Swampy Survey Slides Swampy Survey Slides
Lesson 1: Beyond Base and Height
Trigonometry Unit
The Swampy Situation
"You just bought a triangular plot of land. To calculate the price, you need the area. But there's a problem..."
The Problem:
The center of the plot is an impassable swamp. You can measure the boundaries and the corners, but you cannot measure the height across the middle.
h = ? base = 300m side a = 210m 35° 🐊
How do we find the area if we can't measure the height?
01
Review: The Old Reliable
Standard Area Formula
\( Area = \frac{1}{2}bh \)
Works for all triangles.
Requires a perpendicular height.
Useless if height is unknown or unmeasurable.
The Surveyor's toolkit
Usually, a surveyor can easily measure:
Boundary Sides
Corner Angles
Goal: Write the area formula using only sides and angles.
02
The Derivation
Step 1: The Basic Formula
\( Area = \frac{1}{2} \cdot base \cdot height \)
\( Area = \frac{1}{2} \cdot c \cdot h \)
Step 2: Find "h" using SOH
In the right triangle on the left:
\( \sin(A) = \frac{h}{b} \)
\( h = b \cdot \sin(A) \)
A B C h c (base) b a
Step 3: Substitute
\( Area = \frac{1}{2}bc \cdot \sin(A) \)
03
The Sine Area Formula
\( \frac{1}{2}ab \sin(C) \)
\( \frac{1}{2}bc \sin(A) \)
\( \frac{1}{2}ac \sin(B) \)
X Y Z 42° 12 m 15 m
Check Your Logic:
1. Identify the two sides: 12 and 15.
2. Identify the included angle: 42°.
3. Plug into the formula:
\( Area = \frac{1}{2}(12)(15)\sin(42°) \)
\( Area \approx 60.22 \text{ m}^2 \)
Formula Forge Worksheet Formula Forge Worksheet
Geodetic Survey Division • Technical Report 1.1
Surveyor:
Field Date:
Part 1: Field Derivation
A standard area survey requires a height measurement (\(h\)). If the interior of the triangle is inaccessible (e.g., a swamp), we must substitute the height with measurable perimeter variables.
1. Standard Formula:
\( Area = \frac{1}{2} \cdot \text{base} \cdot \text{height} \)
2. Trigonometric Substitution:
Using Right Triangle Trigonometry (SOH), express the height (\(h\)) in terms of side \(b\) and angle \(A\):
3. The Surveyor's Sine Formula:
Substitute your expression for \(h\) into the standard formula to create the final area formula:
\( Area = \)
A B C b h c
Survey Diagram 1.A
Part 2: Precision Calculations
Calculate the area for the following plots. Round all final results to two decimal places. Include units.
1
Plot Alpha (SAS)
Side a = 14 km
Side b = 18 km
Angle C = 54°
2
Plot Beta (Diagram)
48° 22.5 m 30.0 m
3
The Isosceles Estate
Two sides of an isosceles triangular park are 120m long. The angle between them is 35°. Find the area.
4
Trap Check!
Area = 1/2 ab sin(C). Which of these cannot be solved with one step of the formula?
Sides 10, 12; angle 40° between them.
Three sides given, no angles given.
Side 5, 8; opposite angle of 8 is 30°.
Part 3: Surveyor's Dilemma
Scenario:
"A client presents a triangle with sides of 10cm and 15cm. They claim the area is 100 sq cm. Is this physically possible? Use the Sine Area Formula to justify your answer."
Formula Forge Answer Key Formula Forge Answer Key
Geodetic Survey Division • Internal Use Only
Official Key
Part 1: Field Derivation
2. Trigonometric Substitution:
\( \sin(A) = \frac{h}{b} \implies h = b \cdot \sin(A) \)
3. The Surveyor's Sine Formula:
\( Area = \frac{1}{2}bc \cdot \sin(A) \)
Part 2: Precision Calculations
1. Plot Alpha
\( A = \frac{1}{2}(14)(18)\sin(54^\circ) \)
2. Plot Beta
\( A = \frac{1}{2}(22.5)(30)\sin(48^\circ) \)
3. Isosceles Estate
\( A = \frac{1}{2}(120)(120)\sin(35^\circ) \)
4. Trap Check!
Sides 10, 12; angle 40° between them.
Three sides given, no angles given.
Side 5, 8; opposite angle of 8 is 30°.
*SAS requirement: The angle MUST be between the sides.*
Part 3: Surveyor's Dilemma
Verdict: NOT PHYSICALLY POSSIBLE
Justification: The maximum area for a triangle with sides \(a\) and \(b\) occurs when \(\sin(C)\) is at its maximum value of 1 (a 90° angle).
