Irregular Quadrilateral Area Calculator

How it works

Four side lengths on their own do not fix a quadrilateral: hinge the corners and the shape flexes, changing its area. One more measurement pins it down. The easiest to take with a tape is a diagonal, which cuts the shape into two triangles; each triangle's area follows from its three sides by Heron's formula, and the quadrilateral's area is their sum.

If you know two opposite angles instead, Bretschneider's formula (1842) gives the area directly from the four sides and the half-sum of those angles. It is the four-sided cousin of Heron's formula: when the angles add to 180° the shape can be inscribed in a circle and the formula collapses to Brahmagupta's, and for a rectangle it collapses to length × width.

The calculator also reports both diagonals. In the diagonal method the second diagonal is found by laying the known one on an axis and locating the two far corners; in the angles method the diagonal BD comes from the law of cosines at corner A, and the other from the same construction. Because angle A already determines the shape, angle C must be the one that figure actually has; the page does not check that. If C does not match, the diagonal AC shown belongs to the figure fixed by A alone.

Formula

perimeter = a + b + c + d          s = perimeter ÷ 2
diagonal AC = e:  area = heron(a, b, e) + heron(c, d, e)
                  heron(p, q, r) = √(t(t − p)(t − q)(t − r)),  t = (p + q + r) ÷ 2
angles A and C:   area = √((s − a)(s − b)(s − c)(s − d) − a·b·c·d·cos²((A + C) ÷ 2))
                  BD = √(a² + d² − 2·a·d·cos A)

Example

A lot with sides a = 5, b = 6, c = 7 and d = 8 ft and a diagonal AC of 7 ft. Triangle ABC has sides 5, 6, 7: t = 9, area √(9 × 4 × 3 × 2) = √216 = 14.6969 sq ft. Triangle ACD has sides 7, 8, 7: t = 11, area √(11 × 4 × 3 × 4) = √528 = 22.9783 sq ft. The lot is 14.6969 + 22.9783 = 37.6752 sq ft with a perimeter of 26 ft, and the other diagonal BD works out to 10.9234 ft.

The same four sides with angle A = 90° and the matching angle C = 92.73°: s = 13, (13 − 5)(13 − 6)(13 − 7)(13 − 8) = 1,680, and 5 × 6 × 7 × 8 × cos²(91.365°) = 0.9533, so the area is √(1,680 − 0.9533) = 40.9762 sq ft. A right angle at A opens the shape wider than the 7 ft diagonal did, so the area is larger.

Assumptions and limitations

• The quadrilateral is simple (no crossing edges) and the sides are entered in order around it: a, then b, then c, then d.
• In the diagonal method the diagonal must be AC, joining the corner between d and a to the corner between b and c. The other diagonal pairs the sides differently and gives a different answer.
• The diagonal method also needs AC to run inside the shape: the shape is convex, or its one inward-pointing corner is at B or D. If the inward corner is at A or C, AC runs outside the shape. The real area is then the larger triangle minus the smaller one, but this calculator adds them, so the result is too large. Use the angles method or the polygon-from-coordinates calculator for such a shape.
• In the angles method, A and C must be the two opposite corners of the same figure. The four sides and angle A already determine angle C; a pair that does not belong together gives a number that is not the area of any real quadrilateral.
• Lengths are in the unit you choose and the area is in that unit squared. The unit labels the results; nothing is converted.

Frequently asked questions