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Bretschneider's formula

A quadrilateral

In geometry, Bretschneider's formula is the following expression for the area of a quadrilateral,

$ \text{area}=\sqrt{(T-p)(T-q)(T-r)(T-s)-pqrs\cos^2\left(\tfrac{A+C}{2}\right)} $

Here, $ p,q,r,s $ are the sides of the quadrilateral, $ T $ is half the perimeter, and $ A,C $ are two opposite angles.

Bretschneider's formula works on any quadrilateral regardless of whether it is cyclic or not.

Proof of Bretschneider's formula

Denote the area of the quadrilateral by $ S $ . Then we have

$ S=\text{area}(\triangle ADB)+\text{area}(\triangle BDC)=\tfrac12ps\sin(A)+\tfrac12qr\sin(C) $

Therefore

$ 4S^2=(ps)^2\sin^2(A)+(qr)^2\sin^2(C)+2pqrs\sin(A)\sin(C) $

The cosine law implies that

$ p^2+s^2-2ps\cos(A)=q^2+r^2-2qr\cos(C) $

because both sides equal the square of the length of the diagonal $ BD $ . This can be rewritten as

$ \tfrac14(q^2+r^2-p^2-s^2)^2=(ps)^2\cos^2(A)+(qr)^2\cos^2(C)-2pqrs\cos(A)\cos(C) $

Substituting this in the above formula for $ 4S^2 $ yields

$ 4S^2+\tfrac14(q^2+r^2-p^2-s^2)^2=(ps)^2+(qr)^2-2pqrs\cos(A+C) $

This can be written as

$ 16S^2=(r+s+p-q)(r+s+q-p)(r+p+q-s)(s+p+q-r)-16pqrs\cos^2\left(\tfrac{A+C}{2}\right) $

Introducing the semiperimeter

$ T=\frac{p+q+r+s}{2} $

the above becomes

$ 16S^2=16(T-p)(T-q)(T-r)(T-s)-16pqrs\cos^2\left(\tfrac{A+C}{2}\right) $

and Bretschneider's formula follows.

Related formulas

Bretschneider's formula generalizes Brahmagupta's formula for the area of a cyclic quadrilateral, which in turn generalizes Heron's formula for the area of a triangle.

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