Bretschneider's formula: Difference between revisions

Revision as of 21:52, 14 August 2020

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 $[ABCD]=\frac{1}{4} \cdot \sqrt{4p^2q^2-(b^2+d^2-a^2-c^2)^2}$ .

It can be derived with vector geometry.

Proof

Suppose a quadrilateral has sides $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ such that $\vec{a} + \vec{b} + \vec{c} + \vec{d} = \vec{0}$ and that the diagonals of the quadrilateral are $\vec{p} = \vec{b} + \vec{c} = -\vec{a} - \vec{d}$ and $\vec{q} = \vec{a} + \vec{b} = -\vec{c} - \vec{d}$ . The area of any such quadrilateral is $\frac{1}{2} |\vec{p} \times \vec{q}|$ .

$K = \frac{1}{2} |\vec{p} \times \vec{q}|$

Lagrange's Identity states that $|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2=|\vec{a}\times\vec{b}|^2 \implies \sqrt{|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2}=|\vec{a}\times\vec{b}|$ . Therefore:

$K = \frac{1}{2} \sqrt{|\vec{p}|^2|\vec{q}|^2 - (\vec{p} \cdot \vec{q})^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - (2 \vec{p} \cdot \vec{q})^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [2 (\vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b})]^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [2 \vec{b} \cdot (\vec{a} + \vec{b}) + 2 \vec{c} \cdot (\vec{a} + \vec{b})]^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [-2 \vec{b} \cdot (\vec{c} + \vec{d}) + 2 \vec{c} \cdot (\vec{a} + \vec{b})]^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [-2 \vec{b} \cdot \vec{c} - 2 \vec{b} \cdot \vec{d} + 2 \vec{a} \cdot \vec{c} + 2 \vec{b} \cdot \vec{c}]^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [2 \vec{a} \cdot \vec{c} - 2 \vec{b} \cdot \vec{d}]^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - ([(\vec{a} + \vec{c})\cdot(\vec{a} + \vec{c}) - \vec{a}\cdot\vec{a} - \vec{c}\cdot\vec{c}] - [(\vec{b} + \vec{d})\cdot(\vec{b} + \vec{d}) - \vec{b}\cdot\vec{b} - \vec{d}\cdot\vec{d}])^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{c}|^2 + (\vec{a} + \vec{c})\cdot(\vec{a} + \vec{c}) - (\vec{b} + \vec{d})\cdot(\vec{b} + \vec{d})]^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{c}|^2 + |\vec{a} + \vec{c}|^2 - |\vec{b} + \vec{d}|^2]^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{c}|^2 + |\vec{a} + \vec{c}|^2 - |-(\vec{a} + \vec{c})|^2]^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{c}|^2]^2}$

Then if $a, b, c, d$ represent $|\vec{a}|, |\vec{b}|, |\vec{c}|, |\vec{d}|$ (and are thus the side lengths) while $p, q$ represent $|\vec{p}|, |\vec{q}|$ (and are thus the diagonal lengths), the area of a quadrilateral is:

$K = \frac{1}{4} \sqrt{4p^2q^2 - (b^2 + d^2 - a^2 - c^2)^2}$

@@ Line 1: / Line 1: @@
 Suppose we have a [[quadrilateral]] with [[edge]]s of length <math>a,b,c,d</math> (in that order) and [[diagonal]]s of length <math>p, q</math>. '''Bretschneider's formula''' states that the [[area]]
-<math>[ABCD]=\frac{1}{4}*\sqrt{4p^2q^2-(b^2+d^2-a^2-c^2)^2}</math>.
+<math>[ABCD]=\frac{1}{4} \cdot \sqrt{4p^2q^2-(b^2+d^2-a^2-c^2)^2}</math>.
 It can be derived with [[vector]] [[geometry]].

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Bretschneider's formula: Difference between revisions

Revision as of 21:52, 14 August 2020

Proof

See Also