矩形的最大面积可以外接给给定的大小为 LxW 的矩形

给定一个尺寸为L和W的矩形。任务是找到一个矩形的最大面积，该矩形可以围绕一个给定的尺寸为L和W 的矩形。

例子：

Input: L = 10, W = 10
Output: 200

Input: L = 18, W = 12
Output: 450

方法：下面是尺寸L和W的给定矩形EFGH 。我们必须找到外接矩形EFGH的矩形ABCD的面积。

在上图中：
如果$\angle ABC = X$然后$\angle CGF = 90 - X$因为 GCF 是直角三角形。
所以，
$\angle HGD = 180 - \angle FGH - \angle CGF$
=> $\angle HGD = 180 - 90 - (90 - X)$
=> $\angle HGD = X$
相似地，
$\angle EHA = X$
$\angle FEB = X$
现在，矩形 ABCD 的面积由下式给出：

Area = AB * AD 
Area = (AE + EB)*(AH + HD) …..(1) 
 

According to the projection rule: 
AE = L*sin(X) 
EB = W*cos(X) 
AH = L*cos(X) 
HD = W*sin(X) 
 

将上述投影的值代入等式(1)，我们有：

$Area = (AE + EB)*(AH + HD)$
$Area = (L*sin(X)+ W*cox(X))*(L*cos(X) + W*sin(X))$
$Area = ((L^{2} + W^{2})*sin(X)*cos(X) + L*W)$
$Area = (\frac{(L^{2} + W^{2})*sin(2X)}{2} + L*W)$
Now to maximize the area, the value of sin(2X) must be maximum i.e., 1. 
Therefore after substituting sin(2X) as 1 we have, 
$Area = (\frac{(L^{2} + W^{2})}{2} + L*W)$
$Area = (\frac{(L+W)^{2}}{2})$
 

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find area of rectangle
// inscribed another rectangle of
// length L and width W
double AreaofRectangle(int L, int W)
{
     
    // Area of rectangle
    double area = (W + L) * (W + L) / 2;
     
    // Return the area
    return area;
}
 
// Driver Code
int main()
{
     
    // Given dimensions
    int L = 18;
    int W = 12;
     
    // Function call
    cout << AreaofRectangle(L, W);
    return 0;
}
 
// This code is contributed by Princi Singh

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
     
// Function to find area of rectangle
// inscribed another rectangle of
// length L and width W
static double AreaofRectangle(int L, int W)
{
     
    // Area of rectangle
    double area = (W + L) * (W + L) / 2;
     
    // Return the area
    return area;
}
     
// Driver Code
public static void main(String args[])
{
     
    // Given dimensions
    int L = 18;
    int W = 12;
     
    // Function call
    System.out.println(AreaofRectangle(L, W));
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program for the above approach
 
# Function to find area of rectangle
# inscribed another rectangle of
# length L and width W
def AreaofRectangle(L, W):
   
  # Area of rectangle
  area =(W + L)*(W + L)/2
 
# Return the area
  return area
 
# Driver Code
if __name__ == "__main__":
 
  # Given Dimensions
  L = 18
  W = 12
 
  # Function Call
  print(AreaofRectangle(L, W))

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to find area of rectangle
// inscribed another rectangle of
// length L and width W
static double AreaofRectangle(int L, int W)
{
     
    // Area of rectangle
    double area = (W + L) * (W + L) / 2;
     
    // Return the area
    return area;
}
     
// Driver Code
public static void Main(String []args)
{
     
    // Given dimensions
    int L = 18;
    int W = 12;
     
    // Function call
    Console.Write(AreaofRectangle(L, W));
}
}
 
// This code is contributed by shivanisinghss2110

Javascript

      // JavaScript program for the above approach
 
      // Function to find area of rectangle
      // inscribed another rectangle of
      // length L and width W
      function AreaofRectangle(L, W) {
        // Area of rectangle
        var area = parseFloat(((W + L) * (W + L)) / 2).toFixed(1);
 
        // Return the area
        return area;
      }
 
      // Driver Code
      // Given dimensions
      var L = 18;
      var W = 12;
 
      // Function call
      document.write(AreaofRectangle(L, W));
    

输出：

450.0

时间复杂度： O(1)
辅助空间： O(1)

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