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Solve any triangle — find the area with Heron’s formula or base-height, the hypotenuse, perimeter and angles from just three sides.
Triangles are fully determined by surprisingly little information — three sides, or a base and height, are enough to unlock the area, perimeter, and every angle. Follow these steps to solve any triangle with the tool above.
A handy cross-check: for any right triangle, the area also equals half the product of the legs — ½ × 3 × 4 = 6, matching Heron exactly. When two independent formulas agree, the answer is solid.
A triangle is the simplest polygon: three vertices (corner points), three sides (straight segments joining them), and three interior angles that always sum to exactly 180°. That rigidity is their superpower — unlike a quadrilateral, which can flex, a triangle’s shape is locked by its side lengths, which is why triangles brace bridges, roofs, and cranes.
Triangles are classified two ways. By sides: equilateral (all three equal, all angles 60°), isosceles (two equal sides, base angles equal), scalene (all different, like 3-4-5). By angles: right (one 90° angle; the side opposite it is the hypotenuse, the other two are legs), acute (all angles under 90°), obtuse (one angle over 90°). The 3-4-5 triangle is both scalene and right — the most famous integer-sided (“Pythagorean triple”) triangle in history.
Three theorems do the heavy lifting. The Pythagorean theorem (a² + b² = c²) links the sides of every right triangle. Heron’s formula finds area from three sides alone via the semiperimeter s = (a+b+c)/2: A = √(s(s−a)(s−b)(s−c)). The law of cosines (c² = a² + b² − 2ab·cos C) generalizes Pythagoras to any triangle and yields each angle from the sides. Underlying everything is the triangle inequality: the sum of any two sides must exceed the third, or the “triangle” collapses flat.
The geometry is ancient. Euclid’s Elements (~300 BCE) systematized triangle proofs; Heron of Alexandria (~60 CE) published his area formula; Pythagoras’ school explored the theorem centuries earlier, though Babylonians knew 3-4-5 a thousand years before that. Surveyors have triangulated land since Egypt’s re-marking of Nile flood boundaries, and GPS still works by solving triangles — trilateration from satellites.
Triangles are everywhere today: construction (trusses, A-frames), navigation (bearings and triangulation), computer graphics (every 3D model is a mesh of triangles), physics (force and velocity vectors add tip-to-tail), and design (sails, ramps, gables). Master the triangle and you master the atom of shape.
Four formulas cover nearly every triangle problem:
Sides a = 3, b = 4, c = 5. First the triangle inequality: 3+4 > 5 ✓, 3+5 > 4 ✓, 4+5 > 3 ✓. Semiperimeter s = (3+4+5)/2 = 6. Area = √(6(6−3)(6−4)(6−5)) = √(6×3×2×1) = √36 = 6. Perimeter = 12. Right-triangle check: 3² + 4² = 9 + 16 = 25 = 5² ✓, so the angle opposite side 5 is 90°. The other angles: sin A = 3/5 → A ≈ 36.87°; sin B = 4/5 → B ≈ 53.13°; check 36.87 + 53.13 + 90 = 180° ✓.
Sides 5, 12, 13. Right-triangle check: 5² + 12² = 25 + 144 = 169 = 13² ✓ — right-angled. Via legs: A = ½ × 5 × 12 = 30. Via Heron: s = (5+12+13)/2 = 15; A = √(15×10×3×2) = √900 = 30. Both routes agree, as they must. Angles: 90°, arcsin(5/13) ≈ 22.62°, arcsin(12/13) ≈ 67.38°; sum 180° ✓.
A sailmaker cuts a triangular sail with sides 7 m, 8 m, and 9 m. How much cloth is needed? Triangle inequality: 7+8 > 9 ✓ (and the other two pairs). Semiperimeter s = (7+8+9)/2 = 12. Heron: A = √(12(12−7)(12−8)(12−9)) = √(12×5×4×3) = √720 ≈ 26.83 m². Add 10% cutting waste: 26.83 × 1.1 ≈ 29.5 m² to order. Perimeter = 24 m tells you the edging (bolt rope) length. One formula, three purchasing numbers — area, waste allowance, and edging — from three tape-measure readings.
A 5-meter ladder stands with its base 3 m from the wall. How high does it reach, and is the angle safe? The wall, ground, and ladder form a right triangle with hypotenuse 5 and one leg 3. Height = √(5² − 3²) = √(25−9) = √16 = 4 m — another 3-4-5 triangle in the wild. Angle with the ground: sin θ = 4/5 → θ ≈ 53.13°; with the wall, 36.87°.
