01 · Exact area
Two sides are 4 cm and 9 cm, with included angle 60°. Find the area.
Hint
sin 60° = √3/2.
Worked solution
K = ½ × 4 × 9 × √3/2 = 9√3 cm²
Understand · explore · practise
Triangle areas
Find triangle areas using ½ab sin C, including obtuse angles, three given sides, unknown angles, exact values and maximum-area problems.
Before you startSine and cosine rules, triangle heights and simple algebra
01 / From height to area
Use the angle between the two sides.
A triangle with base c and perpendicular height h has area K = ½ch. If another side b meets the base at angle A, its perpendicular height is b sin A.
K = ½bc sin A
= ½ca sin B = ½ab sin C
In each version, the angle lies between the two sides you multiply.
The formula works for obtuse angles too: sin(180° − A) = sin A gives the positive height even when the foot lies outside the base. Area has squared units.
The perpendicular height sets the areaChoose and compare
b = 6, c = 8, A = 60°. Height = 3√3 and area = 12√3 ≈ 20.785. Supplementary angles give the same area.
Watch the height explain the sine area formula
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Calculate an area
An exact sine can give an exact area.
For sides 6 cm and 8 cm enclosing 30°:
K = ½ × 6 × 8 × 1/2 = 12 cm²
Enclosing 150° gives the same area because sin 150° = 1/2, although the third side and the shape differ. At 60°, the area is 12√3 cm². Keep this exact value if the question asks for an exact answer.
03 / Find the angle
Area alone may leave two possible shapes.
If sides 6 and 8 enclose an angle A and their area is 12, then sin A = 2K/(bc) = 1/2. The candidates are 30° and 150°.
If the third side is required to be the longest, test it with the cosine rule. At 30°, a² = 100 − 48√3, so a < 8 and that candidate fails. At 150°, a² = 100 + 48√3, so a > 8 and it succeeds.
Do not assume the largest angle must be obtuse. For example, two sides of length 6 enclosing 80° have a third side greater than 6, even though the largest angle is acute.
04 / Three given sides
Find the included angle or a perpendicular height first.
A triangle has sides 5, 5 and 6. Its altitude bisects the 6-unit base, giving h = √(5² − 3²) = 4 and area 12.
You can also use the cosine rule for the angle between the equal sides:
cos A = (25 + 25 − 36)/50 = 7/25
sin A = √(1 − 49/625) = 24/25
K = ½ × 5 × 5 × 24/25 = 12
The positive square root is correct because every interior triangle angle has positive sine. For a general SSS triangle, this method combines the cosine rule with the area formula without rounding an intermediate angle.
05 / Maximum area
Fixed sides give their largest area at a right angle.
For fixed positive b and c, sin A ≤ 1. Hence K ≤ ½bc, with equality at A = 90°. Sides 6 and 8 can enclose at most 24 square units.
When the side lengths vary too
Two sides are x + 1 and 7 − x, their included angle is 30°, and −1 < x < 7. Then:
K = ¼(x + 1)(7 − x)
= ¼(−x² + 6x + 7)
= 4 − ¼(x − 3)²
The maximum is 4 at x = 3. The two sides are then both 4. Check that the maximising x is inside the permitted interval.
06 / Unknown lengths
Area conditions can give a quadratic.
Two sides are x and x + 4, their included angle is 30°, and the area is 8 square units.
8 = ½x(x + 4)sin 30°
x² + 4x − 32 = 0
(x + 8)(x − 4) = 0
Only x = 4 is a positive length. If a question gives a perimeter, express one side in terms of another first, then combine the appropriate area or cosine-rule equation with that constraint.
07 / Your turn
Check the included angle and squared units.
Leave exact values as fractions or surds where possible.
02 · Obtuse triangle
Two sides are 5 m and 12 m, with included angle 150°. Find the area.
Hint
sin 150° = 1/2.
Worked solution
K = ½ × 5 × 12 × 1/2 = 15 m²
03 · Two angles
Two sides are 5 and 8 and the area is 10. Find both possible included angles.
Hint
Rearrange the area formula for the sine.
Worked solution
sin A = 20/40 = 1/2
A = 30° or 150°
04 · Impossible area
Can sides 4 and 7 enclose a triangle of area 15?
Hint
What is their maximum possible area?
Worked solution
K ≤ ½ × 4 × 7 = 14
No: the proposed area exceeds the maximum.
05 · Three sides
A triangle has sides 10, 10 and 12. Find its exact area.
Hint
Drop an altitude to the 12-unit base.
Worked solution
h = √(100 − 36) = 8
K = ½ × 12 × 8 = 48
06 · Find a side parameter
Sides x and x + 2 enclose 30°. Their area is 6. Find x.
Hint
The quadratic must give positive lengths.
Worked solution
x(x + 2)/4 = 6
(x + 6)(x − 4) = 0 ⇒ x = 4
07 · Varying sides
Sides x and 10 − x enclose 30°, with 0 < x < 10. Find the maximum area.
Hint
Complete the square.
Worked solution
K = (10x − x²)/4
= 25/4 − (x − 5)²/4
Maximum 25/4 at x = 5.
08 · Largest does not mean obtuse
An equilateral triangle has side 6. Is each largest angle obtuse? Find its area.
Hint
All three angles are tied for largest.
Worked solution
A = B = C = 60°
K = ½ × 6 × 6 × √3/2 = 9√3
No. Every angle is acute.
08 / Recap
Area uses a perpendicular height or an included-angle sine.
- Use ½ × base × perpendicular height when that is easiest.
- In ½bc sin A, A must be between b and c.
- Two supplementary included angles can give the same area.
- With three sides, use cosine first and retain exact values where possible.
- Check a proposed area against the maximum ½bc.
Section 1 of 8 · From height to area