CodeCogs

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

Interface

#include <codecogs/maths/geometry/area/right_triangle.h>

using namespace Maths::Geometry::Area;

inline double right_triangle(double a, double b, double c, double h)

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 $[BB_1C_1C]$ .

Solution

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

$BC = a \qquad AC = b \qquad AB = c$

(1)

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:

$\frac{B_1C_1}{BC} = \frac{b-h}{b} \qquad \Rightarrow \qquad B_1C_1 = a\frac{b-h}{B}$

(2)

Thus:

$\mathcal{A}_{[BB_1C_1C]} = (BC + B_1C_1)\frac{h}{2} = \frac{ah}{2}\left( 2 - \frac{h}{b} \right)$

(3)

To conclude, the solution of the problem is:

$\mathcal{A}_{[BB_1C_1C]} = ah - \frac{ah^2}{2b}$

(4)

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;
}

Output

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

Parameters

first side of the triangle (BC)

second side of the triangle (AC)

third side of the triangle (AB)

the distance between line $d$ and $\mathrm{Ox}$

Returns

The value of the desired area.

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

$a^2 + b^2 = c^2.$

(5)

Interactive Calculator

Result –