# Area of largest semicircle that can be drawn inside a square

Given a square of side a, the task is to find the area of the largest semi-circle that can be drawn inside the square.

Examples:

```Input: a = 3
Output: 4.84865

Input: a = 4
Output: 8.61982
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach
The semicircle of maximal area inscribed in the square has its diameter parallel to a diagonal, and its radius rmax is given as:

• Since the figure is symmetrical in the diagonal BD, angle QPB = 45°.
`OY = r cos 45 = r/ √2`
• Hence
```a = AB
= r + r/√2
= r(1 + 1/√2)
```
• Thus
```r = a / (1 + 1/√2)
= a*√2 / (√2 + 1)
```
• Rationalizing the denominator, we obtain
`r = a*√2*(√2-1)`
• Thus
```r = a*2 - a √2
= a*(2-√2)
```
• Therfore,
```Area of the required semicircle
= pi * r2/2
= 3.14*(a*(2-√2))2 / 2
```

Below is the implementation of the above approach:

## CPP

 `// C++ program to find Area of ` `// semicircle in a square ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find area of semicircle ` `float` `find_Area(``float` `a) ` `{ ` `    ``float` `R = a * (2.0 - ``sqrt``(2)); ` `    ``float` `area = 3.14 * R * R / 2.0; ` `    ``return` `area; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``// side of a square ` `    ``float` `a = 4; ` ` `  `    ``// Call Function to find ` `    ``// the area of semicircle ` `    ``cout << ``" Area of semicircle = "` `         ``<< find_Area(a); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find Area of ` `// semicircle in a square ` `class` `GFG { ` ` `  `    ``// Function to find area of semicircle ` `    ``static` `float` `find_Area(``float` `a) ` `    ``{ ` `        ``float` `R = a * (``float``)(``2.0` `- Math.sqrt(``2``)); ` `        ``float` `area = (``float``)((``3.14` `* R * R) / ``2.0``); ` `        ``return` `area; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args) ` `    ``{ ` `        ``// side of a square ` `        ``float` `a = ``4``; ` `     `  `        ``// Call Function to find ` `        ``// the area of semicircle ` `        ``System.out.println(``" Area of semicircle = "` `+ find_Area(a)); ` `    ``} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 program to find Area of ` `# semicircle in a square ` `from` `math ``import` `sqrt ` ` `  `# Function to find area of semicircle ` `def` `find_Area(a) : ` ` `  `    ``R ``=` `a ``*` `(``2.0` `-` `sqrt(``2``)); ` `    ``area ``=` `3.14` `*` `R ``*` `R ``/` `2.0``; ` `     `  `    ``return` `area; ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"` `: ` ` `  `    ``# side of a square ` `    ``a ``=` `4``; ` ` `  `    ``# Call Function to find ` `    ``# the area of semicircle ` `    ``print``(``"Area of semicircle ="``,find_Area(a)); ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# program to find Area of ` `// semicircle in a square ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// Function to find area of semicircle ` `    ``static` `float` `find_Area(``float` `a) ` `    ``{ ` `        ``float` `R = a * (``float``)(2.0 - Math.Sqrt(2)); ` `        ``float` `area = (``float``)((3.14 * R * R) / 2.0); ` `        ``return` `area; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main (``string``[] args) ` `    ``{ ` `        ``// side of a square ` `        ``float` `a = 4; ` `     `  `        ``// Call Function to find ` `        ``// the area of semicircle ` `        ``Console.WriteLine(``" Area of semicircle = "` `+ find_Area(a)); ` `    ``} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```Area of semicircle = 8.61982
```

Reference: http://www.qbyte.org/puzzles/p153s.html

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