Triangle Area Formulas

In Grade 8 you learned that the area of a triangle equals half the product of a base and the corresponding height: $S = \tfrac{1}{2}ah_a$ . Using trigonometry and the circumradius, we can derive several powerful new formulas.

Two Sides and the Included Angle

⚡Theorem — Theorem 6.1

The area of a triangle equals half the product of two sides and the sine of the included angle:

$\boxed{S = \frac{1}{2}ab\sin\gamma}$

Proof. Draw altitude $BD$ from $B$ to $AC$ . Then $S = \tfrac{1}{2} \cdot AC \cdot BD = \tfrac{1}{2} b \cdot BD$ . In triangle $BDC$ : $BD = a\sin\gamma$ (using the sine definition for obtuse angles too). Therefore $S = \tfrac{1}{2}ab\sin\gamma$ . $\blacktriangleleft$

Heron’s Formula

⚡Theorem — Theorem 6.2 — Heron's Formula

The area of a triangle with sides $a$ , $b$ , $c$ and semiperimeter $p = \dfrac{a+b+c}{2}$ is:

$\boxed{S = \sqrt{p(p-a)(p-b)(p-c)}}$

✎Example — Using Heron's Formula

The sides of a triangle are $17$ cm, $65$ cm, and $80$ cm. Find the area, the smallest altitude, and the inradius and circumradius.

Solution. $p = \frac{17+65+80}{2} = 81$ cm.

$S = \sqrt{81 \cdot 64 \cdot 16 \cdot 1} = 9 \cdot 8 \cdot 4 = 288 \text{ cm}^2$

The smallest altitude is to the longest side $c = 80$ : $h_c = \frac{2S}{c} = \frac{576}{80} = 7.2$ cm.

Inradius: $r = \frac{S}{p} = \frac{288}{81} = \frac{32}{9}$ cm.

Circumradius: $R = \frac{abc}{4S} = \frac{17 \cdot 65 \cdot 80}{4 \cdot 288} = \frac{5525}{72}$ cm. $\blacktriangleleft$

Area via Circumradius

⚡Theorem — Theorem 6.3

$S = \frac{abc}{4R}$

where $R$ is the circumradius. Equivalently, $R = \dfrac{abc}{4S}$ .

Area via Inradius

⚡Theorem — Theorem 6.4

$S = pr$

where $p$ is the semiperimeter and $r$ is the inradius. Equivalently, $r = \dfrac{S}{p}$ .

Circumscribed Polygon

⚡Theorem — Theorem 6.5

The area of a polygon circumscribed about a circle of radius $r$ equals:

$S = pr$

where $p$ is its semiperimeter.

Parallelogram and Quadrilateral

⚡Theorem — Theorem 6.6 — Parallelogram Area

The area of a parallelogram with adjacent sides $a$ , $b$ and included angle $\alpha$ is:

$S = ab\sin\alpha$

⚡Theorem — Theorem 6.7 — Convex Quadrilateral Area

The area of a convex quadrilateral with diagonals $d_1$ , $d_2$ and the angle $\varphi$ between them is:

$S = \frac{1}{2}d_1 d_2 \sin\varphi$

Summary of Formulas

Formula	When to use
$S = \tfrac{1}{2}ab\sin\gamma$	Two sides + included angle known
$S = \sqrt{p(p-a)(p-b)(p-c)}$	All three sides known
$S = \tfrac{abc}{4R}$	All sides + circumradius known
$S = pr$	Semiperimeter + inradius known