Special right triangle

Set square

The side lengths of a 45°–45°–90° triangle

In plane geometry, constructing the diagonal of a square results in a triangle whose three angles are in the ratio 1 : 1 : 2, adding up to 180° or π radians. Hence, the angles respectively measure 45° (π/4), 45° (π/4), and 90° (π/2). The sides in this triangle are in the ratio 1 : 1 : √2, which follows immediately from the Pythagorean theorem.

Of all right triangles, the 45°–45°–90° degree triangle has the smallest ratio of the hypotenuse to the sum of the legs, namely √2/2.[1]^:p.282,p.358 and the greatest ratio of the altitude from the hypotenuse to the sum of the legs, namely √2/4.[1]^:p.282

Triangles with these angles are the only possible right triangles that are also isosceles triangles in Euclidean geometry. However, in spherical geometry and hyperbolic geometry, there are infinitely many different shapes of right isosceles triangles.

Set square

The side lengths of a 30°–60°–90° triangle

This is a triangle whose three angles are in the ratio 1 : 2 : 3 and respectively measure 30° (π/6), 60° (π/3), and 90° (π/2). The sides are in the ratio 1 : √3 : 2.

The proof of this fact is clear using trigonometry. The geometric proof is:

Draw an equilateral triangle ABC with side length 2 and with point D as the midpoint of segment BC. Draw an altitude line from A to D. Then ABD is a 30°–60°–90° triangle with hypotenuse of length 2, and base BD of length 1.

The fact that the remaining leg AD has length √3 follows immediately from the Pythagorean theorem.

The 30°–60°–90° triangle is the only right triangle whose angles are in an arithmetic progression. The proof of this fact is simple and follows on from the fact that if α, α + δ, α + 2δ are the angles in the progression then the sum of the angles 3α + 3δ = 180°. After dividing by 3, the angle α + δ must be 60°. The right angle is 90°, leaving the remaining angle to be 30°.

There are several Pythagorean triples which are well-known, including those with sides in the ratios:

3:	4	:5
5:	12	:13
8:	15	:17
7:	24	:25
9:	40	:41

The 3 : 4 : 5 triangles are the only right triangles with edges in arithmetic progression. Triangles based on Pythagorean triples are Heronian, meaning they have integer area as well as integer sides.

The possible use of the 3 : 4 : 5 triangle in Ancient Egypt, with the supposed use of a knotted rope to lay out such a triangle, and the question whether Pythagoras' theorem was known at that time, have been much debated.[3] It was first conjectured by the historian Moritz Cantor in 1882.[3] It is known that right angles were laid out accurately in Ancient Egypt; that their surveyors did use ropes for measurement;[3] that Plutarch recorded in Isis and Osiris (around 100 AD) that the Egyptians admired the 3 : 4 : 5 triangle;[3] and that the Berlin Papyrus 6619 from the Middle Kingdom of Egypt (before 1700 BC) stated that "the area of a square of 100 is equal to that of two smaller squares. The side of one is ½ + ¼ the side of the other."[4] The historian of mathematics Roger L. Cooke observes that "It is hard to imagine anyone being interested in such conditions without knowing the Pythagorean theorem."[3] Against this, Cooke notes that no Egyptian text before 300 BC actually mentions the use of the theorem to find the length of a triangle's sides, and that there are simpler ways to construct a right angle. Cooke concludes that Cantor's conjecture remains uncertain: he guesses that the Ancient Egyptians probably did know the Pythagorean theorem, but that "there is no evidence that they used it to construct right angles".[3]

The following are all the Pythagorean triple ratios expressed in lowest form (beyond the five smallest ones in lowest form in the list above) with both non-hypotenuse sides less than 256:

11:	60	:61
12:	35	:37
13:	84	:85
15:	112	:113
16:	63	:65
17:	144	:145
19:	180	:181
20:	21	:29
20:	99	:101
21:	220	:221

Isosceles right-angled triangles cannot have sides with integer values, because the ratio of the hypotenuse to either other side is √2, but √2 cannot be expressed as a ratio of two integers. However, infinitely many almost-isosceles right triangles do exist. These are right-angled triangles with integral sides for which the lengths of the non-hypotenuse edges differ by one.[5][6] Such almost-isosceles right-angled triangles can be obtained recursively,

a₀ = 1, b₀ = 2

a_n = 2b_n−1 + a_n−1

b_n = 2a_n + 5b_n−1

a_n is length of hypotenuse, n = 1, 2, 3, .... Equivalently,

${\displaystyle ({\tfrac {x-1}{2}})^{2}+({\tfrac {x+1}{2}})^{2}=y^{2}}$

where {x, y} are the solutions to the Pell equation x² − 2y² = −1, with the hypotenuse y being the odd terms of the Pell numbers 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378... (sequence A000129 in the OEIS).. The smallest Pythagorean triples resulting are:[7]

3 :	4	: 5
20 :	21	: 29
119 :	120	: 169
696 :	697	: 985
4,059 :	4,060	: 5,741
23,660 :	23,661	: 33,461
137,903 :	137,904	: 195,025
803,760 :	803,761	: 1,136,689
4,684,659 :	4,684,660	: 6,625,109

Alternatively, the same triangles can be derived from the square triangular numbers.[8]

A Kepler triangle is a right triangle formed by three squares with areas in geometric progression according to the golden ratio.

The Kepler triangle is a right triangle whose sides are in a geometric progression. If the sides are formed from the geometric progression a, ar, ar² then its common ratio r is given by r = √φ where φ is the golden ratio. Its sides are therefore in the ratio 1 : √φ : φ. Thus, the shape of the Kepler triangle is uniquely determined (up to a scale factor) by the requirement that its sides be in a geometric progression.

The 3–4–5 triangle is the unique right triangle (up to scaling) whose sides are in an arithmetic progression.[9]

The sides of a pentagon, hexagon, and decagon, inscribed in congruent circles, form a right triangle

Let a = 2 sin π/10 = −1 + √5/2 = 1/φ be the side length of a regular decagon inscribed in the unit circle, where φ is the golden ratio. Let b = 2 sin π/6 = 1 be the side length of a regular hexagon in the unit circle, and let c = 2 sin π/5 = ${\displaystyle {\sqrt {\tfrac {5-{\sqrt {5}}}{2}}}}$ be the side length of a regular pentagon in the unit circle. Then a² + b² = c², so these three lengths form the sides of a right triangle.[10] The same triangle forms half of a golden rectangle. It may also be found within a regular icosahedron of side length c: the shortest line segment from any vertex V to the plane of its five neighbors has length a, and the endpoints of this line segment together with any of the neighbors of V form the vertices of a right triangle with sides a, b, and c.[11]