Given a rectangle of dimensions L and W. The task is to find the maximum area of a rectangle that can be circumscribed about a given rectangle with dimensions L and W.
Examples:
Input: L = 10, W = 10
Output: 200Input: L = 18, W = 12
Output: 450
Approach: Let below is the given rectangle EFGH of dimensions L and W. We have to find the area of rectangle ABCD which is circumscribing rectangle EFGH.
In the above figure:
If 
then 
as GCF is right angled triangle.
Therefore, 
=> 
=> 
Similarly, 
Now, The area of rectangle ABCD is given by:
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)
Substituting the value of the above projections in equation (1) we have:
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,
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> 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 Wdef AreaofRectangle(L, W): # Area of rectangle area =(W + L)*(W + L)/2# Return the area return area# Driver Codeif __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
<script> // 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)); </script> |
450.0
Time Complexity: O(1)
Auxiliary Space: O(1)
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