A rectangular prism has dimensions: length = 8 m, width = 5 m, height = 3 m. Find TSA.
$SA = 2(lw + lh + wh)$
$= 2(8 \times 5 + 8 \times 3 + 5 \times 3)$
$= 2(40 + 24 + 15)$
$= 2(79) = 158 \text{ m}^2$
$SA = 2(lw + lh + wh)$
$= 2(8 \times 5 + 8 \times 3 + 5 \times 3)$
$= 2(40 + 24 + 15)$
$= 2(79) = 158 \text{ m}^2$
Example 3: A cylinder has radius 4 cm and height 10 cm. Find the total surface area.
$SA = 2\pi r(r + h) = 2 \times 3.14 \times 4(4 + 10)$
$= 6.28 \times 4 \times 14 = 351.68 \text{ cm}^2$
$SA = 2\pi r(r + h) = 2 \times 3.14 \times 4(4 + 10)$
$= 6.28 \times 4 \times 14 = 351.68 \text{ cm}^2$
Example 4: A sphere has a radius of 7 inches. Find its surface area.
$SA = 4\pi r^2 = 4 \times 3.14 \times 7^2 = 4 \times 3.14 \times 49 = 615.44 \text{ in}^2$
$SA = 4\pi r^2 = 4 \times 3.14 \times 7^2 = 4 \times 3.14 \times 49 = 615.44 \text{ in}^2$
Remember:
• Volume is in cubic units ($\text{cm}^3$, $\text{m}^3$)
• Surface Area is in square units ($\text{cm}^2$, $\text{m}^2$)
• For cylinder and cone, you need both radius and height
• Volume is in cubic units ($\text{cm}^3$, $\text{m}^3$)
• Surface Area is in square units ($\text{cm}^2$, $\text{m}^2$)
• For cylinder and cone, you need both radius and height
4. Precision
Precision: The smallest unit or increment to which a measurement is made
Unit of Precision: The smallest place value of a measurement
Example: 5.7 cm has precision to the nearest 0.1 cm (tenths place)
Unit of Precision: The smallest place value of a measurement
Example: 5.7 cm has precision to the nearest 0.1 cm (tenths place)
Determining Precision:
The precision is determined by the rightmost digit in a measurement
Examples:
• 15 cm → Precision: 1 cm (ones place)
• 15.3 cm → Precision: 0.1 cm (tenths place)
• 15.35 cm → Precision: 0.01 cm (hundredths place)
• 2,500 m → Precision: 100 m (hundreds place)
• 0.005 g → Precision: 0.001 g (thousandths place)
The precision is determined by the rightmost digit in a measurement
Examples:
• 15 cm → Precision: 1 cm (ones place)
• 15.3 cm → Precision: 0.1 cm (tenths place)
• 15.35 cm → Precision: 0.01 cm (hundredths place)
• 2,500 m → Precision: 100 m (hundreds place)
• 0.005 g → Precision: 0.001 g (thousandths place)
| Measurement | Unit of Precision | Place Value |
|---|---|---|
| 8 m | 1 m | Ones |
| 8.0 m | 0.1 m | Tenths |
| 8.75 kg | 0.01 kg | Hundredths |
| 12,300 km | 100 km | Hundreds |
| 0.0045 L | 0.0001 L | Ten-thousandths |
Example 1: What is the precision of 35.6 meters?
Answer: The rightmost digit is in the tenths place, so precision = 0.1 m
Answer: The rightmost digit is in the tenths place, so precision = 0.1 m
Example 2: Which measurement is more precise: 12 cm or 12.5 cm?
Answer: 12.5 cm is more precise (0.1 cm vs 1 cm)
Answer: 12.5 cm is more precise (0.1 cm vs 1 cm)
Example 3: A scale measures to the nearest gram. What is the precision?
