A right triangle is triangle with an angle of 
(
radians). The sides 
, 
,
and 
of such a triangle satisfy the Pythagorean
theorem
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(1)
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where the largest side is conventionally denoted 
and is called the hypotenuse.
The other two sides of lengths 
and 
are called legs, or sometimes catheti.
The favorite A-level math exam question of the protagonist Christopher in the novel The
Curious Incident of the Dog in the Night-Time asks for proof that a triangle
with sides of the form 
,
, and 
where 
is a right triangle, and that the converse does not hold
(Haddon 2003, pp. 214 and 223-226).
The side lengths 
of a right triangle form a so-called Pythagorean
triple. A triangle that is not a right triangle
is sometimes called an oblique triangle. Special
cases of the right triangle include the isosceles
right triangle (middle figure) and 30-60-90
triangle (right figure).
For any three similar shapes of area 
on the sides of a right triangle,
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(2)
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which is equivalent to the Pythagorean theorem.
For a right triangle with sides 
, 
,
and hypotenuse 
, the area is simply
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(3)
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The inradius can be found by equating the area of the triangle 
with the sum of the areas of the three triangles 
, 
, and 
having the inradii as altitudes, giving
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(4)
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Solving for 
then gives
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(5)
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This can also be written in the equivalent forms
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(6)
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(7)
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The hypotenuse of a right triangle is a diameter of the triangle's circumcircle, so the circumradius is given by
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(8)
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A primitive right triangle is a right triangle having integer sides 
, 
,
and 
such that 
,
where 
is the greatest common divisor. The set
of values 
is then known as a primitive Pythagorean
triple.
For a right triangle with integer side lengths, any primitive Pythagorean triple can be written
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(9)
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(10)
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(11)
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Using these, equation (6) becomes
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(12)
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(13)
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which is an integer whenever 
and 
are integers (Ogilvy and Anderson 1988, p. 68).
Given a right triangle 
,
draw the altitude 
from the right angle 
. Then the triangles 
and 
are similar.
In a right triangle, the midpoint of the hypotenuse is equidistant from the three polygon vertices
(Dunham 1990). This can be proved as follows. Given 
, let 
be the midpoint of 
(so that 
). Draw 
, then since 
is similar to 
, it follows that 
. Since both 
and 
are right triangles and the corresponding legs are
equal, the hypotenuses are also equal, so we have
and the theorem is proved.
In addition, the triangle median 
and altitude 
of a triangle 
are reflections about the angle
bisector 
of 
iff 
is a right triangle (G. McRae, pers. comm., May
1, 2006).
Fermat showed how to construct an arbitrary number of equiareal nonprimitive right triangles. An analysis of Pythagorean triples
demonstrates that the right triangle generated by a triple 
has common area
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(Beiler 1966, pp. 126-127). The only extremum of this function occurs at 
.
Since 
for 
,
the smallest area shared by three nonprimitive right
triangles is given by 
,
which results in an area of 840 and corresponds to the triplets (24, 70, 74), (40,
42, 58), and (15, 112, 113) (Beiler 1966, p. 126). One can also find quartets
of right triangles with the same area. The quartet
having the smallest known area is (111, 6160, 6161), (231, 2960, 2969), (518, 1320,
1418), (280, 2442, 2458), with area 
(Beiler 1966, p. 127). Guy (1994) gives additional
information.
It is also possible to find sets of three and four right triangles having the same perimeter (Beiler 1966, pp. 131-132).
In a given right triangle, an infinite sequence of squares that alternately lie on the hypotenuse and longest leg can be constructed,
as illustrated above. These create a sequence of increasingly smaller similar right
triangles. Let the original triangle have legs of lengths 
and 
and hypotenuse of length 
. Also define
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(15)
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 |  | ![1/2sqrt(2[c^2-(a+b)c+ab]).](/images/equations/RightTriangle/Inline88.svg) |
(16)
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Then the sides of the 
square are of length
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(17)
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Number the upper left triangle as 1, and then the remainder by following the "strip" of triangles at adjoining vertices. Then the side lengths of these triangles are
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(18)
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(19)
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(20)
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The inradii of the corresponding triangles can be found from
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(21)
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giving
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A Sangaku problem from 1913 in the Miyagi Prefecture asks for the relationships between the first, third, and fifth inradii (Rothman 1998). This can be solved using elementary trigonometry as well as the explicit equations given above, and has solution
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(23)
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