Area of a Triangle
The area of a triangle equals one-half times its base times its height, representing the region enclosed by its three sides.
Formula
A = \frac{1}{2} \times b \times h
Definition
The area of a triangle with base $b$ and perpendicular height $h$ is half of base times height, $A = (1/2)bh$, since a triangle is always exactly half of a rectangle with the same base and height (even for obtuse triangles, where the altitude foot falls outside the triangle). When only the three side lengths $a$, $b$, $c$ are known, Heron's formula finds the area from the semi-perimeter $s$: $A = \sqrt{s(s-a)(s-b)(s-c)}$. In coordinates, for vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$, the signed area is $A = (1/2)\left|\det\begin{bmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{bmatrix}\right|$, and in 3-D, if two sides are vectors $u$ and $v$, the area is $(1/2)|u \times v|$.
Example
A triangle with a base of $10$ cm and a height of $6$ cm has $A = (1/2) \times 10 \times 6 = 30$ cm². The $5$-$12$-$13$ right triangle gives the same answer both ways: $A = (1/2)(5)(12) = 30$ directly, or via Heron's formula with $s = 15$: $A = \sqrt{15 \cdot 10 \cdot 3 \cdot 2} = \sqrt{900} = 30$. For vertices $(0,0)$, $(4,0)$, $(1,3)$: $A = (1/2)\left|\det\begin{bmatrix}0&0&1\\4&0&1\\1&3&1\end{bmatrix}\right| = (1/2)|0(0-3) - 0(4-1) + 1(12-0)| = (1/2)(12) = 6$.
Key Insight
Draw a rectangle around any triangle using its base and height, and the triangle always fills exactly half of it, which is where the $(1/2)$ in the formula comes from. Heron's formula is remarkable because it finds the area using only side lengths, without needing the height, making it especially useful in coordinate geometry and surveying. The determinant formula for triangle area is the foundation of barycentric coordinates, computational geometry algorithms, and the shoelace formula for polygon areas, which extends the idea to any polygon.