Dado un rombo de diagonales a y b , que contiene una circunferencia inscrita. La tarea es encontrar el área de ese círculo en términos de a y b.
Ejemplos:
Input: l = 5, b = 6 Output: 11.582 Input: l = 8, b = 10 Output: 30.6341
Enfoque: A partir de la figura, vemos que el radio del círculo inscrito es también una altura h=OH del triángulo rectángulo AOB . Para encontrarlo, usamos ecuaciones para el área del triángulo:
Área AOB = 1/2 * (a/2) * (b/2) = ab/8 = 12 canales
donde c = AB , es decir, una hipotenusa. Asi que,
r = h = ab/4c = ab/4√(a^2/4 + b^2/4) = ab/2√(a^2+b^2)
y por lo tanto el area del circulo es
A = Π * r^2 = Π a^2 b^2 /4(a2 + b2)
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ Program to find the area of the circle // which can be inscribed within the rhombus #include <bits/stdc++.h> using namespace std; // Function to find the area // of the inscribed circle float circlearea(float a, float b) { // the diagonals cannot be negative if (a < 0 || b < 0) return -1; // area of the circle float A = (3.14 * pow(a, 2) * pow(b, 2)) / (4 * (pow(a, 2) + pow(b, 2))); return A; } // Driver code int main() { float a = 8, b = 10; cout << circlearea(a, b) << endl; return 0; }
Java
// Java Program to find the area of the circle // which can be inscribed within the rhombus public class GFG { // Function to find the area // of the inscribed circle public static float circlearea(double a, double b) { // the diagonals cannot be negative if (a < 0 || b < 0) return -1 ; //area of the circle float A = (float) ((3.14 * Math.pow(a, 2) * Math.pow(b, 2)) / (4 * (Math.pow(a, 2) + Math.pow(b, 2)))) ; return A ; } // Driver code public static void main(String[] args) { float a = 8, b = 10 ; System.out.println(circlearea(a, b)); } // This code is contributed by ANKITRAI1 }
Python 3
# Python 3 Program to find the area of the circle # which can be inscribed within the rhombus # Function to find the area # of the inscribed circle def circlearea(a, b): # the diagonals cannot be negative if (a < 0 or b < 0): return -1 # area of the circle A = ((3.14 * pow(a, 2) * pow(b, 2))/ (4 * (pow(a, 2) + pow(b, 2)))) return A # Driver code if __name__ == "__main__": a = 8 b = 10 print( circlearea(a, b)) # This code is contributed by ChitraNayal
C#
// C# Program to find the area of the circle // which can be inscribed within the rhombus using System; public class GFG { // Function to find the area // of the inscribed circle public static float circlearea(double a, double b) { // the diagonals cannot be negative if (a < 0 || b < 0) return -1 ; //area of the circle float A = (float) ((3.14 * Math.Pow(a, 2) * Math.Pow(b, 2)) / (4 * (Math.Pow(a, 2) + Math.Pow(b, 2)))) ; return A ; } // Driver code public static void Main() { float a = 8, b = 10 ; Console.WriteLine(circlearea(a, b)); } // This code is contributed by inder_verma.. }
PHP
<?php // PHP Program to find the area // of the circle which can be // inscribed within the rhombus // Function to find the area // of the inscribed circle function circlearea($a, $b) { // the diagonals cannot be negative if ($a < 0 || $b < 0) return -1; // area of the circle $A = (3.14 * pow($a, 2) * pow($b, 2)) / (4 * (pow($a, 2) + pow($b, 2))); return $A; } // Driver code $a = 8; $b = 10; echo circlearea($a, $b); // This code is contributed by anuj_67 ?>
Javascript
<script> // javascript Program to find the area of the circle // which can be inscribed within the rhombus // Function to find the area // of the inscribed circle function circlearea(a , b) { // the diagonals cannot be negative if (a < 0 || b < 0) return -1 ; //area of the circle var A = ((3.14 * Math.pow(a, 2) * Math.pow(b, 2)) / (4 * (Math.pow(a, 2) + Math.pow(b, 2)))) ; return A ; } // Driver code var a = 8, b = 10 ; document.write(circlearea(a, b).toFixed(4)); // This code is contributed by Amit Katiyar </script>
30.6341
Complejidad de tiempo: O(log a ) + O(log b ), donde log n es el tiempo requerido por la función pow
Espacio auxiliar: O(1), ya que no se requiere espacio adicional
Publicación traducida automáticamente
Artículo escrito por IshwarGupta y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA