Trapezoid formula reference
a and b are the parallel bases, c and d are the legs, h is perpendicular height, A is area, P is perimeter and m is the midsegment. Special-shape formulas apply only under the stated assumption.
| Find | Formula | Valid for |
|---|---|---|
| Area | A = (a+b)h/2 | Every trapezoid |
| Height | h = 2A/(a+b) | Every trapezoid when area and bases are known |
| Either base | a = 2A/h−b; b = 2A/h−a | Every trapezoid when area, height and the other base are known |
| Perimeter | P = a+b+c+d | Every trapezoid |
| A missing side from perimeter | a = P−b−c−d, with equivalent rearrangements for b, c or d | Every trapezoid, subject to geometric validity |
| Midsegment | m = (a+b)/2 | Every trapezoid |
| Isosceles leg and height | L = √(h²+((a−b)/2)²); h = √(L²−((a−b)/2)²) | Isosceles only |
| Isosceles diagonal | p = q = √(h²+((a+b)/2)²) | Isosceles only |
| Right trapezoid | one leg = h; other leg = √(h²+(a−b)²) | Right only |
| General height from four sides | s=a−b; x=(c²−d²+s²)/(2s); h=√(c²−x²) | General, unequal bases, supported orientation |
Why the area rearrangements work
Starting with A=(a+b)h/2, multiply by 2 and divide by a+b to get h=2A/(a+b). Dividing by h instead gives a+b=2A/h, so subtract the known base to find the missing one. For four-side geometry, dropping perpendiculars from the top base creates two right triangles; subtracting their Pythagorean equations produces the projection x shown in the table.