Given H (Hypotenuse) and A (area) of a right angled triangle, find the dimensions of right angled triangle such that the hypotenuse is of length H and its area is A. If no such triangle exists, print “Not Possible”.
Examples:
Input : H = 10, A = 24 Output : P = 6.00, B = 8.00 Input : H = 13, A = 36 Output : Not Possible
Approach:
Before moving to exact solution, let’s do some of mathematical calculations related to properties of Right-angled triangle.
Suppose H = Hypotenuse, P = Perpendicular, B = Base and A = Area of right angled triangle.
We have some sort of equations as :
P^2 + B^2 = H^2 P * B = 2 * A (P+B)^2 = P^2 + B^2 + 2*P*B = H^2 + 4*A (P+B) = sqrt(H^2 + 4*A) ----1 (P-B)^2 = P^2 + B^2 - 2*P*B = H^2 - 4*A mod(P-B) = sqrt(H^2 - 4*A) ----2 from equation (2) we can conclude that if H^2 < 4*A then no solution is possible. Further from (1)+(2) and (1)-(2) we have : P = (sqrt(H^2 + 4*A) + sqrt(H^2 - 4*A) ) / 2 B = (sqrt(H^2 + 4*A) - sqrt(H^2 - 4*A) ) / 2
Below is the implementation of above approach:
C++
// CPP program to find dimensions of// Right angled triangle#include <bits/stdc++.h>using namespace std;// function to calculate dimensionvoid findDimen(int H, int A){ // P^2+B^2 = H^2 // P*B = 2*A // (P+B)^2 = P^2+B^2+2*P*B = H^2+4*A // (P-B)^2 = P^2+B^2-2*P*B = H^2-4*A // P+B = sqrt(H^2+4*A) // |P-B| = sqrt(H^2-4*A) if (H * H < 4 * A) { cout << "Not Possible\n"; return; } // sqrt value of H^2 + 4A and H^2- 4A double apb = sqrt(H * H + 4 * A); double asb = sqrt(H * H - 4 * A); // Set precision cout.precision(2); cout << "P = " << fixed << (apb - asb) / 2.0 << "\n"; cout << "B = " << (apb + asb) / 2.0;}// driver functionint main(){ int H = 5; int A = 6; findDimen(H, A); return 0;} |
Java
// Java program to find dimensions of// Right angled triangleclass GFG { // function to calculate dimension static void findDimen(int H, int A) { // P^2+B^2 = H^2 // P*B = 2*A // (P+B)^2 = P^2+B^2+2*P*B = H^2+4*A // (P-B)^2 = P^2+B^2-2*P*B = H^2-4*A // P+B = sqrt(H^2+4*A) // |P-B| = sqrt(H^2-4*A) if (H * H < 4 * A) { System.out.println("Not Possible"); return; } // sqrt value of H^2 + 4A and H^2- 4A double apb = Math.sqrt(H * H + 4 * A); double asb = Math.sqrt(H * H - 4 * A); System.out.println("P = " + Math.round(((apb - asb) / 2.0) * 100.0) / 100.0); System.out.print("B = " + Math.round(((apb + asb) / 2.0) * 100.0) / 100.0); } // Driver function public static void main(String[] args) { int H = 5; int A = 6; findDimen(H, A); }}// This code is contributed by Anant Agarwal. |
Python3
# Python code to find dimensions# of Right angled triangle# importing the math package # to use sqrt functionfrom math import sqrt# function to find the dimensionsdef findDimen( H, A): # P ^ 2 + B ^ 2 = H ^ 2 # P * B = 2 * A # (P + B)^2 = P ^ 2 + B ^ 2 + 2 * P*B = H ^ 2 + 4 * A # (P-B)^2 = P ^ 2 + B ^ 2-2 * P*B = H ^ 2-4 * A # P + B = sqrt(H ^ 2 + 4 * A) # |P-B| = sqrt(H ^ 2-4 * A) if H * H < 4 * A: print("Not Possible") return # sqrt value of H ^ 2 + 4A and H ^ 2- 4A apb = sqrt(H * H + 4 * A) asb = sqrt(H * H - 4 * A) # printing the dimensions print("P = ", "%.2f" %((apb - asb) / 2.0)) print("B = ", "%.2f" %((apb + asb) / 2.0)) # driver codeH = 5 # assigning value to HA = 6 # assigning value to AfindDimen(H, A) # calling function# This code is contributed by "Abhishek Sharma 44" |
C#
// C# program to find dimensions of// Right angled triangleusing System;class GFG { // function to calculate dimension static void findDimen(int H, int A) { // P^2+B^2 = H^2 // P*B = 2*A // (P+B)^2 = P^2+B^2+2*P*B = H^2+4*A // (P-B)^2 = P^2+B^2-2*P*B = H^2-4*A // P+B = sqrt(H^2+4*A) // |P-B| = sqrt(H^2-4*A) if (H * H < 4 * A) { Console.WriteLine("Not Possible"); return; } // sqrt value of H^2 + 4A and H^2- 4A double apb = Math.Sqrt(H * H + 4 * A); double asb = Math.Sqrt(H * H - 4 * A); Console.WriteLine("P = " + Math.Round( ((apb - asb) / 2.0) * 100.0) / 100.0); Console.WriteLine("B = " + Math.Round( ((apb + asb) / 2.0) * 100.0) / 100.0); } // Driver function public static void Main() { int H = 5; int A = 6; findDimen(H, A); }}// This code is contributed by vt_m. |
PHP
<?php// PHP program to find dimensions // of Right angled triangle// function to calculate dimensionfunction findDimen($H, $A){ // P^2+B^2 = H^2 // P*B = 2*A // (P+B)^2 = P^2+B^2+2*P*B = H^2+4*A // (P-B)^2 = P^2+B^2-2*P*B = H^2-4*A // P+B = sqrt(H^2+4*A) // |P-B| = sqrt(H^2-4*A) if ($H * $H < 4 * $A) { echo "Not Possible\n"; return; } // sqrt value of H^2 + 4A and // H^2- 4A $apb = sqrt($H * $H + 4 * $A); $asb = sqrt($H * $H - 4 * $A); echo "P = " , $fixed , ($apb - $asb) / 2.0 , "\n"; echo "B = " , ($apb + $asb) / 2.0;} // Driver Code $H = 5; $A = 6; findDimen($H, $A); // This code is contributed by vt_m.?> |
Javascript
<script>// java script program to find dimensions// of Right angled triangle// function to calculate dimensionfunction findDimen(H, A){ // P^2+B^2 = H^2 // P*B = 2*A // (P+B)^2 = P^2+B^2+2*P*B = H^2+4*A // (P-B)^2 = P^2+B^2-2*P*B = H^2-4*A // P+B = sqrt(H^2+4*A) // |P-B| = sqrt(H^2-4*A) if (H * H < 4 * A) { document.write( "Not Possible"); return; } // sqrt value of H^2 + 4A and // H^2- 4A let apb = Math.sqrt(H * H + 4 * A); let asb = Math.sqrt(H * H - 4 * A); document.write( "P = " +((apb - asb) / 2.0).toFixed(2), "<br>"); document.write( "B = " +((apb + asb) / 2.0).toFixed(2));} // Driver Code let H = 5; let A = 6; findDimen(H, A);// This code is contributed by Gottumukkala Bobby</script> |
Output:
P = 3.00 B = 4.00
Time complexity : O(log(n)) since using inbuilt sqrt functions
Auxiliary Space : O(1)
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