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Area of largest triangle that can be inscribed within a rectangle

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Given a rectangle of length L     and breadth B     . The task is to find the area of the biggest triangle that can be inscribed in it.
Examples
 

Input: L = 5, B = 4
Output: 10

Input: L = 3, B = 2
Output: 3


 


 


From the figure, it is clear that the largest triangle that can be inscribed in the rectangle, should stand on the same base & has height raising between the same parallel sides of the rectangle.
So, the base of the triangle = B 
Height of the triangle = L
Therefore Area, 
 

A = (L*B)/2


Note: It should also be clear that if base of the triangle = diagonal of rectangle, still the area of triangle so obtained = lb/2 as diagonal of a rectangle divides it into 2 triangles of equal area.
Below is the implementation of the above approach:
 

C++

// C++ Program to find the biggest triangle
// which can be inscribed within the rectangle
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the area
// of the triangle
float trianglearea(float l, float b)
{
 
    // a and b cannot be negative
    if (l < 0 || b < 0)
        return -1;
 
    // area of the triangle
    float area = (l * b) / 2;
    return area;
}
 
// Driver code
int main()
{
    float l = 5, b = 4;
    cout << trianglearea(l, b) << endl;
    return 0;
}

                    

Java

// Java Program to find the biggest triangle
// which can be inscribed within the rectangle
import java.util.*;
 
class GFG
{
    // Function to find the area
    // of the triangle
    static float trianglearea(float l, float b)
    {
     
        // a and b cannot be negative
        if (l < 0 || b < 0)
            return -1;
     
        // area of the triangle
        float area = (l * b) / 2;
        return area;
    }
     
    // Driver code
    public static void main(String args[])
    {
        float l = 5, b = 4;
         
        System.out.println(trianglearea(l, b));
    }
}

                    

Python3

# Python3 Program to find the
# biggest triangle which can be
# inscribed within the rectangle
 
# Function to find the area
# of the triangle
def trianglearea(l, b) :
 
    # a and b cannot be negative
    if (l < 0 or b < 0) :
        return -1
 
    # area of the triangle
    area = (l * b) / 2
    return area
 
# Driver code
l = 5
b = 4
print(trianglearea(l, b))
 
# This code is contributed
# by Yatin Gupta

                    

C#

// C# Program to find the biggest
// triangle which can be inscribed
// within the rectangle
using System;
 
class GFG
{
// Function to find the area
// of the triangle
static float trianglearea(float l,
                          float b)
{
 
    // a and b cannot be negative
    if (l < 0 || b < 0)
        return -1;
 
    // area of the triangle
    float area = (l * b) / 2;
    return area;
}
 
// Driver code
public static void Main()
{
    float l = 5, b = 4;
     
    Console.WriteLine(trianglearea(l, b));
}
}
 
// This code is contributed
// by inder_verma

                    

PHP

<?php
// PHP Program to find the biggest
// triangle which can be inscribed
// within the rectangle
 
// Function to find the area
// of the triangle
function trianglearea($l, $b)
{
 
    // a and b cannot be negative
    if ($l < 0 or $b < 0)
        return -1;
 
    // area of the triangle
    $area = ($l * $b) / 2;
    return $area;
}
 
// Driver code
$l = 5; $b = 4;
echo trianglearea($l, $b);
 
// This code is contributed
// by inder_verma
?>

                    

Javascript

<script>
 
// javascript Program to find the biggest triangle
// which can be inscribed within the rectangle
 
// Function to find the area
// of the triangle
function trianglearea( l,  b)
{
 
    // a and b cannot be negative
    if (l < 0 || b < 0)
        return -1;
 
    // area of the triangle
    let area = (l * b) / 2;
    return area;
}
 
// Driver code
 
    let l = 5, b = 4;
    document.write( trianglearea(l, b) );
 
// This code contributed by aashish1995
 
</script>

                    

Output: 
10

 

Time Complexity: O(1) since performing constant operations

Auxiliary Space: O(1), since no extra space has been taken.



Last Updated : 28 Aug, 2022
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