Answer: Precision = 1 gram
Answer: Precision = 1 gram
Key Points:
• More decimal places = greater precision
• Precision refers to the measuring instrument's capability
• Smaller unit of precision = more precise measurement
• More decimal places = greater precision
• Precision refers to the measuring instrument's capability
• Smaller unit of precision = more precise measurement
5. Greatest Possible Error
Greatest Possible Error (GPE): The maximum amount by which a measurement could differ from the actual value
Also called: Margin of Error, Tolerance
Also called: Margin of Error, Tolerance
Greatest Possible Error Formula:
$$\text{GPE} = \frac{\text{Unit of Precision}}{2}$$
or
$$\text{GPE} = \frac{\text{Precision}}{2}$$
$$\text{GPE} = \frac{\text{Unit of Precision}}{2}$$
or
$$\text{GPE} = \frac{\text{Precision}}{2}$$
Measurement Range:
If a measurement is $m$ with GPE = $e$, then:
$$\text{Actual Value} = m \pm e$$
Minimum Value: $m - e$
Maximum Value: $m + e$
If a measurement is $m$ with GPE = $e$, then:
$$\text{Actual Value} = m \pm e$$
Minimum Value: $m - e$
Maximum Value: $m + e$
| Measurement | Precision | GPE | Range |
|---|---|---|---|
| 15 cm | 1 cm | 0.5 cm | 14.5 cm to 15.5 cm |
| 8.3 m | 0.1 m | 0.05 m | 8.25 m to 8.35 m |
| 2.75 kg | 0.01 kg | 0.005 kg | 2.745 kg to 2.755 kg |
| 120 mm | 10 mm | 5 mm | 115 mm to 125 mm |
Example 1: A length is measured as 25 cm to the nearest centimeter. Find the GPE and the range.
Precision: 1 cm
GPE: $\frac{1}{2} = 0.5$ cm
Range: $25 - 0.5$ to $25 + 0.5$ = 24.5 cm to 25.5 cm
Precision: 1 cm
GPE: $\frac{1}{2} = 0.5$ cm
Range: $25 - 0.5$ to $25 + 0.5$ = 24.5 cm to 25.5 cm
Example 2: A weight is measured as 3.8 kg to the nearest tenth of a kilogram. Find the GPE.
Precision: 0.1 kg
GPE: $\frac{0.1}{2} = 0.05$ kg
Range: 3.75 kg to 3.85 kg
Precision: 0.1 kg
GPE: $\frac{0.1}{2} = 0.05$ kg
Range: 3.75 kg to 3.85 kg
Example 3: A distance is measured as 45.25 meters. What is the GPE?
Precision: 0.01 m (hundredths place)
GPE: $\frac{0.01}{2} = 0.005$ m
Range: 45.245 m to 45.255 m
Precision: 0.01 m (hundredths place)
GPE: $\frac{0.01}{2} = 0.005$ m
Range: 45.245 m to 45.255 m
Important:
• GPE is always half the unit of precision
• The actual value lies within: [measurement − GPE, measurement + GPE]
• Used to express uncertainty in measurements
• GPE is always half the unit of precision
• The actual value lies within: [measurement − GPE, measurement + GPE]
• Used to express uncertainty in measurements
6. Minimum and Maximum Area and Volume
Concept: When measurements have errors, the calculated area or volume will also have a range (minimum and maximum values)
For Area (Rectangle):
• Maximum Area: Use maximum length × maximum width
$$A_{\text{max}} = (l + \text{GPE}_l) \times (w + \text{GPE}_w)$$
• Minimum Area: Use minimum length × minimum width
$$A_{\text{min}} = (l - \text{GPE}_l) \times (w - \text{GPE}_w)$$
• Maximum Area: Use maximum length × maximum width
$$A_{\text{max}} = (l + \text{GPE}_l) \times (w + \text{GPE}_w)$$
• Minimum Area: Use minimum length × minimum width
$$A_{\text{min}} = (l - \text{GPE}_l) \times (w - \text{GPE}_w)$$
For Volume (Rectangular Prism):
• Maximum Volume: Use maximum dimensions
$$V_{\text{max}} = (l + \text{GPE}_l) \times (w + \text{GPE}_w) \times (h + \text{GPE}_h)$$
• Minimum Volume: Use minimum dimensions
$$V_{\text{min}} = (l - \text{GPE}_l) \times (w - \text{GPE}_w) \times (h - \text{GPE}_h)$$
• Maximum Volume: Use maximum dimensions
$$V_{\text{max}} = (l + \text{GPE}_l) \times (w + \text{GPE}_w) \times (h + \text{GPE}_h)$$
• Minimum Volume: Use minimum dimensions
$$V_{\text{min}} = (l - \text{GPE}_l) \times (w - \text{GPE}_w) \times (h - \text{GPE}_h)$$
General Rule:
• Maximum: Use upper bounds of all measurements
• Minimum: Use lower bounds of all measurements
• Measured: Use the given measurements
• Maximum: Use upper bounds of all measurements
• Minimum: Use lower bounds of all measurements
• Measured: Use the given measurements
Example 1: A rectangle has length 10 cm and width 6 cm, both measured to the nearest cm. Find the maximum and minimum area.
