# 婆羅摩笈多公式

（重定向自婆罗摩笈多公式

## 基本形式

${\displaystyle {\sqrt {(s-a)(s-b)(s-c)(s-d)}}}$

${\displaystyle s={\frac {a+b+c+d}{2}}}$

## 证明

${\displaystyle ={\frac {1}{2}}pq\sin A+{\frac {1}{2}}rs\sin C.}$

${\displaystyle {\mbox{Area}}={\frac {1}{2}}pq\sin A+{\frac {1}{2}}rs\sin A}$
${\displaystyle ({\mbox{Area}})^{2}={\frac {1}{4}}\sin ^{2}A(pq+rs)^{2}}$
${\displaystyle 4({\mbox{Area}})^{2}=(1-\cos ^{2}A)(pq+rs)^{2}\,}$
${\displaystyle 4({\mbox{Area}})^{2}=(pq+rs)^{2}-cos^{2}A(pq+rs)^{2}.\,}$

${\displaystyle \triangle ADB}$ ${\displaystyle \triangle BDC}$ 利用余弦定理，我们有：

${\displaystyle DB^{2}=p^{2}+q^{2}-2pq\cos A=r^{2}+s^{2}-2rs\cos C.\,}$

${\displaystyle 2\cos A(pq+rs)=p^{2}+q^{2}-r^{2}-s^{2}.\,}$

${\displaystyle 4({\mbox{Area}})^{2}=(pq+rs)^{2}-{\frac {1}{4}}(p^{2}+q^{2}-r^{2}-s^{2})^{2}}$
${\displaystyle 16({\mbox{Area}})^{2}=4(pq+rs)^{2}-(p^{2}+q^{2}-r^{2}-s^{2})^{2},\,}$

${\displaystyle (2(pq+rs)+p^{2}+q^{2}-r^{2}-s^{2})(2(pq+rs)-p^{2}-q^{2}+r^{2}+s^{2})\,}$
${\displaystyle =[(p+q)^{2}-(r-s)^{2}][(r+s)^{2}-(p-q)^{2}]\,}$
${\displaystyle =(p+q+r-s)(p+q+s-r)(p+r+s-q)(q+r+s-p).\,}$

${\displaystyle 16({\mbox{Area}})^{2}=16(T-p)(T-q)(T-r)(T-s).\,}$

${\displaystyle {\mbox{Area}}={\sqrt {(T-p)(T-q)(T-r)(T-s)}}.}$

## 更特殊情況

${\displaystyle {\sqrt {abcd}}}$

### 证明

${\displaystyle S={\sqrt {(p-a)(p-b)(p-c)(p-d)}}}$

${\displaystyle p={\frac {a+b+c+d}{2}}}$

${\displaystyle a+c=b+d}$

${\displaystyle p-a={\frac {b+c+d-a}{2}}={\frac {a+c+c-a}{2}}=c}$

${\displaystyle p-b=d}$ ${\displaystyle p-c=a}$ ${\displaystyle p-d=b}$

${\displaystyle S={\sqrt {abcd}}}$

## 一般情況

${\displaystyle {\sqrt {(p-a)(p-b)(p-c)(p-d)-abcd\cos ^{2}\theta }}}$

${\displaystyle K={\sqrt {(s-a)(s-b)(s-c)(s-d)-\textstyle {1 \over 4}(ac+bd+pq)(ac+bd-pq)}}\,}$

## 相關定理

1. ^ J. L. Coolidge, "A Historically Interesting Formula for the Area of a Quadrilateral", American Mathematical Monthly, 46 (1939) pp. 345-347.