Non-Right Triangles

 

 

 

The base is:

7 − 1 = 6 units

The height is:

6 − 2 = 4 units

Area = ½ × 6 × 4

Area = 12 square units

Using the coordinate formula:

Area = ½|1(2−6) + 7(6−2) + 4(2−2)|

Area = ½|−4 + 28 + 0|

Area = 12 square units

The coordinate method is especially useful when the triangle is tilted and its base or height is not immediately obvious.

Choosing the Most Efficient Method

Learning Target 4: Solving Practical Area Problems

Triangle area calculations are used in surveying, architecture, construction, engineering, land management, and design.

Real-world problems often require us to interpret measurements, select an appropriate formula, and convert units.

Application 1: Measuring a Triangular Field

A farmer owns a triangular field.

Two sides measure 120 m and 175 m, and the included angle is 68°.

Calculate the area of the field.

 

Step 1: Identify the formula.

Area = ½ab sin C

Step 2: Substitute.

Area = ½ × 120 × 175 × sin 68°

Area = 10,500 × sin 68°

Area ≈ 9,735.43 m²

Step 3: Convert to hectares.

1 hectare = 10,000 m²

Area ≈ 9,735.43 / 10,000

Area ≈ 0.974 hectares

This is useful when estimating the amount of seed, fertilizer, or irrigation needed for the field.

Application 2: Designing a Triangular Glass Window

An architect designs a triangular glass window with two sides measuring 2.4 m and 3.1 m.

The angle between the sides is 55°.

Calculate the area of glass required.

Area = ½ × 2.4 × 3.1 × sin 55°

Area = 3.72 × sin 55°

Area ≈ 3.05 m²

Suppose the glass costs $85 per square metre.

Estimated material cost:

Cost = 3.047... × 85

Cost ≈ $259.03

The actual purchase cost may be higher because cutting, waste, installation, and framing are not included.

Application 3: A Triangular Roof Panel

A triangular roof panel has sides of 5 m, 7 m, and 9 m.

Calculate its area.

Since all three sides are known, Heron's formula is appropriate.

s = (5 + 7 + 9) / 2

s = 10.5 m

Area = √[10.5(10.5−5)(10.5−7)(10.5−9)]

Area = √(10.5 × 5.5 × 3.5 × 1.5)

Area ≈ 17.41 m²

If roof covering is sold in panels covering 2 m² each:

17.41 / 2 ≈ 8.71

The minimum theoretical number of panels is 9.

In practice, additional panels may be required because of cutting and waste.

Application 4: A Triangular Sail

A sailboat has a triangular sail.

Two edges of the sail measure 6 m and 8 m, and the angle between them is 70°.

Calculate the sail area.

Area = ½ × 6 × 8 × sin 70°

Area = 24 × sin 70°

Area ≈ 22.55 m²

The sail area influences the amount of wind force the sail can experience, although the actual force also depends on wind speed, sail shape, orientation, and aerodynamic conditions.

Application 5: Land Surveying with Coordinates

A surveyor records the following positions in metres:

A = (0, 0)

B = (80, 20)

C = (30, 90)

Find the area enclosed by the three points.

Using the coordinate formula:

Area = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|

Substitute:

Area = ½|0(20−90) + 80(90−0) + 30(0−20)|

Area = ½|0 + 7200 − 600|

Area = ½ × 6600

Area = 3300 m²

This method is particularly useful for mapping irregular parcels of land using GPS or surveying coordinates.

Application 6: Finding the Maximum Area of a Triangular Design

An engineer has two support beams measuring 8 m and 12 m.

The angle between them can be adjusted.

What is the maximum possible triangular area?

Area = ½ × 8 × 12 × sin θ

Area = 48 sin θ

Since the maximum value of sin θ is 1:

Maximum area = 48 × 1

Maximum area = 48 m²

This occurs when:

θ = 90°

The result demonstrates an important optimization principle: when two side lengths are fixed, their included angle should be 90° to maximize the triangle's area.

Learning Target 5: Justifying Triangle Area Solutions

A complete mathematical solution should do more than provide a numerical answer.

It should explain:

• Why a particular formula is appropriate.

• How the known measurements relate to the formula.

• Why the calculated area is reasonable.

• Whether the answer has the correct units.

• Whether the result can be verified independently.

