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G7.11 Area of a Triangle Using $\frac12ab\sin C$

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01 — Explanation

Why do I need a new area formula?

$\text{Area}=\tfrac12\times\text{base}\times\text{height}$ only works when you know the perpendicular height. For a triangle where you only know two sides and the angle between them, that formula is given on the formula sheet instead:

\text{Area}=\frac12 ab\sin C

where $C$ is the angle included between sides $a$ and $b$ .

02 — Explanation

Where does this formula come from?

Drop a perpendicular height $h$ from one vertex onto the base $a$ . In the right-angled triangle this creates, $\sin C=\dfrac hb$ , so $h=b\sin C$ . Substituting into $\tfrac12\times a\times h$ gives $\tfrac12 ab\sin C$ directly.

03 — Explanation

Does it still work if the included angle is obtuse?

Yes — because $\sin(180°-C)=\sin C$ , the formula is valid for any angle between $0°$ and $180°$ . You don't need to do anything special; just enter the obtuse angle directly into the calculator.

Exam Tip: The angle you use must be the one squeezed between the two given sides (an "SAS" setup). Using a different angle in the triangle is the most common error — physically trace the two sides back to where they meet before substituting.

04 — Explanation

How do I use the formula to find a missing side or angle?

Substitute everything you know, then rearrange.

Missing side: Area $=30\text{ cm}^2$ , sides $12$ and $x$ , included angle $38°$ .

30=\tfrac12(12)(x)\sin38° \Rightarrow x=\frac{30}{6\sin38°}\approx8.12\text{ cm}

Missing angle: Area $=37\text{ cm}^2$ , sides $11$ and $7$ .

37=\tfrac12(11)(7)\sin X \Rightarrow \sin X=\frac{37}{38.5}\approx0.961 \Rightarrow X\approx74.0°

Exam Tip: $\sin X=k$ has two solutions between $0°$ and $180°$ — the calculator's answer, and $180°$ minus it. If the question says the angle is obtuse, you must use $180°-74.0°\approx106.0°$ instead.

05 — Explanation

How do I find the area when I only know all three sides (SSS)?

The area formula needs an angle, so first find one using the cosine rule, then feed it into the area formula.

Worked exam-style question: A triangle has sides $7.4$ , $4.8$ , $4.4$ cm. Find its area.

Find the angle $B$ between the $7.4$ and $4.8$ sides (opposite side $4.4$ ):

\cos B=\frac{7.4^2+4.8^2-4.4^2}{2(7.4)(4.8)}=\frac{58.44}{71.04}\approx0.8227 \Rightarrow B\approx34.65°

Use the unrounded angle in the area formula: $\text{Area}=\tfrac12(7.4)(4.8)\sin(34.65°)\approx10.1\text{ cm}^2$ .

Exam Tip: Carry the full, unrounded angle into the area calculation (store it with the calculator's Ans button) — rounding the angle first before squaring it into the final formula is a common way to lose the last accuracy mark.

06 — Explanation

How do I find the area of a segment of a circle?

Subtract the triangle (formed by the two radii and the chord) from the full sector.

\text{Area}_{\text{segment}} = \frac{\theta}{360°}\pi r^2 - \frac12 r^2\sin\theta

For $r=7$ cm, $\theta=135°$ : sector $=\tfrac{135}{360}\pi(49)\approx57.7$ , triangle $=\tfrac12(49)\sin135°\approx17.3$ , segment $\approx40.4\text{ cm}^2$ .

07 — Common mistakes

Common mistakes to avoid

Using a non-included angle with the two given sides.
Calculator left in radian mode — a negative or wildly wrong area is the giveaway.
Forgetting the $\tfrac12$ in front of the formula.
Ignoring an "obtuse" instruction and giving the acute calculator answer instead of $180°$ minus it.