\( \text{Max Area} = \frac{1}{2}(10)(15)\sin(90^\circ) = 75 \text{ cm}^2 \).
Since 100 cm² exceeds the absolute maximum area possible with those side lengths, the claim is false.
Beyond Base Teacher Guide Teacher Facilitation Guide
Lesson 1: Beyond Base and Height
Timeframe
50-60 Minutes
Math Standard
G-SRT.D.9
Complexity
Moderate (Derivation)
Instructional Sequence
1
The Hook: The Swamp Plot (5-10 mins)
Display the "Swampy Situation" slide. Ask students why they can't just use \( \frac{1}{2}bh \). Emphasize that height is a "virtual" line that often isn't measurable in the physical world. Let them brainstorm how they might find height using a corner angle.
2
The Derivation (15-20 mins)
Guide students through Part 1 of the "Formula Forge" worksheet while using the "Derivation" slide. Key moment: Showing that \( h = b \sin(A) \) is just "Opposite = Hypotenuse × \(\sin(\theta)\)". Many students will struggle with the variable substitution—remind them we are just "swapping one label for a equivalent set of labels."
3
SAS Identification (10 mins)
Perform a "Trap Check" poll. Show triangles with ASA, SSA, and SSS given. Ask: "Can we use the formula right now?" Students must realize the angle MUST be the 'included' angle between the two given sides.
4
The Surveyor's Dilemma (10 mins)
The closing reasoning task (Part 3 of worksheet). This pushes students beyond rote calculation. If \( \sin(C) \) is maxed at 1, then \( \frac{1}{2}ab \) is the absolute max area. It's a great "Aha!" moment for boundary conditions.
Support Strategies
Provide a "Cheat Sheet" showing the three versions of the formula for \( \triangle ABC \).
Have students color-code the "sandwich" (Side-Angle-Side) on their worksheets.
Extension Strategies
Ask: "If we have AAS, what extra step do we need to use the formula?" (Finding the missing side via Law of Sines).
Challenge them to find the area of a parallelogram using a similar derivation.
Misconception Alert
Students often calculate \( \frac{1}{2}ab \cdot C \) (forgetting the sine function entirely) or try to take the sine of the side lengths. Keep an eye out for calculator settings—ensure they are in DEGREE mode.
SAS Specialist Slides Lesson 2: SAS Specialist
Working
Backwards
Yesterday, we found Area. Today, we have the Area—but we're missing the blueprints for the sides and angles.
"If a surveyor knows a plot is exactly 500 acres and knows two boundaries are 1,000m and 1,200m... can they determine the corner angle?"
?
Solving for the Angle
The Goal: Isolate \( \sin(C) \)
\( Area = \frac{1}{2}ab \sin(C) \)
Multiply by 2...
\( 2 \cdot Area = ab \sin(C) \)
Divide by ab...
\( \frac{2 \cdot Area}{ab} = \sin(C) \)
Don't forget the Inverse Sine ! \( C = \sin^{-1}(\dots) \)
Example 1:
Area = 45 cm²
\( \sin(\theta) = \frac{2 \cdot 45}{10 \cdot 12} \)
\( \sin(\theta) = \frac{90}{120} = 0.75 \)
\( \theta = \sin^{-1}(0.75) \approx 48.6^\circ \)
Solving for a Side
The Formula Swap:
To find a missing side \(b\) when you have Area, side \(a\), and angle \(C\):
\( b = \frac{2 \cdot Area}{a \cdot \sin(C)} \)
Think like a Surveyor:
Area given? Yes.
One side given? Yes.
Included angle given? Yes.
If all checks pass, you have exactly enough information to define the triangle's boundaries.
The Ambiguous Area
Wait! Two possible answers?
Remember the unit circle? Two different angles have the same sine value between 0° and 180°.
\( \sin(\theta) = \sin(180^\circ - \theta) \)
Example: \(\sin(30^\circ) = \sin(150^\circ) = 0.5\)
Unless we have a diagram, we might have an acute triangle or an obtuse one with the same area!
30° 150° Same Base,
Area Ambiguity Activity Requisition Form 2.B
Area Ambiguity
Subject: Reverse Area Engineering
Project ID: SAS-REVERSE
NAME:
Phase 1: Missing Boundary Recovery
A survey team was forced to retreat by high winds. They recorded the total area and some dimensions but missed one boundary. Use algebra to recover the missing side length. Round to 1 decimal place.
