Lesson 17 ~35 min Unit 3 · Geometry +85 XP

Area of Non-Right Triangles

Half the product of two sides and the sine of the included angle. Find the area of ANY triangle with one simple formula -- then level up with Heron's formula when you know all three sides.

Today's hook: A triangular paddock has sides 80 m and 120 m with an included angle of 60°. Without any new formulas, can you find its area? Think about what you know about right triangles and how you might use it.

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Think First

warm-up

A triangular paddock has sides 80 m and 120 m with an included angle of 60°. You know that area = ½ base x height, but there is no right angle marked. How could you find the height? What would the area be if the angle were 90° instead? How does the actual area compare?

Record your answer in your workbook.

The Big Idea

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The area of any triangle can be found with just two sides and the included angle: $A = \frac{1}{2}ab \sin C$. This formula generalises the familiar $\frac{1}{2} \times \text{base} \times \text{height}$ by expressing the height as $a \sin C$.

When you know all three sides but no angles, Heron's formula comes to the rescue: $A = \sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{a+b+c}{2}$ is the semi-perimeter. These two formulas cover every possible scenario for finding a triangle's area.

$A = \frac{1}{2}ab \sin C$

Height = a sin C

The geometric basis of the formula.

Heron for SSS

When you know three sides, no angles needed.

Max at 90°

sin C is largest when C = 90°.

What You'll Master

objectives

Know

$A = \frac{1}{2}ab \sin C$ for any triangle
Heron's formula for area given three sides

Understand

How the formula generalises the right triangle area formula
When to use each formula

Can Do

Calculate the area of any triangle given appropriate information
Choose the most efficient area formula
Solve real-world problems involving triangular areas

Words You Need

vocabulary

Area formula$A = \frac{1}{2}ab \sin C$. Works for any triangle when two sides and the included angle are known.

Heron's formula$A = \sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{a+b+c}{2}$. Works when all three sides are known.

Semi-perimeterHalf the perimeter: $s = \frac{a+b+c}{2}$.

Included angleThe angle between two known sides.

Hectare10,000 square metres. A common unit for land area.

GeneralisationA broader rule that includes a simpler rule as a special case.

Spot the Trap

heads-up

Wrong: Using the formula with a non-included angle. If you know sides $a$ and $b$, you MUST use angle $C$ (between them).

Right: The angle in the formula must be the included angle between the two known sides.

Wrong: Using Heron's formula without calculating the semi-perimeter first.

Right: Always calculate $s = \frac{a+b+c}{2}$ first, then substitute into Heron's formula.

The Area Formula

+5 XP

The formula $A = \frac{1}{2}ab \sin C$ works because the height from vertex $B$ to side $b$ is exactly $h = a \sin C$. Substituting into $\frac{1}{2} \times \text{base} \times \text{height}$ gives the formula directly.

This formula is remarkably powerful. It works for acute, obtuse, and right angles alike. When $C = 90°$, $\sin 90° = 1$ and the formula becomes $A = \frac{1}{2}ab$ -- the familiar right triangle area formula. When $C$ approaches 0° or 180°, the area approaches zero (the triangle collapses into a line).

$A = \frac{1}{2}ab \sin C$

Any two sides work

Use whichever pair you know.

Angle must be included

Between the two known sides.

Units are squared

cm → cm², m → m².

Heron's Formula

+5 XP

When you know all three sides but no angles, Heron's formula finds the area without needing to find any angles first. It is elegant but requires careful arithmetic.

Step 1: Calculate the semi-perimeter $s = \frac{a+b+c}{2}$.
Step 2: Calculate $s-a$, $s-b$, and $s-c$.
Step 3: Multiply $s(s-a)(s-b)(s-c)$.
Step 4: Take the square root. All three values $(s-a)$, $(s-b)$, $(s-c)$ must be positive for a valid triangle. If any is zero or negative, the three sides cannot form a triangle.

$A = \sqrt{s(s-a)(s-b)(s-c)}$

Calculate s first

Always find the semi-perimeter.

Check validity

All of s-a, s-b, s-c must be positive.

Exact when possible

Leave as sqrt if it simplifies nicely.

Choosing the Right Formula

+5 XP

Different given information calls for different formulas. Choosing efficiently saves time and reduces errors. Here is your decision guide.

Base and height known → $A = \frac{1}{2}bh$.
Two sides and included angle → $A = \frac{1}{2}ab \sin C$.
All three sides → Heron's formula.
Two sides and non-included angle → Use sine rule to find included angle, then area formula.
Right triangle → $A = \frac{1}{2} \times \text{leg}_1 \times \text{leg}_2$.

Given → Match pattern → Apply formula

SAS is fastest

Half ab sin C needs no intermediate steps.

SSS uses Heron

Or find an angle with cosine rule first.

Right triangle?

Use the simple half base times height.

Common Pitfalls

heads-up

Using a non-included angle

Putting angle A into $\frac{1}{2}ab \sin C$ when sides $a$ and $b$ are known. This gives a completely wrong answer because the height calculation assumes the angle is between the two sides.