Step 1: Find GPE
Precision = 1 cm, so GPE = 0.5 cm
Step 2: Find ranges
Length: 9.5 cm to 10.5 cm
Width: 5.5 cm to 6.5 cm
Step 3: Calculate areas
Measured Area: $10 \times 6 = 60 \text{ cm}^2$
Maximum Area: $10.5 \times 6.5 = 68.25 \text{ cm}^2$
Minimum Area: $9.5 \times 5.5 = 52.25 \text{ cm}^2$
Step 1: Find GPE
Precision = 1 cm, so GPE = 0.5 cm
Step 2: Find ranges
Length: 9.5 cm to 10.5 cm
Width: 5.5 cm to 6.5 cm
Step 3: Calculate areas
Measured Area: $10 \times 6 = 60 \text{ cm}^2$
Maximum Area: $10.5 \times 6.5 = 68.25 \text{ cm}^2$
Minimum Area: $9.5 \times 5.5 = 52.25 \text{ cm}^2$
Example 2: A box has dimensions 5 m × 4 m × 3 m, each measured to the nearest meter. Find the maximum and minimum volume.
GPE: $\frac{1}{2} = 0.5$ m
Ranges:
Length: 4.5 m to 5.5 m
Width: 3.5 m to 4.5 m
Height: 2.5 m to 3.5 m
Measured Volume: $5 \times 4 \times 3 = 60 \text{ m}^3$
Maximum Volume: $5.5 \times 4.5 \times 3.5 = 86.625 \text{ m}^3$
Minimum Volume: $4.5 \times 3.5 \times 2.5 = 39.375 \text{ m}^3$
GPE: $\frac{1}{2} = 0.5$ m
Ranges:
Length: 4.5 m to 5.5 m
Width: 3.5 m to 4.5 m
Height: 2.5 m to 3.5 m
Measured Volume: $5 \times 4 \times 3 = 60 \text{ m}^3$
Maximum Volume: $5.5 \times 4.5 \times 3.5 = 86.625 \text{ m}^3$
Minimum Volume: $4.5 \times 3.5 \times 2.5 = 39.375 \text{ m}^3$
Example 3: A square has side length 8.0 cm (measured to nearest 0.1 cm). Find maximum and minimum area.
GPE: $\frac{0.1}{2} = 0.05$ cm
Range: 7.95 cm to 8.05 cm
Measured Area: $8.0^2 = 64 \text{ cm}^2$
Maximum Area: $8.05^2 = 64.8025 \text{ cm}^2$
Minimum Area: $7.95^2 = 63.2025 \text{ cm}^2$
GPE: $\frac{0.1}{2} = 0.05$ cm
Range: 7.95 cm to 8.05 cm
Measured Area: $8.0^2 = 64 \text{ cm}^2$
Maximum Area: $8.05^2 = 64.8025 \text{ cm}^2$
Minimum Area: $7.95^2 = 63.2025 \text{ cm}^2$
Key Steps:
1. Determine the precision of each measurement
2. Calculate the GPE for each dimension
3. Find the upper and lower bounds
4. Use maximum bounds for maximum area/volume
5. Use minimum bounds for minimum area/volume
1. Determine the precision of each measurement
2. Calculate the GPE for each dimension
3. Find the upper and lower bounds
4. Use maximum bounds for maximum area/volume
5. Use minimum bounds for minimum area/volume
7. Percent Error
Percent Error: A measure of how accurate a measurement is, expressed as a percentage of the actual value
Also called: Percentage Error, Relative Error (as percent)
Also called: Percentage Error, Relative Error (as percent)
Percent Error Formula:
$$\text{Percent Error} = \left|\frac{\text{Measured Value} - \text{Actual Value}}{\text{Actual Value}}\right| \times 100\%$$
or
$$\text{PE} = \left|\frac{M - A}{A}\right| \times 100\%$$
where $M$ = Measured/Experimental value, $A$ = Actual/Theoretical value
$$\text{Percent Error} = \left|\frac{\text{Measured Value} - \text{Actual Value}}{\text{Actual Value}}\right| \times 100\%$$
or
$$\text{PE} = \left|\frac{M - A}{A}\right| \times 100\%$$
where $M$ = Measured/Experimental value, $A$ = Actual/Theoretical value
Steps to Calculate Percent Error:
Step 1: Find the difference: Measured − Actual
Step 2: Take the absolute value of the difference
Step 3: Divide by the actual value
Step 4: Multiply by 100 to get percentage
Step 5: Add the % symbol
Step 1: Find the difference: Measured − Actual
Step 2: Take the absolute value of the difference
Step 3: Divide by the actual value
Step 4: Multiply by 100 to get percentage
Step 5: Add the % symbol
Example 1: A student measures a length as 24.5 cm when the actual length is 25 cm. Find the percent error.
