The idea, step by step
A height must be perpendicular
Area is half a base multiplied by its matching perpendicular height, not by an arbitrary sloping side. You may choose another base only if you also change to its corresponding height.
In an obtuse triangle the height may meet an extension of the base. A diagonal can divide a quadrilateral into triangles whose areas are added.
Two sides and an included angle
When sides a and b meet at angle C, the height relative to side a is b sinC. This gives an area formula for right, acute and obtuse triangles.
The angle must be between the two selected sides. For fixed sides, the greatest area occurs at a right angle, because sine cannot exceed 1.
A common trap
Using a sloping side as the height.
Instead: The height must be perpendicular to the chosen base. Otherwise use an included-angle area rule.
See the thinking
Worked example 1
Sides of 7 cm and 9 cm enclose 60°. Find the triangle area to one decimal place.
- The known angle is included between the known sides.
- K=½×7×9×sin60°.
- K≈27.3 cm².
Worked example 2
Sides of 6 m and 8 m enclose an acute angle C. The area is 12 m². Find C.
- 12=½×6×8×sinC, so sinC=0.5.
- Inverse sine gives 30°.
- The supplementary angle 150° is rejected because C is acute.
Check your understanding
Find the area using sides 5 m and 8 m enclosing 90°.
Show answer
20 m²
½×5×8×sin90°=20 m².
Your turn
6 questions · 15 marks · Work at your own pace.
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°. Round to one decimal place.2 marks
- Sides 6 cm and 8 cm enclose an acute angle. Their triangle has area 12 cm². 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°. Explain why ½×6×9×sin35° 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=½×12×7=42 cm².
42 cm² - Find the area using sides 10 m and 13 m enclosing 40°. Round to one decimal place.2 marks
K=½×10×13×sin40°.
K≈41.8 m².
41.8 m² - Sides 6 cm and 8 cm enclose an acute angle. Their triangle has area 12 cm². Find the angle.2 marks
sinC=2×12/(6×8)=0.5.
C=30° because it is acute.
30° - 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=½×10×4+½×10×7.
K=20+35=55 m².
55 m² - Two sides are fixed at 8 cm and 11 cm. Find the greatest possible triangle area and state when it occurs.3 marks
K=44sinC≤44.
The maximum sine value 1 occurs at 90°.
44 cm², at included angle 90°. - Triangle ABC has a=6, b=9 and A=35°. Explain why ½×6×9×sin35° 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.
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 ↗