how to find the base of a triangle
Program to find area of a triangle
Finding area using given sides:
Examples :
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Input : a = 5, b = 7, c = 8 Output : Area of a triangle is 17.320508 Input : a = 3, b = 4, c = 5 Output : Area of a triangle is 6.000000
Area of a triangle can simply be evaluated using following formula.
Area = sqrt(s*(s-a)*(s-b)*(s-c)) where a, b and c are lengths of sides of triangle and s = (a+b+c)/2
C++
#include <bits/stdc++.h>
using namespace std;
float findArea( float a, float b, float c)
{
if (a < 0 || b < 0 || c < 0 ||
(a + b <= c) || a + c <= b ||
b + c <= a)
{
cout << "Not a valid triangle" ;
exit (0);
}
float s = (a + b + c) / 2;
return sqrt (s * (s - a) *
(s - b) * (s - c));
}
int main()
{
float a = 3.0;
float b = 4.0;
float c = 5.0;
cout << "Area is " << findArea(a, b, c);
return 0;
}
C
#include <stdio.h>
#include <stdlib.h>
float findArea( float a, float b, float c)
{
if (a < 0 || b < 0 || c <0 || (a+b <= c) ||
a+c <=b || b+c <=a)
{
printf ( "Not a valid triangle" );
exit (0);
}
float s = (a+b+c)/2;
return sqrt (s*(s-a)*(s-b)*(s-c));
}
int main()
{
float a = 3.0;
float b = 4.0;
float c = 5.0;
printf ( "Area is %f" , findArea(a, b, c));
return 0;
}
Java
class Test
{
static float findArea( float a, float b, float c)
{
if (a < 0 || b < 0 || c < 0 || (a+b <= c) ||
a+c <=b || b+c <=a)
{
System.out.println( "Not a valid triangle" );
System.exit( 0 );
}
float s = (a+b+c)/ 2 ;
return ( float )Math.sqrt(s*(s-a)*(s-b)*(s-c));
}
public static void main(String[] args)
{
float a = 3 .0f;
float b = 4 .0f;
float c = 5 .0f;
System.out.println( "Area is " + findArea(a, b, c));
}
}
Python
def findArea(a,b,c):
if (a < 0 or b < 0 or c < 0 or (a + b < = c) or (a + c < = b) or (b + c < = a) ):
print ( 'Not a valid triangle' )
return
s = (a + b + c) / 2
area = (s * (s - a) * (s - b) * (s - c)) * * 0.5
print ( 'Area of a triangle is %f' % area)
a = 3.0
b = 4.0
c = 5.0
findArea(a,b,c)
C#
using System;
class Test {
static float findArea( float a, float b,
float c)
{
if (a < 0 || b < 0 || c <0 ||
(a + b <= c) || a + c <=b ||
b + c <=a)
{
Console.Write( "Not a valid triangle" );
System.Environment.Exit(0);
}
float s = (a + b + c) / 2;
return ( float )Math.Sqrt(s * (s - a) *
(s - b) * (s - c));
}
public static void Main()
{
float a = 3.0f;
float b = 4.0f;
float c = 5.0f;
Console.Write( "Area is " + findArea(a, b, c));
}
}
PHP
<?php
function findArea( $a , $b , $c )
{
if ( $a < 0 or $b < 0 or
$c < 0 or ( $a + $b <= $c ) or
$a + $c <= $b or $b + $c <= $a )
{
echo "Not a valid triangle" ;
exit (0);
}
$s = ( $a + $b + $c ) / 2;
return sqrt( $s * ( $s - $a ) *
( $s - $b ) * ( $s - $c ));
}
$a = 3.0;
$b = 4.0;
$c = 5.0;
echo "Area is " , findArea( $a , $b , $c );
?>
Javascript
<script>
function findArea( a, b, c)
{
if (a < 0 || b < 0 || c < 0 ||
(a + b <= c) || a + c <= b ||
b + c <= a)
{
document.write( "Not a valid triangle" );
return ;
}
let s = (a + b + c) / 2;
return Math.sqrt(s * (s - a) *
(s - b) * (s - c));
}
let a = 3.0;
let b = 4.0;
let c = 5.0;
document.write( "Area is " + findArea(a, b, c));
</script>
Output :
Area is 6
Time Complexity: O(log2n)
Auxiliary Space: O(1)
Finding area using coordinates:
If we are given coordinates of three corners, we can apply below Shoelace formula for area.
