Mathematics in Ancient and Medieval India c6.s9.32
Mathematics in Ancient and Medieval India c6.s9.32
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c6.s9.32
AC² = AD² + CD² = (AB-BD)² + CD² = AB² + (BD² + CD²)
- 2AB.BD = AB² + BC² - 2 AB.BD or b² = c² + a² - 2.c.BD ∴ BD = (c² + a² - b²)/2c = ½ [ c + (a² - b²)/c ] = x, say and AD = c - x = ½ [ c - (a² - b²)/c ] According to Bhāskara I CD = √(a² - x²) or √(b² - (c - x)²) Now CD = √(a² - x²) = √(a² - ((c² + a² - b²)/2c)²) = √(((2ac)² - (c² + a² - b²))/(4c²))