Area of non-right triangles | Year 10

01 / LEARN

The idea, step by step

Two sides and the angle between them

The usual half-base-times-height formula still explains the area rule. Dropping a perpendicular gives a height such as b sin C, so the area is half the product of the two known sides and the sine of their included angle.

Label the angle between the two selected sides. An angle elsewhere in the triangle cannot be substituted as if it were included.

A = \frac{1}{2} a b sin C

Use the same relationship backwards

To find a missing side, divide by half the known side and the sine of the included angle. To find an angle, first calculate its sine. In a general triangle, both an acute angle and its obtuse supplement may give the same area.

A contextual restriction such as an acute angle can select one answer. Include squared units for area and ordinary units for side length.

A common trap

Using the sine of an angle not between the two known sides.

Instead: Draw rays along the known sides and select the angle where they meet.

02 / WORKED EXAMPLES

See the thinking

Example 1

Find an area

Two sides of a triangle are 6 m and 9 m. Their included angle is 40°. Find the area to 0.1 square metre.

$A = \frac{1}{2} (6) (9) sin 40^{\circ}$ .
The perpendicular height is already accounted for by the sine factor.

17.4 m²

Example 2

Two angles with the same area

A triangle has area 24 cm² and two sides 8 cm and 10 cm. Find all possible included angles to 0.1°.

$24 = \frac{1}{2} (8) (10) sin C$ , so $sin C = 0.6$ .
The acute angle is ${sin}^{- 1} (0.6)$ .
Its supplement is also possible because both angles have positive sine.

36.9° or 143.1°

Check your understanding

Sides 6 cm and 8 cm include 90°. Find the area.

Show answer

24 cm²

$A = \frac{1}{2} (6) (8) sin 90^{\circ} = 24$ cm².

03 / WORKSHEET

Your turn

6 questions · 15 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find the area when sides 10 cm and 12 cm include 30°.
2 marks
Sides 14 m and 9 m include 50°. Find the area to 0.1 m².
2 marks
A triangle has area 15 cm². One side is 6 cm and another side b meets it at 30°. Find b.
3 marks
Sides 5 cm and 10 cm enclose area 12.5 cm². Find both possible included angles.
3 marks
Two sides of a triangular sign are 8 cm and 11 cm. Their included angle is acute and the area is 22 cm². Find that angle.
3 marks
AB = 7 cm, AC = 9 cm and angle B = 40°. Explain why half × 7 × 9 × sin 40° is not justified.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the area when sides 10 cm and 12 cm include 30°.
2 marks
$A = \frac{1}{2} (10) (12) sin 30^{\circ} = 30$ cm².
30 cm²
Sides 14 m and 9 m include 50°. Find the area to 0.1 m².
2 marks
$A = \frac{1}{2} (14) (9) sin 50^{\circ}$ .
48.3 m²
A triangle has area 15 cm². One side is 6 cm and another side b meets it at 30°. Find b.
3 marks
$15 = \frac{1}{2} (6) b sin 30^{\circ} = 1.5 b$ .
$b = 10$ cm.
10 cm
Sides 5 cm and 10 cm enclose area 12.5 cm². Find both possible included angles.
3 marks
$sin C = 12.5 / 25 = 1 / 2$ .
The two angles between 0° and 180° are 30° and 150°.
30° or 150°
Two sides of a triangular sign are 8 cm and 11 cm. Their included angle is acute and the area is 22 cm². Find that angle.
3 marks
$sin C = 22 / 44 = 1 / 2$ .
The acute answer is 30°.
30°
AB = 7 cm, AC = 9 cm and angle B = 40°. Explain why half × 7 × 9 × sin 40° is not justified.
2 marks
The selected sides meet at A, so their included angle is A.
Find A before using those two sides in the area rule.
40° is not the angle between AB and AC.

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Curriculum reference & lesson scope

Australian Curriculum v9 and South Australian school course scope

Right-triangle applications include AC9M10M03. The approved chapter also develops non-right-triangle methods, exact values and trigonometric functions within one normal Year 10 course.

Official curriculum source ↗

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