给定直角三角形的 H（斜边）和 A（面积），求直角三角形的尺寸，使斜边的长度为 H，面积为 A。如果不存在这样的三角形，则打印“不可能”。
例子：
Input : H = 10, A = 24
Output : P = 6.00, B = 8.00
Input : H = 13, A = 36
Output : Not Possible方法：
在进入精确解之前，让我们做一些与直角三角形性质相关的数学计算。
假设H = 斜边， P = 垂直， B = 底， A = 直角三角形的面积。
我们有一些等式：
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下面是上述方法的实现：
C++
// CPP program to find dimensions of
// Right angled triangle
#include
using namespace std;
// function to calculate dimension
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) {
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 function
int main()
{
int H = 5;
int A = 6;
findDimen(H, A);
return 0;
} Java
// Java program to find dimensions of
// Right angled triangle
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) {
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 function
from math import sqrt
# function to find the dimensions
def 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 code
H = 5 # assigning value to H
A = 6 # assigning value to A
findDimen(H, A) # calling function
# This code is contributed by "Abhishek Sharma 44"C#
// C# program to find dimensions of
// Right angled triangle
using 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
Javascript
输出：
P = 3.00
B = 4.00