CASE 001
Target Area: 300 ft²
Side A: 25 ft
Angle C: 38°
Workspace (Find Side B):
CASE 002
Target Area: 1,450 m²
Side B: 62 m
Angle A: 115°
Workspace (Find Side C):
Phase 2: Corner Orientation
The physical boundaries are known, but the compass bearing (angle) between them is lost. Recover the angle. Note: There may be two possible answers if no diagram is provided.
AREA
SIDE 1
SIDE 2
CALCULATED ANGLE(S)
50 in²
10 in
15 in
12 km²
8 km
3 km
Phase 3: The Ambiguity Protocol
A junior surveyor calculates an angle of 28° for a triangular lot. However, the landowner insists the corner is "wide and obtuse."
1. What is the other possible angle?
________°
2. Why do they share the same area?
3. Summary Task:
Explain in one sentence why knowing "Side-Side-Area" is not enough to perfectly describe a single triangle shape.
Area Ambiguity Answer Key Area Ambiguity KEY
Verification Report: SAS-REVERSE
Teacher Reference
Phase 1: Recovery Log
CASE 001 Solutions:
\( 300 = \frac{1}{2}(25)(b)\sin(38^\circ) \)
\( 300 = 12.5 \cdot b \cdot 0.6157 \)
\( 300 = 7.696b \)
b \(\approx\) 39.0 ft
CASE 002 Solutions:
\( 1450 = \frac{1}{2}(62)(c)\sin(115^\circ) \)
\( 1450 = 31 \cdot c \cdot 0.9063 \)
\( 1450 = 28.095c \)
c \(\approx\) 51.6 m
Phase 2: Orientation Log
50 in²
10 in
15 in
\( \sin(\theta) = \frac{100}{150} = 0.6667 \)
12 km²
8 km
3 km
\( \sin(\theta) = \frac{24}{24} = 1 \)
Phase 3: Ambiguity Verdict
1. Other possible angle:
152° (180 - 28)
2. Why do they share area?
Supplementary angles have the same sine value \(\sin(\theta) = \sin(180 - \theta)\). Since the formula depends only on the sine, the resulting area is identical.
3. Summary Task:
"Because of the properties of the sine function, there are two distinct triangles (one acute, one obtuse) that can satisfy the given side-side-area conditions unless a 90° angle is formed."
Herons Secret Slides Ancient Geometry Reborn
Heron's Secret
2,000 years ago, Heron of Alexandria discovered a way to calculate the area of any triangle without ever measuring a single angle.
Requirement: Side - Side - Side (SSS)
The SSS Dilemma
Imagine you have a triangle with sides 7, 8, and 9.
Method A: The Long Way
Use Law of Cosines to find an angle.
Use \( \frac{1}{2}ab \sin(C) \) to find the area.
Result: High chance of rounding errors at each step!
Method B: Heron's Formula
One formula. No trigonometry. No Law of Cosines. High precision.
s
The Semi-Perimeter
Before we find the Area, we need the Semi-Perimeter (s) . It is exactly what it sounds like: half of the perimeter.
\( s = \frac{a + b + c}{2} \)
Side a
10
Side b
12
Side c
14
\( s = \frac{10+12+14}{2} = \frac{36}{2} = 18 \)
The Master Formula
\( Area = \sqrt{s(s-a)(s-b)(s-c)} \)
Algorithm for Success:
1
Add the sides and divide by 2 to find s.
2
Subtract each side from s.
3
Multiply the results together and take the square root.
Three Sides Task The Three Sides Task
Investigation of Heron's Algorithm • Alexandria Research Unit
Investigator:
Date:
Stage 1: The Semi-Perimeter Sweep
Determine the semi-perimeter (\(s\)) and the total Area for each triangular plot. Show your work steps clearly.
Plot Delta
Sides: 5, 12, 13
Step 1: Semi-Perimeter (\(s\))
Step 2: Area (\(A\))
Plot Gamma
Sides: 7, 10, 15
Plot Sigma
Sides: 11, 14, 21
Stage 2: Algorithmic Efficiency
A surveyor needs the area of a triangle with sides a=15, b=22, c=31. They are debating two methods:
Method 1: The Trig Route
Solve for \(\cos(A)\) using Law of Cosines.
Take \(\cos^{-1}\) to find \(\angle A\).
Calculate Area using \( \frac{1}{2}bc \sin(A) \).
Method 2: Heron's Formula
Find \(s = (15+22+31)/2\).
Subtract sides from \(s\).
