Pythagorean theorem
a² + b² = c²
The sum of the squares of the two legs equals the square of the hypotenuse. Solve for any one side given the other two: c = √(a²+b²), a = √(c²−b²), b = √(c²−a²).
REFERENCE SHEET
Every formula you need for right-triangle geometry and trigonometry, in one place — with the reasoning behind each one.
Every formula on this page uses the same labeling convention as the rest of this site: vertices A, B, and C
with the right angle at C; leg a (side BC) opposite angle α at vertex A; leg b
(side AC) opposite angle β at vertex B; hypotenuse c (side AB); and altitude h drawn
from C perpendicular to the hypotenuse. If you'd rather plug in numbers than work through formulas by hand, the
right triangle calculator applies every formula below automatically and shows its own
step-by-step substitution for whichever two values you provide.
The formulas are grouped into six families: the Pythagorean theorem itself; area and perimeter; the three SOH-CAH-TOA trigonometric ratios; the pair of complementary acute angles; the altitude-on-hypotenuse relations (including the lesser-known geometric-mean identities); and the inscribed and circumscribed circle radii. Together they cover everything needed to fully solve any right triangle from any two known values.
a² + b² = c²
The sum of the squares of the two legs equals the square of the hypotenuse. Solve for any one side given the other two: c = √(a²+b²), a = √(c²−b²), b = √(c²−a²).
Area = (1/2)·a·b
Because the two legs are perpendicular, either one can serve as the "base" while the other is the "height" — half their product is the area.
Area = (1/2)·c·h
Equivalent to the standard base-times-height formula using the hypotenuse as the base and the altitude drawn to it as the height.
Area = (1/2)·c²·sin α·cos α
Each leg is the hypotenuse times the sine or cosine of α (a = c·sin α, b = c·cos α), so half their product gives the area. Equivalently, Area = (1/4)·c²·sin 2α.
Area = a² / (2·tan α)
With leg a opposite angle α, the other leg is b = a / tan α. Substituting into (1/2)·a·b gives the area from a single leg and one acute angle.
P = a + b + c
Simply the sum of all three side lengths — the two legs and the hypotenuse.
sin(α) = opposite / hypotenuse = a / c
For angle α at vertex A, the "opposite" side is a (side BC) and the hypotenuse is c. Rearranged: a = c·sin(α), c = a / sin(α).
cos(α) = adjacent / hypotenuse = b / c
For angle α, the "adjacent" leg is b (side AC). Rearranged: b = c·cos(α), c = b / cos(α).
tan(α) = opposite / adjacent = a / b
The tangent of an acute angle is the ratio of the opposite leg to the adjacent leg — useful for finding an angle when only the two legs are known: α = atan(a/b).
α + β = 90°
The two non-right angles of any right triangle always sum to 90°, so knowing one immediately gives the other.
h = √(p·q)
The altitude drawn from the right angle to the hypotenuse splits it into two segments p and q. The altitude is the geometric mean of those segments — a direct consequence of the three triangles formed all being similar.
h = (a·b) / c
Derived by equating the two area formulas (1/2)ab = (1/2)ch and solving for h.
a² = p·c, b² = q·c
Each leg is the geometric mean of the whole hypotenuse and the hypotenuse segment adjacent to that leg — the basis of the "geometric mean (leg) theorem".
r = (a + b − c) / 2
The radius of the circle inscribed in a right triangle, tangent to all three sides. A useful identity for a quick sanity check on any solved triangle.
R = c / 2
The hypotenuse of a right triangle is always a diameter of its circumscribed circle (Thales’s theorem), so the circumradius is simply half the hypotenuse.
The altitude-on-hypotenuse identities look like a separate set of rules, but they all fall out of one
observation: dropping the altitude from the right angle to the hypotenuse creates two smaller right triangles,
and both of them are similar to the original triangle (and to each other), because all three share the same
set of angles — just relabeled. Once you know the three triangles are similar, their corresponding side ratios
must match, and matching those ratios directly produces h² = p·q, a² = p·c, and
b² = q·c, where p and q are the two segments the altitude splits the hypotenuse into
(p + q = c). This is sometimes called the "geometric mean theorem" or "right triangle altitude
theorem," and it's a favorite proof exercise in introductory geometry courses precisely because a single
similarity argument unlocks three separate-looking formulas at once.
Take the well-known 3-4-5 right triangle: legs a = 3, b = 4, hypotenuse
c = 5. The Pythagorean theorem confirms 3² + 4² = 9 + 16 = 25 = 5². The area is
(1/2)(3)(4) = 6, and the perimeter is 3 + 4 + 5 = 12. The altitude to the hypotenuse
is h = (3 × 4)/5 = 2.4, which splits the hypotenuse into segments p = a²/c = 9/5 = 1.8
and q = b²/c = 16/5 = 3.2 — and sure enough, p + q = 1.8 + 3.2 = 5 = c, and
h² = p·q checks out as 2.4² = 5.76 = 1.8 × 3.2. Finally, the inradius is
(3 + 4 − 5)/2 = 1 and the circumradius is 5/2 = 2.5. Every formula on this page is
internally consistent — which is exactly why they're used together throughout this calculator's step-by-step
solutions.
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Right triangle FAQ
Answers to common right-triangle questions.