Area= | 1/2 [ (x1y2 + x2y3 + ... + xn-1yn + xny1) - (x2y1 + x3y2 + ... + xnyn-1 + x1yn) ] |
C++
#include <bits/stdc++.h>
using namespace std;
double polygonArea( double X[], double Y[], int n)
{
double area = 0.0;
int j = n - 1;
for ( int i = 0; i < n; i++)
{
area += (X[j] + X[i]) * (Y[j] - Y[i]);
j = i;
}
return abs (area / 2.0);
}
int main()
{
double X[] = {0, 2, 4};
double Y[] = {1, 3, 7};
int n = sizeof (X)/ sizeof (X[0]);
cout << polygonArea(X, Y, n);
}
Java
import java.io.*;
import java.math.*;
class GFG {
static double polygonArea( double X[], double Y[], int n)
{
double area = 0.0 ;
int j = n - 1 ;
for ( int i = 0 ; i < n; i++)
{
area += (X[j] + X[i]) * (Y[j] - Y[i]);
j = i;
}
return Math.abs(area / 2.0 );
}
public static void main (String[] args)
{
double X[] = { 0 , 2 , 4 };
double Y[] = { 1 , 3 , 7 };
int n = X.length;
System.out.println(polygonArea(X, Y, n));
}
}
Python3
def polygonArea(X,Y, n) :
area = 0.0
j = n - 1
for i in range ( 0 , n) :
area = area + (X[j] + X[i]) * (Y[j] - Y[i])
j = i
return abs (area / / 2.0 )
X = [ 0 , 2 , 4 ]
Y = [ 1 , 3 , 7 ]
n = len (X)
print (polygonArea(X, Y, n))
C#
using System;
class GFG {
static double polygonArea( double []X,
double []Y, int n)
{
double area = 0.0;
int j = n - 1;
for ( int i = 0; i < n; i++)
{
area += (X[j] + X[i]) *
(Y[j] - Y[i]);
j = i;
}
return Math.Abs(area / 2.0);
}
public static void Main ()
{
double []X = {0, 2, 4};
double []Y = {1, 3, 7};
int n = X.Length;
Console.WriteLine(
polygonArea(X, Y, n));
}
}
PHP
<?php
function polygonArea( $X , $Y , $n )
{
$area = 0.0;
$j = $n - 1;
for ( $i = 0; $i < $n ; $i ++)
{
$area += ( $X [ $j ] + $X [ $i ]) *
( $Y [ $j ] - $Y [ $i ]);
$j = $i ;
}
return abs ( $area / 2.0);
}
$X = array (0, 2, 4);
$Y = array (1, 3, 7);
$n = count ( $X );
echo polygonArea( $X , $Y , $n );
?>
Javascript
<script>
function polygonArea(X, Y, n)
{
let area = 0.0;
let j = n - 1;
for (let i = 0; i < n; i++)
{
area += (X[j] + X[i]) * (Y[j] - Y[i]);
j = i;
}
return Math.abs(area / 2.0);
}
let X = [0, 2, 4];
let Y = [1, 3, 7];
let n = X.length;
document.write(polygonArea(X, Y, n));
</script>
Output:
2
Time Complexity: O(n)
Auxiliary Space: O(1)
https://www.youtube.com/watch?v=-fuEL8MEtOc
how to find the base of a triangle
Source: https://www.geeksforgeeks.org/c-program-find-area-triangle/
Posted by: hemphilldaint1979.blogspot.com
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