Area of a Trapezoid
The area of a trapezoid is one-half the sum of its two parallel sides multiplied by the height between them.
Formula
A = \frac{1}{2}(b_1 + b_2) \times h
Definition
A trapezoid (trapezium in British English) has exactly one pair of parallel sides, called bases $b_1$ and $b_2$, separated by perpendicular height $h$; its area is $A = (1/2)(b_1 + b_2)h$, found by averaging the two bases and multiplying by the height. This can be derived by splitting the trapezoid into two triangles sharing the height $h$, and it is a genuine generalization: if $b_1 = b_2$ it becomes the parallelogram formula ($b \times h$), and if $b_2 = 0$ it becomes the triangle formula $(1/2) \cdot b \cdot h$, so one formula covers rectangle, parallelogram, triangle, and trapezoid. The same formula is the simplest case of numerical integration by the trapezoidal rule, which approximates the definite integral of $f$ on $[a,b]$ as $(b-a)(f(a)+f(b))/2$; the midsegment (median) of a trapezoid has length $(b_1+b_2)/2$, so $A = \text{midsegment} \times h$.
Example
A trapezoid with bases $6$ cm and $10$ cm and height $4$ cm has $A = (1/2)(6 + 10)(4) = (1/2)(16)(4) = 32$ cm². A highway cross-section is trapezoidal with top width $20$ m, bottom width $30$ m, and depth $3$ m, giving area $(1/2)(20 + 30)(3) = 75$ m$^2$, which can be multiplied by road length to get the volume of material. Applying the trapezoidal rule to $f(x) = x^2$ on $[0,2]$ gives $(2)(0+4)/2 = 4$, versus the true integral of $8/3 = 2.67$, an error of $O(h^2)$ that still beats the cruder rectangle rule.
Key Insight
Think of the formula as finding the area of a rectangle whose width is the average of the two bases, times the height. The midsegment connection explains why the trapezoidal rule works: the area under a curve is approximated by a trapezoid whose parallel sides are the function values at the endpoints, and the midsegment is the arithmetic mean of those two values.