从给定的斜边和面积中找出直角三角形的所有边|设置 1
原文:https://www . geesforgeks . org/find-sides-直角三角形-给定-斜边-面积/
给定直角三角形的斜边和面积，得到它的底边和高度，如果任何给定斜边和面积的三角形都不可能，打印就不可能。 例:
Input : hypotenuse = 5, area = 6
Output : base = 3, height = 4
Input : hypotenuse = 5, area = 7
Output : No triangle possible with above specification.
我们可以用直角三角形的一个性质来解决这个问题，可以表述如下:
A right angle triangle with fixed hypotenuse attains
maximum area, when it is isosceles i.e. both height
and base becomes equal so if hypotenuse if H, then
by pythagorean theorem,
Base2 + Height2 = H2
For maximum area both base and height should be equal,
b2 + b2 = H2
b = sqrt(H2/2)
Above is the length of base at which triangle attains
maximum area, given area must be less than this maximum
area, otherwise no such triangle will possible.
如果给定的面积小于这个最大面积，我们可以对基底长度做一个二分搜索法，因为增加基底会增加面积，这是一个单调增加的函数，二分搜索法很容易应用。 在下面的代码中，编写了一个获取直角三角形面积的方法，回想一下，对于直角三角形面积为底高，可以使用勾股定理从底边和斜边计算出高度。 以下是上述办法的实施:
C++
// C++ program to get right angle triangle, given
// hypotenuse and area of triangle
#include <bits/stdc++.h>
using namespace std;
// limit for float comparison
#define eps 1e-6
// Utility method to get area of right angle triangle,
// given base and hypotenuse
double getArea(double base, double hypotenuse)
{
double height = sqrt(hypotenuse*hypotenuse - base*base);
return 0.5 * base * height;
}
// Prints base and height of triangle using hypotenuse
// and area information
void printRightAngleTriangle(int hypotenuse, int area)
{
int hsquare = hypotenuse * hypotenuse;
// maximum area will be obtained when base and height
// are equal (= sqrt(h*h/2))
double sideForMaxArea = sqrt(hsquare / 2.0);
double maxArea = getArea(sideForMaxArea, hypotenuse);
// if given area itself is larger than maxArea then no
// solution is possible
if (area > maxArea)
{
cout << "Not possiblen";
return;
}
double low = 0.0;
double high = sideForMaxArea;
double base;
// binary search for base
while (abs(high - low) > eps)
{
base = (low + high) / 2.0;
if (getArea(base, hypotenuse) >= area)
high = base;
else
low = base;
}
// get height by pythagorean rule
double height = sqrt(hsquare - base*base);
cout << base << " " << height << endl;
}
// Driver code to test above methods
int main()
{
int hypotenuse = 5;
int area = 6;
printRightAngleTriangle(hypotenuse, area);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to get right angle triangle, given
// hypotenuse and area of triangle
public class GFG {
// limit for float comparison
final static double eps = (double) 1e-6;
// Utility method to get area of right angle triangle,
// given base and hypotenuse
static double getArea(double base, double hypotenuse) {
double height = Math.sqrt(hypotenuse * hypotenuse - base * base);
return 0.5 * base * height;
}
// Prints base and height of triangle using hypotenuse
// and area information
static void printRightAngleTriangle(int hypotenuse, int area) {
int hsquare = hypotenuse * hypotenuse;
// maximum area will be obtained when base and height
// are equal (= sqrt(h*h/2))
double sideForMaxArea = Math.sqrt(hsquare / 2.0);
double maxArea = getArea(sideForMaxArea, hypotenuse);
// if given area itself is larger than maxArea then no
// solution is possible
if (area > maxArea) {
System.out.print("Not possible");
return;
}
double low = 0.0;
double high = sideForMaxArea;
double base = 0;
// binary search for base
while (Math.abs(high - low) > eps) {
base = (low + high) / 2.0;
if (getArea(base, hypotenuse) >= area) {
high = base;
} else {
low = base;
}
}
// get height by pythagorean rule
double height = Math.sqrt(hsquare - base * base);
System.out.println(Math.round(base) + " " + Math.round(height));
}
// Driver code to test above methods
static public void main(String[] args) {
int hypotenuse = 5;
int area = 6;
printRightAngleTriangle(hypotenuse, area);
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python 3 program to get right angle triangle, given
# hypotenuse and area of triangle
# limit for float comparison
# define eps 1e-6
import math
# Utility method to get area of right angle triangle,
# given base and hypotenuse
def getArea(base, hypotenuse):
height = math.sqrt(hypotenuse*hypotenuse - base*base);
return 0.5 * base * height
# Prints base and height of triangle using hypotenuse
# and area information
def printRightAngleTriangle(hypotenuse, area):
hsquare = hypotenuse * hypotenuse
# maximum area will be obtained when base and height
# are equal (= sqrt(h*h/2))
sideForMaxArea = math.sqrt(hsquare / 2.0)
maxArea = getArea(sideForMaxArea, hypotenuse)
# if given area itself is larger than maxArea then no
# solution is possible
if (area > maxArea):
print("Not possiblen")
return
low = 0.0
high = sideForMaxArea
# binary search for base
while (abs(high - low) > 1e-6):
base = (low + high) / 2.0
if (getArea(base, hypotenuse) >= area):
high =base
else:
low = base
