Difference between revisions of "Brahmagupta's Formula"

Latest revision as of 14:13, 26 July 2024

Brahmagupta's Formula is a formula for determining the area of a cyclic quadrilateral given only the four side lengths, given as follows: $[ABCD] = \sqrt{(s-a)(s-b)(s-c)(s-d)}$ where $a$ , $b$ , $c$ , $d$ are the four side lengths and $s = \frac{a+b+c+d}{2}$ .

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 amnado

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)

Quadrilateral $ABCD$ with side lengths $AB=7, BC=24, CD=20, DA=15$ is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form $\frac{a\pi-b}{c},$ where $a,b,$ and $c$ are positive integers such that $a$ and $c$ have no common prime factor. What is $a+b+c?$ (Source)

@@ Line 30: / Line 30: @@
 === Intermediate ===
 *<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]])
+*Quadrilateral <math>ABCD</math> with side lengths <math>AB=7, BC=24, CD=20, DA=15</math> is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form <math>\frac{a\pi-b}{c},</math> where <math>a,b,</math> and <math>c</math> are positive integers such that <math>a</math> and <math>c</math> have no common prime factor. What is <math>a+b+c?</math> ([[2022 AMC 10A Problems/Problem 15|Source]])
 [[Category:Geometry]]
 [[Category:Theorems]]

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