问题
I can get my code to compile, but it doesn't produce the area that is desired. I'm not sure where I have stumbled.
They want you to have the user enter 6 coordinates (x and y value) for the 3 points of a triangle and get the area. My code is as follows:
import java.util.Scanner;
public class AreaTriangle {
// find the area of a triangle
public static void main (String [] args) {
double side1 = 0;
double side2 = 0;
double side3 = 0;
Scanner input = new Scanner(System.in);
//obtain three points for a triangle
System.out.print("Enter three points for a triangle (x and y intercept): ");
double side1x = input.nextDouble();
double side1y = input.nextDouble();
double side2x = input.nextDouble();
double side2y = input.nextDouble();
double side3x = input.nextDouble();
double side3y = input.nextDouble();
//find length of sides of triangle
side1 = Math.pow(Math.pow((side2x - side1x), 2) + Math.pow((side2y - side1y), 2) * .05, side1);
side2 = Math.pow(Math.pow((side3x - side2x), 2) + Math.pow((side3y - side2y), 2) * .05, side2);
side3 = Math.pow(Math.pow((side1x - side3x), 2) + Math.pow((side1y - side3y), 2) * .05, side3);
double s = (side1 + side2 + side3) / 2;
double area = Math.sqrt(s * (s - side1) * (s - side2) * (s-side3)) * 0.5;
System.out.println("area" + area);
}
}
回答1:
You should try implementing this equation. http://www.mathopenref.com/coordtrianglearea.html
回答2:
@Michael's suggestion is a good one. Following your code, I'd use Pythagoras' Theorem like this:
side1 = Math.sqrt(
Math.pow((side2x - side1x), 2)
+ Math.pow((side2y - side1y), 2));
In your code:
side1 = Math.pow(
Math.pow((side2x - side1x), 2)
+ Math.pow((side2y - side1y), 2) * .05
, side1);
side1
is 0
before the calculation, and almost anything to the power 0
is 1. Therefore side1
ends as 1 regardless of the points.
回答3:
Another way I've discovered is that you can use the cross product to find the area of a triangle. This may be slightly easier for you, since you already have the points. You can turn the three points in to two vectors and take the cross product.
edit: Whoops, forgot to add in the area of a triangle would be half of the cross product, since the cross product would give you the area of a parallelogram formed by the two vectors (and a triangle is half that).
来源:https://stackoverflow.com/questions/14573785/calculate-area-of-triangle-given-3-user-defined-points-beginner