$\text{PE} = \left|\frac{24.5 - 25}{25}\right| \times 100\%$
$= \left|\frac{-0.5}{25}\right| \times 100\%$
$= 0.02 \times 100\% = 2\%$
Answer: 2% error
$\text{PE} = \left|\frac{24.5 - 25}{25}\right| \times 100\%$
$= \left|\frac{-0.5}{25}\right| \times 100\%$
$= 0.02 \times 100\% = 2\%$
Answer: 2% error
Example 2: The actual weight of an object is 50 kg, but a scale reads 52 kg. Calculate the percent error.
$\text{PE} = \left|\frac{52 - 50}{50}\right| \times 100\%$
$= \left|\frac{2}{50}\right| \times 100\%$
$= 0.04 \times 100\% = 4\%$
Answer: 4% error
$\text{PE} = \left|\frac{52 - 50}{50}\right| \times 100\%$
$= \left|\frac{2}{50}\right| \times 100\%$
$= 0.04 \times 100\% = 4\%$
Answer: 4% error
Example 3: A clock shows 3:15 PM when the actual time is 3:12 PM. Find the percent error in minutes.
Measured = 195 minutes (from midnight)
Actual = 192 minutes (from midnight)
$\text{PE} = \left|\frac{195 - 192}{192}\right| \times 100\%$
$= \left|\frac{3}{192}\right| \times 100\%$
$\approx 1.56\%$
Answer: 1.56% error
Measured = 195 minutes (from midnight)
Actual = 192 minutes (from midnight)
$\text{PE} = \left|\frac{195 - 192}{192}\right| \times 100\%$
$= \left|\frac{3}{192}\right| \times 100\%$
$\approx 1.56\%$
Answer: 1.56% error
Interpreting Percent Error:
• 0%: Perfect measurement (no error)
• 1-5%: Very good measurement
• 5-10%: Acceptable in many cases
• >10%: Significant error, measurement needs improvement
Note: We use absolute value so percent error is always positive
• 0%: Perfect measurement (no error)
• 1-5%: Very good measurement
• 5-10%: Acceptable in many cases
• >10%: Significant error, measurement needs improvement
Note: We use absolute value so percent error is always positive
8. Percent Error: Area and Volume
Concept: When dimensions have measurement errors, the calculated area or volume will also have percent error
Percent Error for Area:
$$\text{PE}_{\text{area}} = \left|\frac{\text{Measured Area} - \text{Actual Area}}{\text{Actual Area}}\right| \times 100\%$$
$$\text{PE}_{\text{area}} = \left|\frac{\text{Measured Area} - \text{Actual Area}}{\text{Actual Area}}\right| \times 100\%$$
Percent Error for Volume:
$$\text{PE}_{\text{volume}} = \left|\frac{\text{Measured Volume} - \text{Actual Volume}}{\text{Actual Volume}}\right| \times 100\%$$
$$\text{PE}_{\text{volume}} = \left|\frac{\text{Measured Volume} - \text{Actual Volume}}{\text{Actual Volume}}\right| \times 100\%$$
Important Relationship:
• If each dimension has percent error $e\%$, then:
• Area percent error ≈ $2e\%$ (for small errors)
• Volume percent error ≈ $3e\%$ (for small errors)
Example: If length error = 2%, width error = 2%
Then area error ≈ 4% (approximately)
• If each dimension has percent error $e\%$, then:
• Area percent error ≈ $2e\%$ (for small errors)
• Volume percent error ≈ $3e\%$ (for small errors)
Example: If length error = 2%, width error = 2%
Then area error ≈ 4% (approximately)
Example 1: A rectangle's dimensions are measured as 10 cm × 6 cm, but actual dimensions are 10.2 cm × 6.1 cm. Find the percent error in area.
