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Difference between revisions of "Bretschneider's formula"
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It can be derived with [[vector]] [[geometry]].
 
It can be derived with [[vector]] [[geometry]].
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−
==See Also==
 
−
* [[Brahmagupta's formula]]
 
−
* [[Geometry]]
 
  
 
==The Proof==
 
==The Proof==
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:<math> 4S^2 = (pq)^2\sin^2 A + (rs)^2\sin^2 C + 2pqrs\sin A\sin C. \, </math>
 
:<math> 4S^2 = (pq)^2\sin^2 A + (rs)^2\sin^2 C + 2pqrs\sin A\sin C. \, </math>
  
−
The [[cosine law]] implies that
+
The [[law of cosines]] implies that
 
:<math> p^2 + q^2 -2pq\cos A = r^2 + s^2 -2rs\cos C, \, </math>
 
:<math> p^2 + q^2 -2pq\cos A = r^2 + s^2 -2rs\cos C, \, </math>
 
because both sides equal the square of the length of the diagonal ''BD''. This can be rewritten as
 
because both sides equal the square of the length of the diagonal ''BD''. This can be rewritten as
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:<math>16S^2 = 16(T-p)(T-q)(T-r)(T-s) - 16pqrs \cos^2 \frac{A+C}2</math>
 
:<math>16S^2 = 16(T-p)(T-q)(T-r)(T-s) - 16pqrs \cos^2 \frac{A+C}2</math>
 
and Bretschneider's formula follows.
 
and Bretschneider's formula follows.
 +
 
NOTE TO ALL: this proof was taken from Wikipedia on December the 1st, 2006.
 
NOTE TO ALL: this proof was taken from Wikipedia on December the 1st, 2006.
 +
 +
==See Also==
 +
* [[Brahmagupta's formula]]
 +
* [[Geometry]]

Revision as of 04:11, 2 December 2006

Suppose we have a quadrilateral with edges of length $a,b,c,d$ (in that order) and diagonals of length $p, q$. Bretschneider's formula states that the area $[ABCD]=\frac{1}{4}*\sqrt{4p^2q^2-(b^2+d^2-a^2-c^2)^2}$.

It can be derived with vector geometry.

The Proof

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

$\begin{align} S &= \text{area of } \triangle ADB + \text{area of } \triangle BDC \\
                       &= \tfrac{1}{2}pq\sin A + \tfrac{1}{2}rs\sin C

\end{align}$ (Error compiling LaTeX. Unknown error_msg)

Therefore

$4S^2 = (pq)^2\sin^2 A + (rs)^2\sin^2 C + 2pqrs\sin A\sin C. \,$

The law of cosines implies that

$p^2 + q^2 -2pq\cos A = r^2 + s^2 -2rs\cos C, \,$

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

$\tfrac14 (r^2 + s^2 - p^2 - q^2)^2 = (pq)^2\cos^2 A +(rs)^2\cos^2 C -2 pqrs\cos A\cos C. \,$

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

$4S^2 + \tfrac14 (r^2 + s^2 - p^2 - q^2)^2 = (pq)^2 + (rs)^2 - 2pqrs\cos (A+C). \,$

This can be written as

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

Introducing the semiperimeter

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

the above becomes

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

and Bretschneider's formula follows.

NOTE TO ALL: this proof was taken from Wikipedia on December the 1st, 2006.

See Also

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