Given a right angled triangle with height l, base b & hypotenuse h.We need to find the area of the largest square that can fit in the right angled triangle.
Examples:
Input: l = 3, b = 4, h = 5 Output: 2.93878 The biggest square that can fit inside is of 1.71428 * 1.71428 dimension Input: l = 5, b = 12, h = 13 Output: 12.4567
Considering the above diagram, we see,tanx = l/b.
Here it is also true that, tanx = a/(b-a).
So, l/b = a/(b-a) which means that, a = (l*b)/(l+b)
Below is the required implementation:
C++
// C++ Program to find the area of the biggest square// which can fit inside the right angled triangle#include <bits/stdc++.h>using namespace std;// Function to find the area of the biggest squarefloat squareArea(float l, float b, float h){ // the height or base or hypotenuse // cannot be negative if (l < 0 || b < 0 || h < 0) return -1; // side of the square float a = (l * b) / (l + b); // squaring to get the area return a * a;}// Driver codeint main(){ float l = 5, b = 12, h = 13; cout << squareArea(l, b, h) << endl; return 0;} |
Java
//Java Program to find the area of the biggest square//which can fit inside the right angled trianglepublic class GFG { //Function to find the area of the biggest square static float squareArea(float l, float b, float h) { // the height or base or hypotenuse // cannot be negative if (l < 0 || b < 0 || h < 0) return -1; // side of the square float a = (l * b) / (l + b); // squaring to get the area return a * a; } //Driver code public static void main(String[] args) { float l = 5, b = 12, h = 13; System.out.println(squareArea(l, b, h)); }} |
Python3
# Python 3 Program to find the # area of the biggest square # which can fit inside the right# angled triangle # Function to find the area of the biggest squaredef squareArea(l, b, h) : # the height or base or hypotenuse # cannot be negative if l < 0 or b < 0 or h < 0 : return -1 # side of the square a = (l * b) / (l + b) # squaring to get the area return a * a# Driver Codeif __name__ == "__main__" : l, b, h = 5, 12, 13 print(round(squareArea(l, b, h),4))# This code is contributed by ANKITRAI1 |
C#
// C# Program to find the area of // the biggest square which can // fit inside the right angled triangleusing System;class GFG {// Function to find the area // of the biggest squarestatic float squareArea(float l, float b, float h){// the height or base or hypotenuse// cannot be negativeif (l < 0 || b < 0 || h < 0) return -1;// side of the squarefloat a = (l * b) / (l + b);// squaring to get the areareturn a * a;}// Driver codepublic static void Main() { float l = 5, b = 12, h = 13; Console.WriteLine(squareArea(l, b, h));}}// This code is contributed // by inder_verma.. |
PHP
<?php// PHP Program to find the area // of the biggest square which// can fit inside the right // angled triangle// Function to find the area// of the biggest squarefunction squareArea($l, $b, $h){ // the height or base or // hypotenuse cannot be // negative if ($l < 0 || $b < 0 || $h < 0) return -1; // side of the square $a = ($l * $b) / ($l + $b); // squaring to get the area return $a * $a;}// Driver code$l = 5;$b = 12;$h = 13;echo round(squareArea($l, $b, $h), 4);// This code is contributed by mits?> |
Javascript
<script>// javascript Program to find the area of the biggest square// which can fit inside the right angled triangle// Function to find the area of the biggest squarefunction squareArea(l , b , h){ // the height or base or hypotenuse // cannot be negative if (l < 0 || b < 0 || h < 0) return -1; // side of the square var a = (l * b) / (l + b); // squaring to get the area return a * a;}// Driver code var l = 5, b = 12, h = 13; document.write(squareArea(l, b, h).toFixed(4));// This code contributed by Princi Singh </script> |
Output:
12.4567
Time complexity: O(1)
space complexity: O(1)
Example in c :
Approach
1. Find the length of the hypotenuse of the triangle using the Pythagorean theorem.
2. Divide the hypotenuse by sqrt(2) to get the side length of the largest square that can fit inside the triangle.
3. Calculate the area of the square using the formula area = side_length * side_length.
C
#include <stdio.h>#include <math.h>int main() { double a, b, c, side_length, area; // assume a and b are the lengths of the two legs of the right-angled triangle a = 3.0; b = 4.0; c = sqrt(a * a + b * b); // hypotenuse length side_length = c / sqrt(2); // side length of largest square area = side_length * side_length; // area of largest square printf("The area of the largest square that can fit inside the right-angled triangle is %f\n", area); return 0;} |
C++
#include <bits/stdc++.h>using namespace std;int main() { double a, b, c, side_length, area; // assume a and b are the lengths of the two legs of the right-angled triangle a = 3.0; b = 4.0; c = sqrt(a * a + b * b); // hypotenuse length side_length = c / sqrt(2); // side length of largest square area = side_length * side_length; // area of largest square cout << "The area of the largest square that can fit inside the right-angled triangle is " << area << std::endl; return 0;} |
Java
import java.lang.Math;public class Main { public static void main(String[] args) { double a, b, c, side_length, area; // assume a and b are the lengths of the two legs of the right-angled triangle a = 3.0; b = 4.0; c = Math.sqrt(a * a + b * b); // hypotenuse length side_length = c / Math.sqrt(2); // side length of largest square area = side_length * side_length; // area of largest square System.out.println("The area of the largest square that can fit inside the right-angled triangle is " + area); }} |
Python3
import matha, b = 3.0, 4.0 # assume a and b are the lengths of the two legs of the right-angled trianglec = math.sqrt(a ** 2 + b ** 2) # hypotenuse lengthside_length = c / math.sqrt(2) # side length of largest squarearea = side_length ** 2 # area of largest squareprint("The area of the largest square that can fit inside the right-angled triangle is", area) |
Javascript
let a = 3.0, b = 4.0; // assume a and b are the lengths of the two legs of the right-angled trianglelet c = Math.sqrt(a ** 2 + b ** 2); // hypotenuse lengthlet side_length = c / Math.sqrt(2); // side length of largest squarelet area = side_length ** 2; // area of largest squareconsole.log("The area of the largest square that can fit inside the right-angled triangle is", area); |
C#
using System;class MainClass { static void Main() { double a, b, c, side_length, area; // assume a and b are the lengths of the two legs of // the right-angled triangle a = 3.0; b = 4.0; c = Math.Sqrt(a * a + b * b); // hypotenuse length side_length = c / Math.Sqrt( 2); // side length of largest square area = side_length * side_length; // area of largest square Console.WriteLine( "The area of the largest square that can fit inside the right-angled triangle is " + area); }} |
Output
The area of the largest square that can fit inside the right-angled triangle is 12.500000
The time complexity of this code is O(1), as all the calculations are done in constant time, regardless of the input values.
The auxiliary space of the code is also O(1), as we are only storing a few variables, which do not depend on the size of the input.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