Fix: always verify the angle is between the two sides you are using. Draw a diagram and label carefully.

Forgetting to halve in Heron's formula

Using $s = a+b+c$ instead of $s = \frac{a+b+c}{2}$. This makes the area calculation wildly incorrect.

Fix: write "s = (a+b+c)/2" as the very first step. Double-check before substituting into the square root.

Unit confusion

Mixing centimetres and metres in the same problem, or forgetting that area units are squared (cm² not cm).

Fix: convert all lengths to the same unit before calculating. Write the unit squared in your final answer.

Watch Me Solve It · 3 examples

Watch Me Solve It · Area formula

+15 XP per step

PROBLEM

Find the area of a triangle with sides $a = 8$ cm, $b = 10$ cm, and included angle $C = 50°$.

1

Identify the formula

Two sides and included angle (SAS) → $A = \frac{1}{2}ab \sin C$
2

Substitute

$A = \frac{1}{2}(8)(10) \sin 50°$

$A = 40 \times 0.766$
3

Calculate

$A \approx 30.6$ cm²

Check: if $C = 90°$, area would be 40 cm². Since $50° < 90°$, the actual area should be less. 30.6 < 40 ✓

Answer$A \approx 30.6$ cm²

Watch Me Solve It · Heron's formula

+15 XP per step

PROBLEM

A triangle has sides 7 cm, 8 cm, and 9 cm. Find its area using Heron's formula.

1

Find the semi-perimeter

$s = \frac{7 + 8 + 9}{2} = \frac{24}{2} = 12$

Always calculate s first.
2

Calculate the differences

$s - a = 12 - 7 = 5$

$s - b = 12 - 8 = 4$

$s - c = 12 - 9 = 3$
3

Apply Heron's formula

$A = \sqrt{12 \times 5 \times 4 \times 3} = \sqrt{720}$

$A = 12\sqrt{5} \approx 26.8$ cm²
4

Verify

Using cosine rule: $\cos C = \frac{49 + 64 - 81}{112} = \frac{32}{112} = 0.286$

$C \approx 73.4°$, then $A = \frac{1}{2}(7)(8) \sin 73.4° \approx 26.8$ cm² ✓

Both methods give the same answer.

Answer$A = 12\sqrt{5} \approx 26.8$ cm²

Watch Me Solve It · Practical problem

+15 XP per step

PROBLEM

A triangular paddock has sides 100 m and 150 m with included angle 70°. Find the area and the cost to fence it at $\$15$ per metre.

1

Find the area

$A = \frac{1}{2}(100)(150) \sin 70° = 7500 \times 0.940 \approx 7048$ m²

This is approximately 0.705 hectares.
2

Find the third side

$c^2 = 100^2 + 150^2 - 2(100)(150) \cos 70°$

$c^2 = 10000 + 22500 - 30000 \times 0.342 = 22240$

$c = \sqrt{22240} \approx 149.1$ m
3

Find the perimeter and cost

Perimeter = $100 + 150 + 149.1 = 399.1$ m

Cost = $399.1 \times 15 \approx \$5987$
4

Check reasonableness

If the triangle were right-angled: perimeter $\approx 100 + 150 + 180 = 430$ m

Our perimeter (399 m) is less because the angle is acute. This makes sense.

AnswerArea $\approx 7048$ m²; fencing cost $\approx \$5987$

Copy Into Your Books

Area (SAS)

$A = \frac{1}{2}ab \sin C$
Use when two sides and included angle known
Generalises right triangle formula

Heron's Formula

$s = \frac{a+b+c}{2}$
$A = \sqrt{s(s-a)(s-b)(s-c)}$
Use when all three sides known

Right Triangle

$A = \frac{1}{2} \times \text{leg}_1 \times \text{leg}_2$
Simplest case
Special case of half ab sin C

Unit Conversions

1 hectare = 10,000 m²
1 km² = 100 hectares
Area units are always squared

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Area drill

4 problems

Four quick problems on triangle area. Work each, then reveal the answer.

1 Find the area of a triangle with sides 12 cm and 15 cm and included angle 40°.

$A = \frac{1}{2}(12)(15) \sin 40° = 90 \times 0.643 \approx 57.9$ cm².$57.9$ cm²
2 A triangle has sides 5 cm, 6 cm, and 7 cm. Find its area using Heron's formula.

$s = \frac{5+6+7}{2} = 9$. $A = \sqrt{9 \times 4 \times 3 \times 2} = \sqrt{216} = 6\sqrt{6} \approx 14.7$ cm².$14.7$ cm²
3 Find the area of a triangle with two sides 10 cm and included angle 120°.

$A = \frac{1}{2}(10)(10) \sin 120° = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3} \approx 43.3$ cm².$43.3$ cm²
4 For a triangle with two fixed sides of 8 cm and 12 cm, what included angle gives the maximum area? What is that maximum area?

Maximum when $\sin C = 1$, so $C = 90°$. Max area = $\frac{1}{2}(8)(12) = 48$ cm².$C = 90°$, max area = $48$ cm²

Complete in your workbook.

