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this unit 1.25
sub units 0.00

right triangle

Computes the area of a trapezium within a right angled triangle with a fixed edge.
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Right Triangle

doubleh )[inline]
This module computes the area of the trapezium formed between a right angled triangle with a fixed edge on a reference line and a line found at a given distance distance from this reference line.

This situation is described by the following image. The area which we want to compute is that of the filled trapezium [BB_1C_1C].



Let \mathrm{xOy} be an orthogonal coordinate system and let \triangle ABC be a right angled triangle (\angle C = 90^{\circ}) so that BC \subset \mathrm{Ox} and where a, b, c \in \mathbb{R}_+^* are fixed numbers. Also let d \parallel \mathrm{Ox} so that the distance from line d to \mathrm{Ox} is h \in \mathbb{R}_+ and AB \cap d = \{B_1\}, AC \cap d = \{C_1\}.

Obviously \triangle AB_1C_1  \sim \triangle ABC, which implies: Thus: To conclude, the solution of the problem is:

Example 1

#include <codecogs/geometry/area/right_triangle.h>
#include <stdio.h>
int main()
  // the lengths of the sides
  double a = 3.0, b = 4.0, c = 5.0;
  // display the lengths of the sides
  printf("a = %.1lf\nb = %.1lf\nc = %.1lf\n\n", a, b, c);
  // display the area for different values of h
  for (double h = 0.1; h < 1.09; h += 0.1)
    printf("h = %.1lf   Area = %.2lf\n", h, Geometry::Area::right_triangle(a, b, c, h));
  return 0;


a = 3.0
b = 4.0
c = 5.0
h = 0.1   Area = 0.30
h = 0.2   Area = 0.59
h = 0.3   Area = 0.87
h = 0.4   Area = 1.14
h = 0.5   Area = 1.41
h = 0.6   Area = 1.66
h = 0.7   Area = 1.92
h = 0.8   Area = 2.16
h = 0.9   Area = 2.40
h = 1.0   Area = 2.62


The values of the sides must be Pythagorean numbers, i.e. satisfying the equality:


afirst side of the triangle (BC)
bsecond side of the triangle (AC)
cthird side of the triangle (AB)
hthe distance between line d and \mathrm{Ox}


The value of the desired area.


Eduard Bentea (September 2006)
Source Code

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