Method 1: Explain Why the Formula Applies

Suppose a triangle has two sides of 9 cm and 14 cm, with an included angle of 65°.

The sine area formula is appropriate because we know two sides and the angle between them.

Area = ½ × 9 × 14 × sin 65°

Area ≈ 57.10 cm²

A complete justification might state:

"The sine area formula is appropriate because two sides and their included angle are known. The calculated area is positive and is expressed in square centimetres."

Method 2: Check the Maximum Possible Area

For two fixed sides a and b:

Area = ½ab sin C

Because:

0 < sin C ≤ 1 for 0° < C < 180°

the area must satisfy:

0 < Area ≤ ½ab

For sides of 9 cm and 14 cm:

Maximum area = ½ × 9 × 14

Maximum area = 63 cm²

Our calculated area of 57.10 cm² is below the maximum, so it passes this check.

If a calculation produced 80 cm², we would know immediately that an error had occurred.

Method 3: Verify Using Another Formula

Consider a triangle with sides 13 cm, 14 cm, and 15 cm.

Using Heron's formula, we found:

Area = 84 cm²

We can verify this using trigonometry.

First calculate the angle C opposite the 15 cm side:

cos C = (13² + 14² − 15²) / (2 × 13 × 14)

cos C = (169 + 196 − 225) / 364

cos C = 140 / 364 = 5/13

Therefore:

sin C = √[1 − (5/13)²]

sin C = 12/13

Now calculate area:

Area = ½ × 13 × 14 × (12/13)

Area = 84 cm²

Both methods agree exactly.

This provides strong verification.

Method 4: Check Units

If the side lengths are measured in metres, the area must be measured in square metres.

For example:

Area = ½ × 5 m × 8 m × sin 40°

Area ≈ 12.86 m²

The sine function has no units, so multiplying two lengths produces squared units.

An answer of 12.86 m would be incorrect because metres measure length, not area.

Method 5: Evaluate the Context

Suppose a triangular garden has two sides of 4 m and 6 m.

A student calculates an area of 25 m².

The maximum possible area is:

½ × 4 × 6 = 12 m²

Therefore, 25 m² is impossible.

The answer should be rejected even before checking the detailed calculation.

This demonstrates why mathematical reasoning and estimation are important parts of problem-solving.

Common Mistakes

Mistake 1: Using an Angle That Is Not Included

Incorrect:

Area = ½ab sin A

when A is not the angle between sides a and b.

Correct:

Area = ½ab sin C

The angle must be between the two selected sides.

Mistake 2: Forgetting the Factor of One-Half

Incorrect:

Area = ab sin C

Correct:

Area = ½ab sin C

Forgetting ½ doubles the answer.

Mistake 3: Using a Sloping Side as the Perpendicular Height

The height must be perpendicular to the base.

If a sloping side makes an angle θ with the base, the height may be calculated using:

h = side × sin θ

Mistake 4: Using Heron's Formula Without Calculating the Semiperimeter

Heron's formula uses:

s = (a + b + c) / 2

It does not use the full perimeter.

Mistake 5: Using the Wrong Calculator Mode

When angles are measured in degrees, the calculator must be in degree mode.

For example:

sin 30° = 0.5

Mistake 6: Giving Area in Linear Units

Incorrect:

Area = 45 cm

Correct:

Area = 45 cm²

Area always requires square units.

Mistake 7: Rounding Intermediate Values Too Early

If you first calculate an unknown side or angle, retain the unrounded value when using it to find the area.

Rounding intermediate values can introduce unnecessary errors.

Mistake 8: Assuming Every Triangle Requires Trigonometry

If the base and perpendicular height are already known, use:

Area = ½bh

There is no need to calculate an angle.

The most efficient method is usually the one that uses the given information directly.

Key Terms

• Area: The amount of two-dimensional space enclosed by a shape.

• Base: A selected side of a triangle used when calculating its area.

• Perpendicular height: The shortest distance from a vertex to the line containing the opposite side.

• Included angle: The angle between two specified sides.

• Sine area formula: A trigonometric formula for finding area using two sides and their included angle.

• Heron's formula: A formula for finding the area of a triangle using its three side lengths.

• Semiperimeter: Half the perimeter of a triangle.

• Law of Sines: A relationship connecting the sides of a triangle with the sines of their opposite angles.

