OF THE AREA OF A TRIANGLE AND OF THE INSCRIBED CIRCLE

Excerpts

OF THE AREA OF A TRIANGLE AND OF THE INSCRIBED CIRCLE
A triangle is equal to the rectangle contained by its semi-perimeter and the radius of the inscribed circle.
OF THE AREA OF A TRIANGLE AND OF THE INSCRIBED CIRCLE
The two tangents from each angle to the inscribed circle are equal: hence, if three tangents, one from each angle, be taken, their sum is the semi-perimeter, and therefore a tangent from one of the angles, together with the side opposite that angle, is equal to the semi-perimeter.
OF THE AREA OF A TRIANGLE AND OF THE INSCRIBED CIRCLE
Let two of the sides of the triangle $ABC$ be produced, and a circle described touching the two produced sides and the third side.
OF THE AREA OF A TRIANGLE AND OF THE INSCRIBED CIRCLE
This word is often spelled “*escribed*” improperly.
OF THE AREA OF A TRIANGLE AND OF THE INSCRIBED CIRCLE
This most useful proposition was known to the Greeks of Alexandria, and by them communicated to the Arabians
OF THE AREA OF A TRIANGLE AND OF THE INSCRIBED CIRCLE
whence the area can be calculated in square units when the lengths of the sides are given numerically in units.
OF THE AREA OF A TRIANGLE AND OF THE INSCRIBED CIRCLE
Then, numerically, the Area = rs.

Equations

OF THE AREA OF A TRIANGLE AND OF THE INSCRIBED CIRCLE
\text{Then, numerically, the Area} = rs.
The area of the triangle equals the radius of the inscribed circle times the semi-perimeter.
OF THE AREA OF A TRIANGLE AND OF THE INSCRIBED CIRCLE
Ab = Ac = s - a
The two tangent lengths from vertex A to the inscribed circle are equal, and each equals the semi-perimeter minus the side a opposite A.
OF THE AREA OF A TRIANGLE AND OF THE INSCRIBED CIRCLE
Bc = Ba = s-b
The two tangent lengths from vertex B to the inscribed circle are equal, and each equals the semi-perimeter minus the side b opposite B.
OF THE AREA OF A TRIANGLE AND OF THE INSCRIBED CIRCLE
Ca = Cb = s - c
The two tangent lengths from vertex C to the inscribed circle are equal, and each equals the semi-perimeter minus the side c opposite C.
OF THE AREA OF A TRIANGLE AND OF THE INSCRIBED CIRCLE
(s - a) \alpha = r s = \text{the area}.
The area of the triangle equals both the excircle radius on side a times s minus a, and the inradius times the semi-perimeter.
OF THE AREA OF A TRIANGLE AND OF THE INSCRIBED CIRCLE
s(s-b) : \Delta :: \Delta : (s-a)(s-c)
The area is a mean proportional between s(s-b) and (s-a)(s-c).
OF THE AREA OF A TRIANGLE AND OF THE INSCRIBED CIRCLE
\Delta^2 = s(s-a)(s-b)(s-c)
The square of the area of a triangle equals the semi-perimeter times the three differences between the semi-perimeter and each side.
OF THE AREA OF A TRIANGLE AND OF THE INSCRIBED CIRCLE
r^2 = \dfrac{(s-a)(s-b)(s-c)}{s}
The square of the inscribed circle's radius equals the product of the three differences between the semi-perimeter and the sides, divided by the semi-perimeter.

Problems

No exercises in this chapter.