Triangles: area and perimeter

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The elementary table lists a general triangle, a right triangle, and an equilateral triangle, each with its own $A$ and $C$ . Three names look like three topics. They are one area formula with extra freezes. $C$ is the perimeter (often listed as circumference). This math handbook writes the included angle $γ$ in radians; a right angle is $π /2$ . See measuring angles.

Base, height, and the included angle

Take $a$ as a base, $b$ as the adjacent side, $h$ as the corresponding height, and $γ$ as the included angle:

$A = \frac{1}{2} ah = \frac{1}{2} ab sin γ, C = a + b + c .$

The two expressions for $A$ are the same fact: $h = b sin γ$ . Drag a vertex. No extra freeze: the dashed height and $γ$ both move. Let $s := C /2$ be the semi-perimeter. Then the three sides alone determine the same area (Heron's formula):

$A = s (s - a) (s - b) (s - c) .$

Extra — a general triangle, no right angle frozen. Drag a vertex. The dashed segment is the height to base a; γ is the included angle at A.

Freeze a right angle

Keep that formula and freeze $γ = π /2$ at $A$ . Then $sin γ = 1$ , so

$A = \frac{1}{2} ab, C = a + b + c .$

The third vertex stays on the perpendicular through $A$ . Drag it off that line and the construction refuses: the extra freeze is the right angle, not a new area rule.

Let $c$ be the hypotenuse, $a$ the opposite leg, and $b$ the adjacent leg to an acute angle $α$ . Then

$a = c sin α, b = c cos α, a = b tan α,$

and the sides satisfy the theorem of Pythagoras:

$a^{2} + b^{2} = c^{2} .$

Those identities pay the same freeze: $γ = π /2$ .

Extra — the right angle stays at A. Drag B for the adjacent side b. Drag C on the perpendicular for the opposite side a. α sits at B.

Keep the right angle and move it to the third vertex. Sit $C$ on the circle with diameter $A B$ : the angle at $C$ stays right (an extra freeze on where the right angle sits). Drop a perpendicular from $C$ to the hypotenuse. It meets $A B$ in two segments of lengths $p$ and $q$ . The altitude $h$ to the hypotenuse obeys

$h^{2} = pq .$

Drag $C$ on that circle. Drop the diameter constraint and the angle at $C$ is no longer right; $h^{2} = pq$ is gone.

Extra — C stays on the circle with diameter AB, so the right angle is at C. Drag C. The dashed altitude h splits the hypotenuse into p and q.

Three equal sides

Go back to the general formula and freeze the sides instead of an angle: every side has length $a$ . Then the included angle is $π /3$ and $sin (π /3) = 3 /2$ , so

$A = \frac{3}{4} a^{2}, C = 3 a .$

That is an equilateral triangle. Drag $B$ ; the third vertex is the image of $B$ under a rotation of $π /3$ about $A$ , so the three sides stay equal. Drop equal sides and you are back at $\frac{1}{2} ab sin γ$ .

Extra — every side stays a. Drag B; the third vertex follows by a rotation of π/3.

Squares and parallelograms from the same table live under quadrilaterals. Regular $n$ -gons live under regular n-gons. Circles, circular sectors, and ellipses live under circles, sectors, and ellipses.