Difference between revisions of "Brahmagupta's Formula"

Revision as of 14:57, 10 March 2022

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)

@@ Line 10: / Line 10: @@
-===Proofs===
+==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:
 <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
+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>

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