Measured Area: $10 \times 6 = 60 \text{ cm}^2$
Actual Area: $10.2 \times 6.1 = 62.22 \text{ cm}^2$
$\text{PE} = \left|\frac{60 - 62.22}{62.22}\right| \times 100\%$
$= \left|\frac{-2.22}{62.22}\right| \times 100\%$
$\approx 3.57\%$
Answer: 3.57% error in area
Measured Area: $10 \times 6 = 60 \text{ cm}^2$
Actual Area: $10.2 \times 6.1 = 62.22 \text{ cm}^2$
$\text{PE} = \left|\frac{60 - 62.22}{62.22}\right| \times 100\%$
$= \left|\frac{-2.22}{62.22}\right| \times 100\%$
$\approx 3.57\%$
Answer: 3.57% error in area
Example 2: A cube's side is measured as 5 cm but the actual side is 5.1 cm. Find the percent error in volume.
Measured Volume: $5^3 = 125 \text{ cm}^3$
Actual Volume: $5.1^3 = 132.651 \text{ cm}^3$
$\text{PE} = \left|\frac{125 - 132.651}{132.651}\right| \times 100\%$
$= \left|\frac{-7.651}{132.651}\right| \times 100\%$
$\approx 5.77\%$
Answer: 5.77% error in volume
Measured Volume: $5^3 = 125 \text{ cm}^3$
Actual Volume: $5.1^3 = 132.651 \text{ cm}^3$
$\text{PE} = \left|\frac{125 - 132.651}{132.651}\right| \times 100\%$
$= \left|\frac{-7.651}{132.651}\right| \times 100\%$
$\approx 5.77\%$
Answer: 5.77% error in volume
Example 3: A cylinder has measured radius 4 cm and height 10 cm. Actual values are radius 4.05 cm and height 10.1 cm. Find percent error in volume.
Measured Volume: $\pi r^2 h = 3.14 \times 4^2 \times 10 = 502.4 \text{ cm}^3$
Actual Volume: $\pi r^2 h = 3.14 \times 4.05^2 \times 10.1 = 520.26 \text{ cm}^3$
$\text{PE} = \left|\frac{502.4 - 520.26}{520.26}\right| \times 100\%$
$\approx 3.43\%$
Answer: 3.43% error in volume
Measured Volume: $\pi r^2 h = 3.14 \times 4^2 \times 10 = 502.4 \text{ cm}^3$
Actual Volume: $\pi r^2 h = 3.14 \times 4.05^2 \times 10.1 = 520.26 \text{ cm}^3$
$\text{PE} = \left|\frac{502.4 - 520.26}{520.26}\right| \times 100\%$
$\approx 3.43\%$
Answer: 3.43% error in volume
Example 4: Using GPE - A square has side 8 cm (measured to nearest cm). Find the maximum percent error in area.