• Law of Cosines: A relationship connecting the sides of a triangle and the cosine of an included angle.

• Coordinate area formula: A method for calculating area from the coordinates of three vertices.

• Optimization: The process of finding a maximum or minimum value under given conditions.

• Verification: Checking whether a mathematical result satisfies the original information and constraints.

• Square units: Units used to measure area, such as cm² or m².

• Triangulation: A measurement technique that uses triangles to determine unknown distances or positions.

Key Takeaways

• The basic triangle area formula is Area = ½bh.

• The sine area formula is Area = ½ab sin C, where C is the included angle.

• The sine area formula works for acute, right, and obtuse triangles.

• Missing sides or angles can be calculated using the Law of Sines or Law of Cosines before finding area.

• Heron's formula calculates area directly from three side lengths.

• Coordinate geometry provides another method when the vertices are known.

• Choosing an efficient method depends on the given measurements.

• For two fixed side lengths, the maximum area occurs when the included angle is 90°.

• Area must always be expressed in square units.

• A strong mathematical solution includes appropriate formulas, accurate calculations, clear reasoning, and verification.

Check Your Understanding

These questions assess all five learning targets, including calculating areas, finding missing measurements, comparing methods, solving practical problems, and justifying answers.

Part A: Calculating Area Using Trigonometry

Question 1

A triangle has two sides measuring 8 cm and 11 cm. The included angle is 50°.

Calculate its area to two decimal places.

Question 2

A triangle has sides a = 12 m and b = 15 m, with included angle C = 120°.

Calculate its area and explain why the formula works for an obtuse triangle.

Part B: Determining Missing Measurements

Question 3

Triangle ABC has:

A = 40°, B = 60°, a = 10 cm

Calculate the area of the triangle.

Question 4

A triangle has side lengths 6 cm, 8 cm, and 10 cm.

Use the Law of Cosines to determine the largest angle, then calculate the area.

Part C: Comparing Methods

Question 5

Choose the most efficient method for each situation.

a) Base = 12 cm, perpendicular height = 7 cm

b) Two sides of 9 cm and 13 cm, included angle = 45°

c) Three sides of 8 cm, 10 cm, and 12 cm

d) Vertices at (1, 1), (7, 1), and (4, 5)

Explain each choice.

Part D: Practical Applications

Question 6

A triangular garden has two sides measuring 18 m and 24 m, with an included angle of 65°.

Calculate the area of the garden.

Question 7

A triangular metal plate has sides measuring 5 m, 7 m, and 8 m.

Calculate its area using Heron's formula.

Question 8

An architect designs a triangular window with two sides of 2 m and 3 m.

What is the maximum possible area of the window? At what included angle does this occur?

Part E: Justification and Reasoning

Question 9

A student calculates the area of a triangle with two sides of 6 cm and 10 cm as 40 cm².

Explain why this answer must be incorrect, regardless of the included angle.

Question 10

A triangle has sides of 13 cm, 14 cm, and 15 cm.

Calculate its area using Heron's formula and explain how you could verify the result using trigonometry.

Check Your Understanding — Fully Worked Answers

Answer 1

Area = ½ab sin C

Area = ½ × 8 × 11 × sin 50°

Area = 44 sin 50°

Area ≈ 33.71 cm²

Answer 2

Area = ½ × 12 × 15 × sin 120°

Area = 90 × sin 120°

Area ≈ 77.94 m²

The formula works because the sine of an obtuse angle between 90° and 180° is positive. The perpendicular height is related to the supplementary acute angle, which has the same sine.

Answer 3

Given:

A = 40°, B = 60°, a = 10 cm

Step 1: Find angle C.

C = 180° − 40° − 60° = 80°

Step 2: Find side b.

b = 10 sin 60° / sin 40°

b ≈ 13.47 cm

Step 3: Find the area.

Area = ½ab sin C

Area = ½ × 10 × (10 sin 60° / sin 40°) × sin 80°

Area ≈ 66.34 cm²

Answer 4

Given sides 6 cm, 8 cm, and 10 cm.