# get height by pythagorean rule
height = math.ceil(math.sqrt(hsquare - base*base))
base = math.floor(base)
print(base,height)
# Driver code to test above methods
if __name__ == '__main__':
hypotenuse = 5
area = 6
printRightAngleTriangle(hypotenuse, area)
# This code is contributed by
# Surendra_Gangwar
C
// C# program to get right angle triangle, given
// hypotenuse and area of triangle
using System;
public class GFG{
// limit for float comparison
static double eps = (double) 1e-6;
// Utility method to get area of right angle triangle,
// given base and hypotenuse
static double getArea(double base1, double hypotenuse) {
double height = Math.Sqrt(hypotenuse * hypotenuse - base1 * base1);
return 0.5 * base1 * height;
}
// Prints base and height of triangle using hypotenuse
// and area information
static void printRightAngleTriangle(int hypotenuse, int area) {
int hsquare = hypotenuse * hypotenuse;
// maximum area will be obtained when base and height
// are equal (= sqrt(h*h/2))
double sideForMaxArea = Math.Sqrt(hsquare / 2.0);
double maxArea = getArea(sideForMaxArea, hypotenuse);
// if given area itself is larger than maxArea then no
// solution is possible
if (area > maxArea) {
Console.Write("Not possible");
return;
}
double low = 0.0;
double high = sideForMaxArea;
double base1 = 0;
// binary search for base
while (Math.Abs(high - low) > eps) {
base1 = (low + high) / 2.0;
if (getArea(base1, hypotenuse) >= area) {
high = base1;
} else {
low = base1;
}
}
// get height by pythagorean rule
double height = Math.Sqrt(hsquare - base1 * base1);
Console.WriteLine(Math.Round(base1) + " " + Math.Round(height));
}
// Driver code to test above methods
static public void Main() {
int hypotenuse = 5;
int area = 6;
printRightAngleTriangle(hypotenuse, area);
}
}
// This code is contributed by 29AjayKumar
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to get right angle triangle,
// given hypotenuse and area of triangle
// limit for float comparison
$eps =.0000001;
// Utility method to get area of right
// angle triangle, given base and hypotenuse
function getArea($base, $hypotenuse)
{
$height = sqrt($hypotenuse * $hypotenuse -
$base * $base);
return 0.5 * $base * $height;
}
// Prints base and height of triangle
// using hypotenuse and area information
function printRightAngleTriangle($hypotenuse,
$area)
{
global $eps;
$hsquare = $hypotenuse * $hypotenuse;
// maximum area will be obtained when base
// and height are equal (= sqrt(h*h/2))
$sideForMaxArea = sqrt($hsquare / 2.0);
$maxArea = getArea($sideForMaxArea,
$hypotenuse);
// if given area itself is larger than
// maxArea then no solution is possible
if ($area > $maxArea)
{
echo "Not possiblen";
return;
}
$low = 0.0;
$high = $sideForMaxArea;
$base;
// binary search for base
while (abs($high - $low) > $eps)
{
$base = ($low + $high) / 2.0;
if (getArea($base, $hypotenuse) >= $area)
$high = $base;
else
$low = $base;
}
// get height by pythagorean rule
$height = sqrt($hsquare - $base * $base);
echo (ceil($base)) ," ",
(floor($height)), "\n";
}
// Driver Code
$hypotenuse = 5;
$area = 6;
printRightAngleTriangle($hypotenuse, $area);
// This code is contributed by Sachin
?>
java 描述语言
<script>
// JavaScript program to get right angle triangle, given
// hypotenuse and area of triangle
// limit for float comparison
let eps = 1e-6;
// Utility method to get area of right angle triangle,
// given base and hypotenuse
function getArea(base, hypotenuse) {
let height = Math.sqrt(hypotenuse * hypotenuse - base * base);
return 0.5 * base * height;
}
// Prints base and height of triangle using hypotenuse
// and area information
function printRightAngleTriangle(hypotenuse, area) {
let hsquare = hypotenuse * hypotenuse;
// maximum area will be obtained when base and height
// are equal (= sqrt(h*h/2))
let sideForMaxArea = Math.sqrt(hsquare / 2.0);
let maxArea = getArea(sideForMaxArea, hypotenuse);
// if given area itself is larger than maxArea then no
// solution is possible
if (area > maxArea) {
document.write("Not possible");
return;
}
let low = 0.0;
let high = sideForMaxArea;
let base = 0;
// binary search for base
while (Math.abs(high - low) > eps) {
base = (low + high) / 2.0;
if (getArea(base, hypotenuse) >= area) {
high = base;
} else {
low = base;
}
}
// get height by pythagorean rule
let height = Math.sqrt(hsquare - base * base);
document.write(Math.round(base) + " " + Math.round(height));
}
// Driver Code
let hypotenuse = 5;
let area = 6;
printRightAngleTriangle(hypotenuse, area);
// This code is contributed by chinmoy1997pal.
</script>
输出:
3 4
还有一个解决方案将在下面的帖子中讨论。 检查给定区域和斜边是否可能成直角 本文由 乌塔什·特里维迪 供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以用contribute.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 contribute@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果发现有不正确的地方，或者想分享更多关于上述话题的信息，请写评论。
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