Cyclic Quadrilateral Area
Enter sides a,b,c,d (cyclic). We compute s and √[(s−a)(s−b)(s−c)(s−d)].
—Ancient wisdom, modern computation. Trigonometry, Kuttaka, cyclic quadrilaterals, and composition identities—reimagined.
Enter sides a,b,c,d (cyclic). We compute s and √[(s−a)(s−b)(s−c)(s−d)].
—Extended Euclid to find one integer solution; parametric family shown.
—Angle → sine/versine; equal-arc step intuition for tables.
—Compose (x² + n y²) and (u² + n v²) → (xu − n yv)² + n(xv + yu)²
—Vedic thought treated number, time, and cosmos as a continuum. Temple architecture, calendar-making, and astronomy all relied on mathematics—hence we present Aryabhata & Brahmagupta together, connecting śāstra with calculational rigor.
Area = √[(s−a)(s−b)(s−c)(s−d)], where s=(a+b+c+d)/2.
(x² + n y²)(u² + n v²) = (xu − n yv)² + n(xv + yu)²
For x² − N y² = 1, composition yields new solutions from known ones.
Linear indeterminate ax + by = c solved via repeated division (extended Euclid style).
c/g and parametrize all integer solutions.π ≈ 3.1416 (close value via verse).
Let a=4, b=5, c=7, d=8 → s=(4+5+7+8)/2=12 → Area=√(8·7·5·4)=√1120≈33.47
Solve 26x+9y=1. gcd=1; extended Euclid → 26(−1)+9(3)=1 ⇒ x=−1, y=3; general: x=−1+9t, y=3−26t.