Calculate Area using \( \sqrt{s(s-a)(s-b)(s-c)} \).
Reflective Analysis:
Which method is less likely to produce a rounding error in the final answer? Why?
Stage 3: The impossible triangle?
"A landowner claims their triangular garden has sides of 10m, 12m, and 25m. What happens when you try to use Heron's Formula on this plot? Explain what this tells you about the triangle."
Three Sides Answer Key Three Sides Task KEY
MASTER LOG
Stage 1: Calculation Verification
Plot Delta
Sides: 5, 12, 13
Semi-Perimeter (s)
\( s = \frac{5+12+13}{2} = \mathbf{15} \)
Area (A)
\( A = \sqrt{15(10)(3)(2)} = \sqrt{900} = \mathbf{30} \)
Plot Gamma
Sides: 7, 10, 15
\( s = \frac{7+10+15}{2} = \mathbf{16} \)
\( A = \sqrt{16(9)(6)(1)} = \sqrt{864} \approx \mathbf{29.39} \)
Plot Sigma
Sides: 11, 14, 21
\( s = \frac{11+14+21}{2} = \mathbf{23} \)
\( A = \sqrt{23(12)(9)(2)} = \sqrt{4968} \approx \mathbf{70.48} \)
Stage 2: Efficiency Verdict
Verdict: Method 2 (Heron's Formula) is superior for precision.
Reasoning: Method 1 requires intermediate steps where an angle is rounded (e.g., 42.54°). When this rounded value is plugged into the sine function, the error compounds. Heron's Formula uses exact side lengths; if the side lengths are integers, the only rounding happens at the final square root step, maintaining maximum accuracy.
Stage 3: The Impossible Garden
Analysis:
If you calculate \(s\): \( s = \frac{10+12+25}{2} = 23.5 \).
Conclusion:
Taking the square root of a negative number is impossible in real geometry. This confirms the Triangle Inequality Theorem : the sum of two sides (10+12=22) must be greater than the third side (25). This triangle cannot exist physically.
Polygon Puzzle Slides Lesson 4: Polygon Puzzles
The Real World
isn't a Triangle
Architects, landscapers, and engineers almost never work with perfect shapes. They work with composite regions.
Irregular Rooms
Complex Land Deeds
Architectural Facades
1
Strategy: Decomposition
Any complex polygon can be broken into non-overlapping triangles by drawing diagonals from a single vertex.
The Golden Rule:
"Break it until you recognize it."
Once broken down, you can use:
Sine Area Formula (SAS)
Heron's Formula (SSS)
T1 T2 T3
2
Practice: Irregular Lot
30m 22m 20m 42° 25m 20.5m
Step 1: Lower Triangle
Area = \(\frac{1}{2}(30)(22)\sin(42^\circ)\)
\(\approx 220.8 \text{ m}^2\)
Step 2: Upper Triangle
SSS (Use Heron's!)
Sides: 20, 25, 20.5
\(\approx 203.4 \text{ m}^2\)
Total Area
424.2 m²
Floor Plan Deconstructor Floor Plan Deconstructor
Project: Irregular Living Space Calculation
Phase 4: Cumulative Area
Architect:
Standard Operating Procedure:
"Decompose the irregular polygon into non-overlapping triangles. For each triangle, determine which formula (SAS or SSS) is applicable based on the given dimensions. Sum the results to find the total square footage."
32 ft 22 ft 28 ft 24 ft 105° P1 P2 P3 P4
Living Room Layout: Spec-402
Triangle 1: Lower Region (P1-P2-P3)
1. Sketch Diagonal & Dimensions:
2. Show Calculation:
Triangle 2: Upper Region (P1-P3-P4)
1. Required Side Length (Diagonal):
Hint: Use Law of Cosines on Triangle 1 first to find the diagonal.
