Area of a triangle in terms of the inscribed circle or incircle radius,

Applications of Trigonometry

Oblique or Scalene Triangle

Area of a triangle, the radius of the circumscribed circle and the radius of the inscribed circle

The radius of the circumscribed circle or circumcircle

Area of a triangle in terms of the inscribed circle or incircle

The radius of the inscribed circle

Oblique or scalene triangle examples

Oblique or Scalene Triangle

Area of a triangle, the radius of the circumscribed circle and the radius of the inscribed circle

Rectangular in the figure below is composed of two pairs of congruent right triangles formed by the given oblique triangle. Therefore, the area of a triangle equals the half of the rectangular area,

In the right triangles in the right diagram,

h_a = b · sing, h_b = c · sina, h_c = a · sinb,

and by plugging into above formulas for the area

the area of a triangle in terms of an angle and the sides adjacent to it.

If, in the above formulas for the area, we substitute each side applying the sine law, that is

obtained is the area of a triangle in terms of a side and all its angles,

The radius of the circumscribed circle or circumcircle

Using known relation, which states that the angle subtended by a chord at the circumference is half the angle subtended at the center, from the right triangle in the below diagram follows,

the radius of the circumscribed circle, or

a = 2R · sina, b = 2R · sinb, c = 2R · sing.

Plugging the sides into A = (1/2) ab sing obtained is

A = 2R² · sina · sinb · sing

the area of a triangle in terms of the radius of circumcircle and angles.

If, in the above formula for the area which includes two adjacent sides and the angle between them, a given angle is substituted as follows

obtained is

the area of a triangle in terms of sides and the radius of the circumcircle.

Area of a triangle in terms of the inscribed circle (or incircle) radius

The oblique triangle ABC in the figure below consists of three triangles, ABO, BCO and ACO with the same altitude r therefore, its area can be written as

where

is the semi-perimeter

then

A = r · s

the area of a triangle in terms of the

semi-perimeter and inradius.

The parts of a triangle denoted as in the diagram relates as follows,

x_A + x_B = c, x_A + x_C = b, x_B + x_C = a then, 2x_A = - (x_B + x_C) + b + c = - a + b + c,

2x_B = a - (x_A + x_C) + c = a - b + c and 2x_C = a + b - (x_A + x_B) + b + c = a + b - c,

again using a + b + c = 2s it follows that x_A = s - a, x_B = s - b and x_C = s - c.

Then, from the right triangles in the diagram,

Equating obtained formulas with the half-angle formulas, as for example

the radius of the inscribed circle.

Plugging given r into the formula for the area of a triangle A = r · s yields

Heron's formula.

Oblique or scalene triangle examples

Example: Determine the area of an isosceles triangle of which, the line segment that joints the midpoint of one of its equal sides by the midpoint of the base equals the half of the radius R of the circumcircle.

Solution: In the similar triangles ABC and CDE, b/2 = R/2 => b = R.

Area of the triangle

Equating obtained formula with the known formula for the area of a triangle in terms of the radius of the

circumcircle

so, the area of the triangle

Example: Given is a triangle with sides, a, b and c, and angles, a, b and g. In the triangle inscribed is a triangle whose vertices lie in foots of the altitudes of the given triangle, as is shown in the figure down.

Determine sides, angles and the area of the inscribed triangle.

Solution: Quadrilateral ODBF is cyclic (since the sum of the opposite angles is 180°) that is around which a circle can be circumscribed.

Thus, angle ODF = angle OBF = 90° - a since they are inscribed angles subtended by the same arc OF.

As well, angle ODE = angle OCE = 90° - a so angle ODF + angle ODE = d = 180° - 2a.

Note that two angles with mutually perpendicular sides are equal.

Also, angle OFD = angle OBD = 90° - g

since they are inscribed angles subtended by the same arc OD.

As well, angle OFE = angle EAO = 90° - g

so angle OFD + angle OFE = j = 180° - 2g.

On a similar way can be proved that e = 180° - 2b.

Triangles, BDF and ABC are similar since, angle DFB = 90° - angle OFD = 90° - (90° - g) = g therefore

Geometry and use of trigonometry contents - B