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# Area of a circle inscribed in a rectangle which is inscribed in a semicircle

• Last Updated : 16 Mar, 2021

Given a semicircle with radius R, which inscribes a rectangle of length L and breadth B, which in turn inscribes a circle of radius r. The task is to find the area of the circle with radius r.
Examples:

```Input : R = 2
Output : 1.57

Input : R = 5
Output : 9.8125``` Approach:

We know the biggest rectangle that can be inscribed within the semicircle has, length, l=√2R/2
Also, the biggest circle that can be inscribed within the rectangle has radius, r=b/2=R/2√2(Please refer
So area of the circle, A=π*r^2=π(R/2√2)^2

## C++

 `// C++ Program to find the area of the circle``// inscribed within the rectangle which in turn``// is inscribed in a semicircle``#include ``using` `namespace` `std;` `// Function to find the area of the circle``float` `area(``float` `r)``{` `    ``// radius cannot be negative``    ``if` `(r < 0)``        ``return` `-1;` `    ``// area of the circle``    ``float` `area = 3.14 * ``pow``(r / (2 * ``sqrt``(2)), 2);``    ``return` `area;``}` `// Driver code``int` `main()``{``    ``float` `a = 5;``    ``cout << area(a) << endl;``    ``return` `0;``}`

## Java

 `// Java Program to find the area of the circle``// inscribed within the rectangle which in turn``// is inscribed in a semicircle` `import` `java.io.*;` `class` `GFG {`  `// Function to find the area of the circle``static` `float` `area(``float` `r)``{` `    ``// radius cannot be negative``    ``if` `(r < ``0``)``        ``return` `-``1``;` `    ``// area of the circle``    ``float` `area = (``float``)(``3.14` `* Math.pow(r / (``2` `* Math.sqrt(``2``)), ``2``));``    ``return` `area;``}` `// Driver code` `    ``public` `static` `void` `main (String[] args) {``            ``float` `a = ``5``;``    ``System.out.println( area(a));``    ``}``}` ` ``// This code is contributed by ajit`

## Python3

 `# Python 3 Program to find the``# area of the circle inscribed``# within the rectangle which in``# turn is inscribed in a semicircle``from` `math ``import` `pow``, sqrt` `# Function to find the area``# of the circle``def` `area(r):``    ` `    ``# radius cannot be negative``    ``if` `(r < ``0``):``        ``return` `-``1` `    ``# area of the circle``    ``area ``=` `3.14` `*` `pow``(r ``/` `(``2` `*` `sqrt(``2``)), ``2``);``    ` `    ``return` `area;` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``a ``=` `5``    ``print``(``"{0:.6}"``.``format``(area(a)))` `# This code is contributed By``# Surendra_Gangwar`

## C#

 `// C# Program to find the area of``// the circle inscribed within the``// rectangle which in turn is``// inscribed in a semicircle``using` `System;` `class` `GFG``{` `// Function to find the area``// of the circle``static` `float` `area(``float` `r)``{` `    ``// radius cannot be negative``    ``if` `(r < 0)``        ``return` `-1;` `    ``// area of the circle``    ``float` `area = (``float``)(3.14 * Math.Pow(r /``                        ``(2 * Math.Sqrt(2)), 2));``    ``return` `area;``}` `// Driver code``static` `public` `void` `Main (String []args)``{``    ``float` `a = 5;``    ``Console.WriteLine(area(a));``}``}` `// This code is contributed``// by Arnab Kundu`

## PHP

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## Javascript

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Output:
`9.8125`

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