Safety verdict: ladder guidance calls for roughly a 75° ground angle (base about one-quarter of the ladder’s length out — here just 1.25 m, not 3 m). At 53° the ladder is far too shallow and could slip out at the base. The fix the calculator implies: move the base to ~1.25 m, giving height √(25−1.5625) ≈ 4.84 m at ≈ 75.5°. Geometry doesn’t just describe the setup — it judges it.
Triangles connect to the wider geometry shelf. The Area Calculator covers circles, rectangles, and other shapes; the Volume Calculator extends area thinking into 3D; the Scientific Calculator supplies the sine, cosine, and square-root buttons behind these formulas; the Percentage Calculator handles waste allowances and grade computations; and the Slope Calculator is right-triangle geometry in disguise (rise over run). Browse all of them in Math Calculators.
This page provides a triangle solver with three input modes: three sides (Heron’s formula), base and height, or two legs of a right triangle. Methods: semiperimeter s=(a+b+c)/2, area A=√(s(s−a)(s−b)(s−c)); A=½bh with perpendicular height; hypotenuse c=√(a²+b²); perimeter P=a+b+c; angles via law of cosines; validity via triangle inequality (each pair-sum exceeds the third). Verified numeric facts: 3-4-5 gives s=6, A=6, P=12, angles ≈36.87°/53.13°/90°, right triangle confirmed by 9+16=25; 5-12-13 gives A=30 by both legs and Heron (s=15, √900), angles ≈22.62°/67.38°/90°; sail 7-8-9 gives A=√720≈26.83 m², P=24 m; ladder 5 m at 3 m out reaches 4 m at ≈53.13° (below the ~75° safety guideline). Angle sum is always 180°.
Sources: Euclid, Elements, Book I (triangle theorems); Heron of Alexandria, Metrica (Heron’s formula); OpenStax, Geometry/Precalculus (trigonometry). All numerical examples were computed by hand and cross-checked against the calculator.
Three numbers on the results panel deserve a second look before you trust them. First, the validity check: the calculator enforces the triangle inequality, but when you work by hand, always confirm the two shorter sides sum to more than the longest — 10 + 12 > 15 clears it, while 1 + 2 < 5 never forms a triangle at all. Second, the angle sum: the three reported angles must total exactly 180° (up to rounding). If yours add to 179.99°, that is just display rounding; if they add to 170°, something is wrong with the inputs.
Third, match sides to angles: the longest side always faces the largest angle, and equal sides face equal angles. In a 3-4-5 triangle the side of length 5 faces the 90° angle — if your results ever show the longest side facing a small angle, recheck which value went where. Fourth, mind your units: area comes out in square units and perimeter in linear units. Entering sides in meters gives square meters of area; mixing meters and centimeters silently corrupts everything, so convert first.
Finally, pick the right input mode for the job. Know all three sides? Heron’s formula mode is the one. Know a base and its perpendicular height? The base–height mode is simpler and exact. Working with a right triangle? The two-legs mode doubles as a Pythagorean check. The modes must agree where they overlap — a 3-4-5 triangle gives area 6 whether you use Heron or half the product of the legs — and any disagreement means an input error, never a math error.
A gardener’s plot is a triangle with sides 10 m, 12 m, and 15 m. Two questions matter: how much fencing to buy, and how much soil to fill it. The perimeter is immediate: 10 + 12 + 15 = 37 m of fencing (plus a little extra for the gate and posts). The area needs Heron: s = 37/2 = 18.5, so A = √(18.5 × 8.5 × 6.5 × 3.5) = √3577.44 ≈ 59.81 m².
At 5 cm of topsoil, that is about 2.99 m³ of soil to order — and here the units lesson bites: the sides went in as meters, so the area is square meters and the soil volume is cubic meters, with no conversion needed. As a bonus, the angles (≈ 41.6°, 52.9°, 85.5°) confirm an acute triangle with the longest side facing the largest angle, exactly as theory demands.
Heron’s formula. For 3-4-5: s = 6, A = √36 = 6.
Check a² + b² = c². 3² + 4² = 25 = 5², so 3-4-5 is right-angled.
c = √(a² + b²); legs 3 and 4 give 5.
Law of cosines; for 3-4-5: 36.9°, 53.1°, 90°.
Any two sides must sum to more than the third.
A = ½ × base × height = ½ × 12 × 7 = 42 square units. This is the simplest area formula — use it whenever you know the base and the perpendicular height.
c² = a² + b² − 2ab·cos(C). For a = 5, b = 7, and included angle C = 60°: c² = 25 + 49 − 2(5)(7)(0.5) = 74 − 35 = 39, so c = √39 ≈ 6.245.
It is a right triangle (5² + 12² = 25 + 144 = 169 = 13²), so one angle is 90°. The others: arcsin(5/13) ≈ 22.62° and arcsin(12/13) ≈ 67.38°. Check: 90 + 22.62 + 67.38 = 180°.