Multiple Choice · 5 questions

Area formula application

+10 XP

A triangle has sides 6 cm and 8 cm with included angle 45°. What is its area?

12 cm²

16.97 cm²

24 cm²

48 cm²

B is correct. $A = \frac{1}{2}(6)(8) \sin 45° = 24 \times \frac{\sqrt{2}}{2} \approx 24 \times 0.707 = 16.97$ cm².

Heron's formula

+10 XP

A triangle has sides 13 cm, 14 cm, and 15 cm. What is its area?

84 cm²

182 cm²

91 cm²

126 cm²

A is correct. $s = \frac{13+14+15}{2} = 21$. $A = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84$ cm².

Maximum area

+10 XP

For a triangle with two fixed sides of 5 cm and 10 cm, what included angle gives the maximum possible area?

0°

45°

90°

180°

C is correct. The area is $A = \frac{1}{2}ab \sin C$, so it is maximised when $\sin C$ is maximised. The maximum value of $\sin C$ is 1, which occurs at $C = 90°$.

Formula selection

+10 XP

A triangle has sides 9 cm, 10 cm, and 11 cm. No angles are given. Which formula should you use to find the area?

$A = \frac{1}{2}bh$

$A = \frac{1}{2}ab \sin C$

Heron's formula

Cosine rule

C is correct. When all three sides are known but no angles, Heron's formula is the direct method. Option D (cosine rule) could be used to find an angle first, then you would use option B, but Heron's formula is more direct.

Practical application

+10 XP

A triangular paddock has sides 80 m and 120 m with included angle 60°. What is its area in hectares?

0.24 hectares

0.42 hectares

0.48 hectares

0.96 hectares

B is correct. $A = \frac{1}{2}(80)(120) \sin 60° = 4800 \times \frac{\sqrt{3}}{2} = 2400\sqrt{3} \approx 4157$ m². Dividing by 10,000: $\approx 0.42$ hectares.

Short Answer · 3 questions

Heron's formula

+15 XP

A triangle has sides 9 cm, 10 cm, and 11 cm. Use Heron's formula to find its exact area and give the answer to one decimal place.

Solution:

$s = \frac{9+10+11}{2} = 15$

$s-a = 6$, $s-b = 5$, $s-c = 4$

$A = \sqrt{15 \times 6 \times 5 \times 4} = \sqrt{1800} = 30\sqrt{2} \approx 42.4$ cm²

Optimisation

+15 XP

A triangular garden has sides 15 m and 20 m with included angle $\theta$. Find the maximum possible area and the angle that produces it.

Solution:

$A = \frac{1}{2}(15)(20) \sin \theta = 150 \sin \theta$

Maximum when $\sin \theta = 1$, so $\theta = 90°$

Maximum area = $150 \times 1 = 150$ m²

Real-world context

+15 XP

A triangular kite has two sides of 50 cm and 70 cm with included angle 55°. Calculate the area of the kite.

Solution:

$A = \frac{1}{2}(50)(70) \sin 55°$

$A = 1750 \times 0.819 = 1433.3$ cm²

Stretch Challenge

+25 XP

A triangle has area 30 cm² with two sides of 10 cm and 12 cm. Find the possible values of the included angle. Then, if the same area were achieved with sides 6 cm and 20 cm, what would the included angle be? What does this tell you about the relationship between side lengths and included angle for a fixed area?

Solution:

Part 1: $30 = \frac{1}{2}(10)(12) \sin C = 60 \sin C$

$\sin C = 0.5$, so $C = 30°$ or $C = 150°$

Part 2: $30 = \frac{1}{2}(6)(20) \sin C = 60 \sin C$

$\sin C = 0.5$, so $C = 30°$ or $C = 150°$

Observation: Different side pairs can give the same area with the same included angle, as long as the product $ab$ is the same. For a fixed area, smaller side pairs require larger angles, and larger side pairs require smaller angles.

Quick Review

all together

$A = \frac{1}{2}ab \sin C$

Works for any triangle with two sides and included angle.

Heron's formula

$A = \sqrt{s(s-a)(s-b)(s-c)}$ for SSS triangles.

Semi-perimeter

$s = \frac{a+b+c}{2}$ -- calculate this first.

Choose wisely

SAS → half ab sin C, SSS → Heron, right → simple half bh.

Maximum area

Occurs when included angle = 90°.

Units matter

Area units are always squared. 1 hectare = 10,000 m².

Interactive Area Calculator

+10 XP

Use the interactive calculator to practice finding areas using both formulas. Switch between SAS and SSS modes to test your understanding.

Badge Wall

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First Step

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Daily Challenge

+15 XP

Find the area of a triangle with sides 11 cm, 13 cm, and 20 cm. Show all working.

Lesson Complete

+20 XP

You have finished Lesson 17 -- Area of Non-Right Triangles. You now know two powerful formulas for finding the area of any triangle, and you can choose the right one for any situation.

11 cards studied

4 drills completed

5 MCQs

3 SAQs

2 area formulas mastered