Area of a Triangle (AA SL)
Most students meet the area of a triangle as \(\tfrac12\times\text{base}\times\text{height}\), but IB questions rarely hand you a height. Instead you're given two sides and the angle between them, and asked to work from there. This page covers the one formula that handles that case, plus the mistakes that cost marks when the given angle isn't the one enclosed by the two sides. It's part of the broader Triangles (Sine/Cosine Rules) topic.
25 questions on this sub-topic.
The area formula
Covered under IB syllabus reference SL3.2: the sine rule \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\), the cosine rule \(c^2=a^2+b^2-2ab\cos C\), and area as \(\tfrac12 ab\sin C\). This sub-topic focuses on the third of those.
Area, no height needed
\(\text{Area}=\tfrac12 ab\sin C\)
Use whenever you know two sides and the angle between them - this is in the formula booklet, so you don't need to memorise it.
C must be the included angle
\(a\) and \(b\) must be the two sides that meet at \(C\). If the angle you're given sits opposite one of the sides instead, this formula doesn't apply directly - you'd need the sine or cosine rule first to find the missing piece.
Need the full syllabus wording, formula-booklet reference table, or GDC guidance? See Triangles (Sine/Cosine Rules).
Worked examples
Find the exact area of a triangle with sides \(a = 6\) cm, \(b = 9\) cm and included angle \(C = 60^\circ\).
Worked solution
Two sides and the included angle: Area \(=\tfrac12 ab\sin C\). M1
Using the exact value \(\sin 60^\circ=\tfrac{\sqrt3}{2}\): \(\tfrac12(6)(9)\sin 60^\circ=27\cdot\tfrac{\sqrt3}{2}\), giving the exact surd form \(\text{Area}=\dfrac{27\sqrt3}{2}\ \text{cm}^2.\) A1
A triangle with sides \(9\) cm and \(12\) cm has area \(40\) cm\(^2\).
Find the acute included angle.
Worked solution
Set up the area equation: \(\tfrac12(9)(12)\sin\theta=40\Rightarrow \sin\theta=\dfrac{80}{108}\approx 0.741.\) M1
\(\sin\theta\approx0.741.\) A1
\(\theta=\sin^{-1}(0.741)\approx 47.8^\circ\) to 3 significant figures. A1
Common mistakes
- Using the wrong angle. \(C\) must be the angle enclosed between the two sides \(a\) and \(b\) you're substituting - plugging in an angle that sits opposite one of those sides gives a completely wrong area.
- Reaching for \(\tfrac12\times\text{base}\times\text{height}\) out of habit. If the question gives two sides and an angle rather than a perpendicular height, \(\tfrac12 ab\sin C\) is the formula that fits the data you actually have.
- Forgetting to sanity-check the angle sum. A triangle's angles must add to \(180^\circ\) - if your three angles don't, a mistake happened earlier (often a wrong degree/radian mode on the GDC).
- Rounding too early. Carry full calculator accuracy through \(\tfrac12ab\sin C\) and only round the final answer - rounding the sine value early can shift the last significant figure of the area.
Ready to practise properly?
24 triangle-area questions, marked instantly like the real exam.
Quick answers
What is the formula for the area of a triangle without a height?
\(\text{Area}=\tfrac12 ab\sin C\), where \(a\) and \(b\) are two sides and \(C\) is the angle between them. It works even when you have no perpendicular height.
Do the two sides have to include the given angle?
Yes. \(C\) must be the angle enclosed between sides \(a\) and \(b\) - using an angle that sits opposite one of the sides instead gives the wrong area.