Difference between revisions of "Brahmagupta's Formula"

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===Proof===
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===Proofs===
 
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:  
 
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>
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<cmath>4[ABCD]^2=\sin^2 B(ab+cd)^2</cmath>
 
Substituting <math>\sin^2B=1-\cos^2B</math> results in  
 
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>
 
<cmath>4[ABCD]^2=(1-\cos^2B)(ab+cd)^2=(ab+cd)^2-\cos^2B(ab+cd)^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>
 
<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)^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>
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<cmath>16[ABCD]^2=(a+b+c-d)(a+b-c+d)(c+d+a-b)(c+d-a+b)</cmath>
 
<cmath>16[ABCD]^2=16(s-a)(s-b)(s-c)(s-d)</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>
 
<cmath>[ABCD]=\sqrt{(s-a)(s-b)(s-c)(s-d)}</cmath>
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Brahmagupta's formula reduces to [[Heron's formula]] by setting the side length <math>{d}=0</math>.
 
Brahmagupta's formula reduces to [[Heron's formula]] by setting the side length <math>{d}=0</math>.
 +
 +
A similar formula which Brahmagupta derived for the area of a general quadrilateral is
 +
<cmath>[ABCD]^2=(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left({\frac{B+D}{2}}\right)</cmath>
 +
<cmath>[ABCD]=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left({\frac{B+D}{2}}\right)}</cmath>
 +
where <math>s=\frac{a+b+c+d}{2}</math> is the [[semiperimeter]] of the quadrilateral. What happens when the quadrilateral is cyclic?
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 +
== Problems ==
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=== Intermediate ===
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*<math>ABCD</math> is a cyclic quadrilateral that has an inscribed circle. The diagonals of <math>ABCD</math> intersect at <math>P</math>. If <math>AB = 1, CD = 4,</math> and <math>BP : DP = 3 : 8,</math> then the area of the inscribed circle of <math>ABCD</math> can be expressed as <math>\frac{p\pi}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Determine <math>p + q</math>. ([[Mock AIME 3 Pre 2005 Problems/Problem 7|Source]])
  
  
 
[[Category:Geometry]]
 
[[Category:Geometry]]
{{stub}}
 
 
[[Category:Theorems]]
 
[[Category:Theorems]]

Latest revision as of 14:23, 30 March 2018

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.


Proofs

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: \[4[ABCD]^2=\sin^2 B(ab+cd)^2\] 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 \[2\cos B(ab+cd)=a^2+b^2-c^2-d^2\] \[4[ABCD]^2=(ab+cd)^2-\frac{1}{4}(a^2+b^2-c^2-d^2)^2\] \[16[ABCD]^2=4(ab+cd)^2-(a^2+b^2-c^2-d^2)^2\] \[16[ABCD]^2=(2(ab+cd)+(a^2+b^2-c^2-d^2))(2(ab+cd)-(a^2+b^2-c^2-d^2))\] \[16[ABCD]^2=(a^2+2ab+b^2-c^2+2cd-d^2)(-a^2+2ab-b^2+c^2+2cd+d^2)\] \[16[ABCD]^2=((a+b)^2-(c-d)^2)((c+d)^2-(a-b)^2)\] \[16[ABCD]^2=(a+b+c-d)(a+b-c+d)(c+d+a-b)(c+d-a+b)\] \[16[ABCD]^2=16(s-a)(s-b)(s-c)(s-d)\] \[[ABCD]=\sqrt{(s-a)(s-b)(s-c)(s-d)}\]

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 ${d}=0$.

A similar formula which Brahmagupta derived for the area of a general quadrilateral is \[[ABCD]^2=(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left({\frac{B+D}{2}}\right)\] \[[ABCD]=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left({\frac{B+D}{2}}\right)}\] where $s=\frac{a+b+c+d}{2}$ is the semiperimeter of the quadrilateral. What happens when the quadrilateral is cyclic?

Problems

Intermediate

  • $ABCD$ is a cyclic quadrilateral that has an inscribed circle. The diagonals of $ABCD$ intersect at $P$. If $AB = 1, CD = 4,$ and $BP : DP = 3 : 8,$ then the area of the inscribed circle of $ABCD$ can be expressed as $\frac{p\pi}{q}$, where $p$ and $q$ are relatively prime positive integers. Determine $p + q$. (Source)
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