Let C be the angle opposite the longest side.

cos C = (6² + 8² − 10²) / (2 × 6 × 8)

cos C = (36 + 64 − 100) / 96

cos C = 0

Therefore:

C = 90°

Area = ½ × 6 × 8 × sin 90°

Area = 24 cm²

Answer 5

a) Base and height formula, because both required measurements are known.

b) Sine area formula, because two sides and their included angle are known.

c) Heron's formula, because all three sides are known.

d) Coordinate formula, because the three vertices are known. In this particular case, the base-height formula is also straightforward.

Answer 6

Area = ½ × 18 × 24 × sin 65°

Area = 216 sin 65°

Area ≈ 195.76 m²

Answer 7

Given:

a = 5 m, b = 7 m, c = 8 m

Semiperimeter:

s = (5 + 7 + 8) / 2 = 10 m

Heron's formula:

Area = √[10(10−5)(10−7)(10−8)]

Area = √(10 × 5 × 3 × 2)

Area = √300

Area ≈ 17.32 m²

Answer 8

Area = ½ × 2 × 3 × sin θ

Area = 3 sin θ

The maximum value of sine is 1, which occurs at θ = 90°.

Maximum area = 3 m²

Included angle = 90°

Answer 9

The maximum possible area is:

Area(max) = ½ × 6 × 10

Area(max) = 30 cm²

Since 40 cm² exceeds the maximum possible area, the student's result is impossible.

The area cannot exceed 30 cm².

Answer 10

Given:

a = 13 cm, b = 14 cm, c = 15 cm

Semiperimeter:

s = (13 + 14 + 15) / 2

s = 21 cm

Heron's formula:

Area = √[21(21−13)(21−14)(21−15)]

Area = √(21 × 8 × 7 × 6)

Area = 84 cm²

To verify, calculate the included angle between the 13 cm and 14 cm sides using the Law of Cosines, then substitute into:

Area = ½ × 13 × 14 × sin C

This also gives 84 cm².

Final Challenge: Designing a Triangular Recreation Area

A town plans to construct a triangular recreation area with sides measuring:

• 80 m

• 110 m

• 130 m

The town needs to determine the total area before estimating landscaping costs.

Calculate:

1. The area using Heron's formula.

2. The angle between the 80 m and 110 m sides.

3. The area again using the sine area formula.

4. The landscaping cost if the price is $12 per square metre.

5. Explain why the two area calculations agree.

Fully Worked Solution

Step 1: Calculate the semiperimeter.

s = (80 + 110 + 130) / 2

s = 160 m

Step 2: Apply Heron's formula.

Area = √[160(160−80)(160−110)(160−130)]

Area = √(160 × 80 × 50 × 30)

Area = √19,200,000

Area ≈ 4381.78 m²

Step 3: Calculate the included angle.

Let θ be the angle between the 80 m and 110 m sides.

Using the Law of Cosines:

cos θ = (80² + 110² − 130²) / (2 × 80 × 110)

cos θ = (6400 + 12100 − 16900) / 17600

cos θ = 1600 / 17600

cos θ = 1/11

θ ≈ 84.78°

Step 4: Verify using the sine area formula.

Area = ½ × 80 × 110 × sin θ

Using cos θ = 1/11:

sin θ = √(1 − 1/121)

sin θ = √120/11

Therefore:

Area = 4400 × √120/11

Area = 400√120

Area ≈ 4381.78 m²

The two methods agree.

Step 5: Calculate the landscaping cost.

Cost = Area × $12

Cost = 4381.780... × 12

Cost ≈ $52,581.36

Final Results

The two area calculations agree because both formulas describe the same geometric property of the triangle, using different measurements.

This provides an independent verification of the solution.

Conclusion

Calculating the area of a triangle is much more than substituting values into a single formula.

By combining geometry and trigonometry, we can determine triangle areas from a variety of measurements, even when the perpendicular height is unknown.

The sine area formula is especially useful when two sides and their included angle are known. The Law of Sines and Law of Cosines allow us to calculate missing measurements, while Heron's formula and coordinate geometry provide alternative methods.

Knowing which method to choose is an important part of mathematical problem-solving. Efficient solutions use the available information directly and avoid unnecessary calculations.

Equally important is the ability to justify and verify results. Comparing methods, checking units, and evaluating whether an answer is reasonable help us develop mathematical confidence.

The ability to calculate and justify triangle areas is essential in surveying, architecture, construction, engineering, land management, and many other practical applications of mathematics.