Triangle area | Year 11 | Ascent Library

01 / LEARN

Understand the idea.

A height must be perpendicular

In $K = \frac{1}{2} b h$ , the height is perpendicular to the chosen base. A sloping side is not a height merely because it rises from that base. Changing the chosen base requires its matching perpendicular height.

For an obtuse triangle, the perpendicular may meet an extension of the base. The same area rule still applies. A quadrilateral can be split along a diagonal, provided each triangular area uses compatible measurements.

K = \frac{1}{2} b h

Two sides and an included angle

If sides $a$ and $b$ enclose angle $C$ , a perpendicular height has magnitude $b sin C$ . Substituting into the base-height rule gives $K = \frac{1}{2} a b sin C$ , including when $C$ is obtuse.

The angle must lie between the two selected sides. For fixed positive side lengths, sine is greatest at $90^{\circ}$ , so this included angle gives the largest possible area. If solving backwards, check acute or obtuse restrictions.

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

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Sides of 7 cm and 9 cm enclose 60°. Find the triangle area to one decimal place.

Worked solution

1
Identify the two sides and their included angle.
$a = 7 cm, b = 9 cm, C = 60^{\circ}$
The angle between the two known sides permits the sine area rule.
2
Substitute into the area formula.
$K = \frac{1}{2} (7) (9) sin 60^{\circ}$
The sine term supplies the perpendicular-height factor.
3
Calculate and round the area.
$K \approx 27.2798 {cm}^{2} \approx 27.3 {cm}^{2}$
The product of two lengths has square-centimetre units, and the question requests one decimal place.

Answer

27.3 cm²

EXAMPLE 02

Worked example 2

Sides of 6 m and 8 m enclose an acute angle C. The area is 12 m². Find C.

Worked solution

1
Set the area equal to the given value.
$12 = \frac{1}{2} (6) (8) sin C = 24 sin C$
The angle C is enclosed by the six- and eight-metre sides.
2
Isolate the sine.
$sin C = \frac{12}{24} = 0.5$
Divide both sides by the positive coefficient twenty-four.
3
Find both triangle-angle candidates.
$C = 30^{\circ} or 150^{\circ}$
Supplementary angles have equal sines within the interval from zero to one hundred eighty degrees.
4
Apply the acute-angle condition.
$C = 30^{\circ}$
The stated angle is acute, so the obtuse candidate is rejected.

Answer

30°

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the area using sides 5 m and 8 m enclosing 90°.

Show answer & reasoning

20 m²

The area is $K = \frac{1}{2} (5) (8) sin 90^{\circ} = 20 m^{2}$ . At a right angle, the two given sides are perpendicular and sine equals one.

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

A triangle has base 12 cm and perpendicular height 7 cm. Find its area.
2 marks
Find the area using sides 10 m and 13 m enclosing $40^{\circ}$ . Round to one decimal place.
2 marks
Sides 6 cm and 8 cm enclose an acute angle. Their triangle has area $12 {cm}^{2}$ . Find the angle.
2 marks
A 10 m diagonal splits a quadrilateral into triangles. Their perpendicular heights to that diagonal, on opposite sides, are 4 m and 7 m. Find the total area.
3 marks
Two sides are fixed at 8 cm and 11 cm. Find the greatest possible triangle area and state when it occurs.
3 marks
Triangle ABC has $a = 6$ , $b = 9$ and $A = 35^{\circ}$ . Explain why $\frac{1}{2} \times 6 \times 9 \times sin 35^{\circ}$ is not justified.
3 marks

Read the worked solutions

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

A triangle has base 12 cm and perpendicular height 7 cm. Find its area.
2 marks
$K = \frac{1}{2} \times 12 \times 7 = 42 {cm}^{2}$ .
$42 {cm}^{2}$
Find the area using sides 10 m and 13 m enclosing $40^{\circ}$ . Round to one decimal place.
2 marks
$K = \frac{1}{2} \times 10 \times 13 \times sin 40^{\circ}$ .
$K \approx 41.8 m^{2}$ .
$41.8 m^{2}$
Sides 6 cm and 8 cm enclose an acute angle. Their triangle has area $12 {cm}^{2}$ . Find the angle.
2 marks
$sin C = \frac{2 \times 12}{6 \times 8} = 0.5$ .
$C = 30^{\circ}$ because it is acute.
$30^{\circ}$
A 10 m diagonal splits a quadrilateral into triangles. Their perpendicular heights to that diagonal, on opposite sides, are 4 m and 7 m. Find the total area.
3 marks
$K = \frac{1}{2} \times 10 \times 4 + \frac{1}{2} \times 10 \times 7$ .
$K = 20 + 35 = 55 m^{2}$ .
$55 m^{2}$
Two sides are fixed at 8 cm and 11 cm. Find the greatest possible triangle area and state when it occurs.
3 marks
$K = 44 sin C \leq 44$ .
The maximum sine value 1 occurs at $90^{\circ}$ .
$44 {cm}^{2}$ , at included angle $90^{\circ}$ .
Triangle ABC has $a = 6$ , $b = 9$ and $A = 35^{\circ}$ . Explain why $\frac{1}{2} \times 6 \times 9 \times sin 35^{\circ}$ is not justified.
3 marks
Sides a and b meet at C, whereas A is opposite a.
Find a matching included angle before using the formula.
A is not the included angle between sides a and b.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 General Mathematics: practical trigonometry, including similarity, right triangles, triangle area, sine/cosine rules and contextual interpretation.

Official curriculum source ↗

In this lesson

Understand the idea.

A height must be perpendicular

Two sides and an included angle

See the method in action.

Worked example 1

Identify the two sides and their included angle.

Substitute into the area formula.

Calculate and round the area.

Worked example 2

Set the area equal to the given value.

Isolate the sine.

Find both triangle-angle candidates.

Apply the acute-angle condition.

Try it yourself.

Check your understanding

Keep practising