D = ________ ft
2. Show Calculation (Heron's):
Subtotal T1 Area:
Subtotal T2 Area:
Cumulative Floor Area
________ sq ft
Floor Plan Answer Key Floor Plan KEY
Project Verification: Spec-402
Teacher Master
Triangle 1: Lower Region (P1-P2-P3)
Method: SAS
\( Area = \frac{1}{2}(32)(22)\sin(105^\circ) \)
\( Area = 352 \cdot 0.9659 \)
Area \(\approx\) 340.00 sq ft
Step 1.5: The Diagonal (P1 to P3)
Required for the next triangle (Law of Cosines):
\( D^2 = 32^2 + 22^2 - 2(32)(22)\cos(105^\circ) \)
D \(\approx\) 43.27 ft
Triangle 2: Upper Region (P1-P3-P4)
Method: Heron's (SSS)
Sides: 24, 28, 43.27
\( s = (24 + 28 + 43.27) / 2 = \mathbf{47.64} \)
\( A = \sqrt{47.64(47.64-24)(47.64-28)(47.64-43.27)} \)
\( A = \sqrt{47.64(23.64)(19.64)(4.37)} \)
\( A = \sqrt{96,654.54} \)
Area \(\approx\) 310.89 sq ft
T1 Area:
340.00
T2 Area:
310.89
TOTAL FLOOR AREA
650.89 sq ft
Note to Teacher: Small variations in rounding the diagonal (43.27 vs 43.3) can lead to final area variations of ±1-2 sq ft. Accept any total area between 648 and 653 sq ft as long as the steps are logically sound.
Tycoon Tactics Slides Final Challenge
Tycoon
Tactics
You are a real estate tycoon evaluating a high-stakes land acquisition. The sellers have provided the deeds, but you need to prove the value.
The Brief
Your firm is bidding on a prime piece of coastline property. To secure funding, your report must include:
The Catch
"The deed only lists the side lengths and one crucial interior angle. You must use the most efficient formula at each step to avoid costly precision errors."
Strategy Checklist
Use SAS When...
...you have two boundaries and the angle where they meet.
\( \frac{1}{2}ab \sin(C) \)
Use SSS When...
...you have all three boundaries of a triangular section.
\( \sqrt{s(s-a)(s-b)(s-c)} \)
The Golden Rule of Surveying
Never use a rounded result as an exact input for a new formula if an exact alternative exists.
Deed Surveyor Challenge Certified Land Deed
Record: L-2049-SUMMIT • Coastline District
OFFICIAL SEAL
Legal Description
"Starting at the Coastal Marker (Point A), proceed 450 feet north to the Ridge Line (Point B). From the Ridge Line, travel 380 feet northeast to the Pine Grove (Point C). The interior angle at the Ridge Line (Angle B) is exactly 112°. From the Pine Grove, return directly to the Coastal Marker along the boundary line of 688 feet. Attached to this triangle is a secondary plot extending from the Pine Grove (Point C) to the Old Oak (Point D), a distance of 290 feet, and from the Old Oak back to the Coastal Marker (Point A), a distance of 510 feet."
Stage 1: Field Sketch
Carefully sketch the property boundaries based on the deed. Label all points (A, B, C, D) and given dimensions.
Stage 2: Area Analysis
Region 1: Plot ABC
Method Selection: SAS or SSS?
Region 2: Plot ACD
Method Selection: SAS or SSS?
Valuation Report
Market Price
\$22.50 / sq ft
Total Square Footage
________
Total Valuation
\$ ________
Bid Ready
Surveyor:
I certify that these calculations were performed with geometric precision using official geodetic formulas.
Deed Surveyor Answer Key Challenge Rubric & Key
Project: Record L-2049-SUMMIT • Coastal Acquisition
Official Key: Valuation Data
Triangle ABC (SAS)
Sides: 450, 380, \(\angle B\) = 112°
Triangle ACD (SSS)
Sides: 688, 290, 510
Total Cumulative Area:
145,780.96 sq ft
Final Market Valuation:
\$3,280,071.60
Assessment Criteria
Criteria Exceptional (5) Proficient (3-4) Developing (1-2) Field Sketch Precise scale representation. All vertices and side lengths labeled correctly according to deed. Mostly accurate sketch. Labels are present but may contain minor orientation errors. Sketch is missing or doesn't reflect the deed description. Labels missing. Formula Selection Correctly identified SAS for ABC and SSS for ACD. Justification is sound. Used correct formulas but didn't explicitly justify why SSS was better for ACD. Used incorrect formula (e.g., trying SAS where angle is unknown). Mathematical Precision Calculations are error-free. Correct semi-perimeter and sine values used. Rounding at final step only. Process is correct. Minor arithmetic or rounding error in sub-calculation. Major calculation errors. Rounding compounded throughout. Financial Logic Summed areas correctly and applied market price to get accurate final valuation. Calculated valuation but used a slightly incorrect subtotal area. Valuation is missing or lacks clear connection to area total.
Facilitator Note: Students often ask if they should find the area of triangle ABD instead. Explain that partitioning the quadrilateral into ABC and ACD is most efficient because ABC's side AC can be found via Law of Cosines, but the deed already gives AC as 688ft, making Heron's Formula (SSS) the fastest and most precise path for the second triangle.