GPE: 0.5 cm
Range: 7.5 cm to 8.5 cm
Measured Area: $8^2 = 64 \text{ cm}^2$
Maximum Area: $8.5^2 = 72.25 \text{ cm}^2$
Minimum Area: $7.5^2 = 56.25 \text{ cm}^2$
Maximum Error from Measured:
Upper: $|72.25 - 64| = 8.25$
Lower: $|56.25 - 64| = 7.75$
Use larger: 8.25
$\text{PE}_{\text{max}} = \frac{8.25}{64} \times 100\% \approx 12.89\%$
Answer: Maximum 12.89% error
GPE: 0.5 cm
Range: 7.5 cm to 8.5 cm
Measured Area: $8^2 = 64 \text{ cm}^2$
Maximum Area: $8.5^2 = 72.25 \text{ cm}^2$
Minimum Area: $7.5^2 = 56.25 \text{ cm}^2$
Maximum Error from Measured:
Upper: $|72.25 - 64| = 8.25$
Lower: $|56.25 - 64| = 7.75$
Use larger: 8.25
$\text{PE}_{\text{max}} = \frac{8.25}{64} \times 100\% \approx 12.89\%$
Answer: Maximum 12.89% error
Key Insights:
• Errors in dimensions magnify in area and volume calculations
• Area has approximately 2× the linear percent error
• Volume has approximately 3× the linear percent error
• Always use absolute values in calculations
• Compare calculated value to actual/true value
• Errors in dimensions magnify in area and volume calculations
• Area has approximately 2× the linear percent error
• Volume has approximately 3× the linear percent error
• Always use absolute values in calculations
• Compare calculated value to actual/true value
Quick Reference Guide
Essential 2D Formulas:
• Rectangle: $A = lw$, $P = 2(l+w)$
• Square: $A = s^2$, $P = 4s$
• Triangle: $A = \frac{1}{2}bh$
• Circle: $A = \pi r^2$, $C = 2\pi r$
• Trapezoid: $A = \frac{1}{2}(b_1 + b_2)h$
• Rectangle: $A = lw$, $P = 2(l+w)$
• Square: $A = s^2$, $P = 4s$
• Triangle: $A = \frac{1}{2}bh$
• Circle: $A = \pi r^2$, $C = 2\pi r$
• Trapezoid: $A = \frac{1}{2}(b_1 + b_2)h$
Essential 3D Formulas:
• Cube: $V = s^3$, $SA = 6s^2$
• Rectangular Prism: $V = lwh$, $SA = 2(lw + lh + wh)$
• Cylinder: $V = \pi r^2h$, $SA = 2\pi r(r+h)$
• Sphere: $V = \frac{4}{3}\pi r^3$, $SA = 4\pi r^2$
• Cone: $V = \frac{1}{3}\pi r^2h$, $SA = \pi r(r+l)$
• Cube: $V = s^3$, $SA = 6s^2$
• Rectangular Prism: $V = lwh$, $SA = 2(lw + lh + wh)$
• Cylinder: $V = \pi r^2h$, $SA = 2\pi r(r+h)$
• Sphere: $V = \frac{4}{3}\pi r^3$, $SA = 4\pi r^2$
• Cone: $V = \frac{1}{3}\pi r^2h$, $SA = \pi r(r+l)$
Measurement Error Formulas:
• Greatest Possible Error: $\text{GPE} = \frac{\text{Precision}}{2}$
• Range: $[\text{Measured} - \text{GPE}, \text{Measured} + \text{GPE}]$
• Maximum Area: Use upper bounds for all dimensions
• Minimum Area: Use lower bounds for all dimensions
• Greatest Possible Error: $\text{GPE} = \frac{\text{Precision}}{2}$
• Range: $[\text{Measured} - \text{GPE}, \text{Measured} + \text{GPE}]$
• Maximum Area: Use upper bounds for all dimensions
• Minimum Area: Use lower bounds for all dimensions
Percent Error Formula:
$$\text{Percent Error} = \left|\frac{\text{Measured} - \text{Actual}}{\text{Actual}}\right| \times 100\%$$
Quick Approximations:
• If linear error = $e\%$, then area error ≈ $2e\%$
• If linear error = $e\%$, then volume error ≈ $3e\%$
$$\text{Percent Error} = \left|\frac{\text{Measured} - \text{Actual}}{\text{Actual}}\right| \times 100\%$$
Quick Approximations:
• If linear error = $e\%$, then area error ≈ $2e\%$
• If linear error = $e\%$, then volume error ≈ $3e\%$
Success Tips:
✓ Always identify the shape before choosing a formula
✓ Write down all given information
✓ Include proper units in your answer (cm², m³, etc.)
✓ For word problems, draw a diagram if possible
✓ Check if your answer makes sense in context
✓ Remember: Volume → cubic units, Surface Area → square units
✓ Use absolute value for percent error calculations
✓ Always identify the shape before choosing a formula
✓ Write down all given information
✓ Include proper units in your answer (cm², m³, etc.)
✓ For word problems, draw a diagram if possible
✓ Check if your answer makes sense in context
✓ Remember: Volume → cubic units, Surface Area → square units
✓ Use absolute value for percent error calculations