The Area Rule

Q: Can the area rule give two answers for an angle?

Yes. Working backwards from an area gives $\sin C$, and both the acute and obtuse angles may be valid.

The area rule finds the area of any triangle from two sides and the angle between them. It often appears as the last step of a two-dimensional problem that also needs the sine or cosine rule.

Proof of the area rule

Proof (examinable, acute-angled triangles). In acute-angled $△ A B C$ , draw the height $A D ⊥ B C$ and let $A D = h$ .

Area $= \frac{1}{2} \times B C \times A D = \frac{1}{2} ah$
In right-angled $△ A D C$ : $sin C = \frac{h}{b}$ , so $h = b sin C$
So Area $= \frac{1}{2} ab sin C$

The proof uses only the definition of $sin$ in a right-angled triangle. It may not assume the sine rule.

Area = ½ab sin C

$Area of △ A B C = \frac{1}{2} ab sin C = \frac{1}{2} b c sin A = \frac{1}{2} a c sin B$

Use it when you know two sides and the angle between them.

Example. Sides 9 cm and 14 cm with an included angle of 38°:

$Area = \frac{1}{2} (9) (14) sin 38° \approx 38, 79 cm^{2}$

Working backwards. If the area is 30 cm² with sides 8 cm and 10 cm, then $sin C = 0, 75$ , so $\hat{C} \approx 48, 59°$ or $\hat{C} \approx 131, 41°$ . Both are possible.

Combining the sine, cosine and area rules

For a problem in two dimensions:

Draw and label the triangle.
Identify what you know: sides, angles, and which pairs are opposite each other.
Choose a rule for each unknown in turn.
Use the area rule once you have two sides and the included angle.

Example. A triangle has sides 13, 14 and 15. The cosine rule gives the angle opposite 14:

$cos B = \frac{1 3 ^{2} + 1 5 ^{2} - 1 4 ^{2}}{2 ( 13 ) ( 15 )} = \frac{33}{65}$

Then $sin B = \frac{56}{65}$ , and the area is $\frac{1}{2} (13) (15) (\frac{56}{65}) = 84$ .

Regular polygons. Split into identical triangles from the centre. A regular hexagon with side 6 is six equilateral triangles: area $= 6 \times \frac{1}{2} (6) (6) sin 60° = 543$ .

Three-dimensional problems are Grade 12 work.

Key terms

Area rule: Area $= \frac{1}{2} ab sin C$ , using two sides and the included angle.
Included angle: The angle between the two sides used.
Regular polygon: A polygon with all sides and all angles equal.

Common mistakes

Using an angle that is not between the two sides.
Forgetting the $\frac{1}{2}$ .
Missing the obtuse angle when working backwards from an area.
Rounding an intermediate side before using it in the area rule.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

routine

Calculate the area of a triangle with sides 9 cm and 14 cm and an included angle of 38°.

Worked solution

Hidden - try the question first

Solution

Area $= \frac{1}{2} (9) (14) sin 38°$
$\approx 38, 79$ cm²

Question 2

5 marks total

routine

A triangle has sides 8 cm and 10 cm and an area of 30 cm². Calculate the two possible sizes of the angle between these sides.

Worked solution

Hidden - try the question first

Solution

$30 = \frac{1}{2} (8) (10) sin C$
$sin C = 0, 75$
$\hat{C} \approx 48, 59°$
or $\hat{C} \approx 131, 41°$
Both are possible, because either angle forms a valid triangle

Question 3

6 marks total

complex

(a) Calculate the area of an equilateral triangle with sides of 6 cm, in simplest surd form. (3)

(b) Hence calculate the area of a regular hexagon with sides of 6 cm. (3)

Worked solution

Hidden - try the question first

Solution

(a) Area $= \frac{1}{2} (9) (14) sin 38°$
$\approx 38, 79$
$= 93$ cm²
(b) A regular hexagon splits into 6 equilateral triangles from the centre
$6 \times 93 = 543$
$\approx 93, 53$ cm²

Question 4

9 marks total

problem-solving

A triangle $A B C$ has sides $A B = 13$ cm, $A C = 14$ cm and $B C = 15$ cm.

(a) Use the cosine rule to show that $cos B = \frac{33}{65}$ . (3)

(b) Without a calculator, show that $sin B = \frac{56}{65}$ . (2)

(d) Calculate the length of the altitude from $B$ to $A C$ . (2)

Worked solution

Hidden - try the question first

Solution

(a) $= \frac{1}{2} (9) (14) sin 38°$
$196 = 394 - 390 cos B$
$\hat{C} \approx 131, 41°$
(b) $sin^{2} B = 1 - \frac{1 089}{4 225} = \frac{3 136}{4 225}$
$sin B = \frac{56}{65}$ ( $B$ is acute)
(c) Area $= \frac{1}{2} (13) (15) (\frac{56}{65})$
$= 84$ cm²
(d) $\frac{1}{2} (14) h = 84$
$h = 12$ cm

FAQs

When do I use the area rule instead of half base times height?

When you do not know the perpendicular height but do know two sides and the angle between them.

Is the proof of the area rule examinable?

Yes, in acute-angled triangles. Draw the height, write the area as half base times height, and replace the height using sin C. Do not use the sine rule in this proof.

Can the area rule give two answers for an angle?

Yes. Working backwards from an area gives $sin C$ , and both the acute and obtuse angles may be valid.

Are three-dimensional problems in Grade 11?

No. Grade 11 keeps to two dimensions. Three-dimensional problems are Grade 12 work.

Proof of the area rule

Area = ½ab sin C

Combining the sine, cosine and area rules

Key terms

Common mistakes

Practice questions

FAQs

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