Difference between revisions of "Brahmagupta's Formula"

Revision as of 18:14, 12 June 2023

Brahmagupta's Formula is a formula for determining the area of a cyclic quadrilateral given only the four side lengths.

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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, and an easy way to remember it is because it is useful

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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 == 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.
+[[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, and an easy way to remember it is because it is useful
 Brahmagupta's formula reduces to [[Heron's formula]] by setting the side length <math>{d}=0</math>.

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