Difference between revisions of "Brahmagupta's Formula"

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(Proof)
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===Proof===
 
===Proof===
{{incomplete|proof}}
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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:
 
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<cmath>4[ABCD]}^2=\sin^2 B(ab+cd)^2</cmath>
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Substituting <math>\sin^2B=1-\cos^2B</math> results in
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<cmath>4[ABCD]^2=(1-\cos^2B)(ab+cd)^2=(ab+cd)^2-\cos^2B(ab+cd)^2</cmath>
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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
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<cmath>2\cos B(ab+cd)=a^2+b^2-c^2-d^2</cmath>
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<cmath>4[ABCD]^2=(ab+cd)^2-\frac{1}{4}(a^2+b^2-c^2-d^2)^2</cmath>
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<cmath>16[ABCD]^2=4(ab+cd)^2-(a^2+b^2-c^2-d^2)^2</cmath>
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<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>
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<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>
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<cmath>16[ABCD]^2=((a+b)^2-(c-d)^2)((c+d)^2-(a-b)^2)</cmath>
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<cmath>16[ABCD]^2=(a+b+c-d)(a+b-c+d)(c+d+a-b)(c+d-b+a)</cmath>
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<cmath>16[ABCD]^2=16(s-a)(s-b)(s-c)(s-d)</cmath>
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<cmath>[ABCD]=\sqrt{(s-a)(s-b)(s-c)(s-d)}</cmath>
  
 
== Similar formulas ==
 
== Similar formulas ==

Revision as of 22:52, 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 ${a}$, ${b}$, ${c}$, ${d}$, the area ${K}$ can be found as:

\[K = \sqrt{(s-a)(s-b)(s-c)(s-d)}\]

where $s=\frac{a+b+c+d}{2}$ is the semiperimeter of the quadrilateral.


Proof

If we draw $AC$, we find that $[ABCD]=\frac{ab\sin B}{2}+\frac{cd\sin D}{2}=\frac{ab\sin B+cd\sin D}{2}$. Since $B+D=180^\circ$, $\sin B=\sin D$. Hence, $[ABCD]=\frac{\sin B(ab+cd)}{2}$. Multiplying by 2 and squaring, we get:

\[4[ABCD]}^2=\sin^2 B(ab+cd)^2\] (Error compiling LaTeX. Unknown error_msg)

Substituting $\sin^2B=1-\cos^2B$ results in \[4[ABCD]^2=(1-\cos^2B)(ab+cd)^2=(ab+cd)^2-\cos^2B(ab+cd)^2\] By the Law of Cosines, $a^2+b^2-2ab\cos B=c^2+d^2-2cd\cos D$. $\cos B=-\cos D$, so a little rearranging gives \[2\cos B(ab+cd)=a^2+b^2-c^2-d^2\] \[4[ABCD]^2=(ab+cd)^2-\frac{1}{4}(a^2+b^2-c^2-d^2)^2\] \[16[ABCD]^2=4(ab+cd)^2-(a^2+b^2-c^2-d^2)^2\] \[16[ABCD]^2=(2(ab+cd)+(a^2+b^2-c^2-d^2))(2(ab+cd)-(a^2+b^2-c^2-d^2))\] \[16[ABCD]^2=(a^2+2ab+b^2-c^2+2cd-d^2)(-a^2+2ab-b^2+c^2+2cd+d^2)\] \[16[ABCD]^2=((a+b)^2-(c-d)^2)((c+d)^2-(a-b)^2)\] \[16[ABCD]^2=(a+b+c-d)(a+b-c+d)(c+d+a-b)(c+d-b+a)\] \[16[ABCD]^2=16(s-a)(s-b)(s-c)(s-d)\] \[[ABCD]=\sqrt{(s-a)(s-b)(s-c)(s-d)}\]

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 ${d}=0$. This article is a stub. Help us out by expanding it.