Difference between revisions of "Brahmagupta's Formula"
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===Proof=== | ===Proof=== | ||
| − | {{ | + | If we draw <math>AC</math>, we find that <math>[ABCD]=\frac{ab\sin B}{2}+\frac{cd\sin D}{2}=\frac{ab\sin B+cd\sin D}{2}</math>. Since <math>B+D=180^\circ</math>, <math>\sin B=\sin D</math>. Hence, <math>[ABCD]=\frac{\sin B(ab+cd)}{2}</math>. Multiplying by 2 and squaring, we get: |
| − | + | <cmath>4[ABCD]}^2=\sin^2 B(ab+cd)^2</cmath> | |
| + | Substituting <math>\sin^2B=1-\cos^2B</math> results in | ||
| + | <cmath>4[ABCD]^2=(1-\cos^2B)(ab+cd)^2=(ab+cd)^2-\cos^2B(ab+cd)^2</cmath> | ||
| + | By the Law of Cosines, <math>a^2+b^2-2ab\cos B=c^2+d^2-2cd\cos D</math>. <math>\cos B=-\cos D</math>, so a little rearranging gives | ||
| + | <cmath>2\cos B(ab+cd)=a^2+b^2-c^2-d^2</cmath> | ||
| + | <cmath>4[ABCD]^2=(ab+cd)^2-\frac{1}{4}(a^2+b^2-c^2-d^2)^2</cmath> | ||
| + | <cmath>16[ABCD]^2=4(ab+cd)^2-(a^2+b^2-c^2-d^2)^2</cmath> | ||
| + | <cmath>16[ABCD]^2=(2(ab+cd)+(a^2+b^2-c^2-d^2))(2(ab+cd)-(a^2+b^2-c^2-d^2))</cmath> | ||
| + | <cmath>16[ABCD]^2=(a^2+2ab+b^2-c^2+2cd-d^2)(-a^2+2ab-b^2+c^2+2cd+d^2)</cmath> | ||
| + | <cmath>16[ABCD]^2=((a+b)^2-(c-d)^2)((c+d)^2-(a-b)^2)</cmath> | ||
| + | <cmath>16[ABCD]^2=(a+b+c-d)(a+b-c+d)(c+d+a-b)(c+d-b+a)</cmath> | ||
| + | <cmath>16[ABCD]^2=16(s-a)(s-b)(s-c)(s-d)</cmath> | ||
| + | <cmath>[ABCD]=\sqrt{(s-a)(s-b)(s-c)(s-d)}</cmath> | ||
== Similar formulas == | == Similar formulas == | ||
Revision as of 21:52, 24 December 2008
Brahmagupta's Formula is a formula for determining the area of a cyclic quadrilateral given only the four side lengths.
Definition
Given a cyclic quadrilateral with side lengths 
, 
, 
, 
, the area 
can be found as:
![\[K = \sqrt{(s-a)(s-b)(s-c)(s-d)}\]](http://latex.artofproblemsolving.com/1/a/3/1a3a0e62995e8ed2d8e6e8a119206ee174dfdb18.png)
where 
is the semiperimeter of the quadrilateral.
Proof
If we draw 
, we find that ![$[ABCD]=\frac{ab\sin B}{2}+\frac{cd\sin D}{2}=\frac{ab\sin B+cd\sin D}{2}$](http://latex.artofproblemsolving.com/9/a/7/9a78a4a4d7823c156744a4ab766dcec6dee7ad68.png)
. Since 
, 
. Hence, ![$[ABCD]=\frac{\sin B(ab+cd)}{2}$](http://latex.artofproblemsolving.com/1/6/8/1684bb2cb4c40be745baf43185216216c383eff1.png)
. Multiplying by 2 and squaring, we get:
\[4[ABCD]}^2=\sin^2 B(ab+cd)^2\] (Error compiling LaTeX. Unknown error_msg)
Substituting 
results in
![\[4[ABCD]^2=(1-\cos^2B)(ab+cd)^2=(ab+cd)^2-\cos^2B(ab+cd)^2\]](http://latex.artofproblemsolving.com/1/f/2/1f28be489b766c55eae6df424d410139c5eb3df5.png)
By the Law of Cosines, 
. 
, so a little rearranging gives
![\[2\cos B(ab+cd)=a^2+b^2-c^2-d^2\]](http://latex.artofproblemsolving.com/7/4/9/749c4e5212d14409e90c0ab2ce8f25a5a809f281.png)
![\[4[ABCD]^2=(ab+cd)^2-\frac{1}{4}(a^2+b^2-c^2-d^2)^2\]](http://latex.artofproblemsolving.com/b/c/e/bcec630c2ff8130c1a61d3c01188d931bdd856ab.png)
![\[16[ABCD]^2=4(ab+cd)^2-(a^2+b^2-c^2-d^2)^2\]](http://latex.artofproblemsolving.com/9/1/f/91f36e4160b82996311964620d4914b61bb5c9e4.png)
![\[16[ABCD]^2=(2(ab+cd)+(a^2+b^2-c^2-d^2))(2(ab+cd)-(a^2+b^2-c^2-d^2))\]](http://latex.artofproblemsolving.com/c/2/2/c22c67c3a85044b92ad940c0aefd4f63ae091e80.png)
![\[16[ABCD]^2=(a^2+2ab+b^2-c^2+2cd-d^2)(-a^2+2ab-b^2+c^2+2cd+d^2)\]](http://latex.artofproblemsolving.com/8/8/2/8826dbf9ed70539e9715150f323d75aaa5921e26.png)
![\[16[ABCD]^2=((a+b)^2-(c-d)^2)((c+d)^2-(a-b)^2)\]](http://latex.artofproblemsolving.com/5/8/7/58740b6212d38b06f97975f3789b42d7d5f18488.png)
![\[16[ABCD]^2=(a+b+c-d)(a+b-c+d)(c+d+a-b)(c+d-b+a)\]](http://latex.artofproblemsolving.com/f/0/9/f0918d965a7ab2cba2461d3f81994df8d8236af7.png)
![\[16[ABCD]^2=16(s-a)(s-b)(s-c)(s-d)\]](http://latex.artofproblemsolving.com/f/9/0/f90de0123e7e0b19af70a51fdb63dbf701933a68.png)
![\[[ABCD]=\sqrt{(s-a)(s-b)(s-c)(s-d)}\]](http://latex.artofproblemsolving.com/e/b/7/eb7b6e6945de987f981c05f35be9b81105e76f09.png)
Similar formulas
Bretschneider's formula gives a formula for the area of a non-cyclic quadrilateral given only the side lengths; applying Ptolemy's Theorem to Bretschneider's formula reduces it to Brahmagupta's formula.
Brahmagupta's formula reduces to Heron's formula by setting the side length 
.
This article is a stub. Help us out by expanding it.
