Triangle Calculator

How to use

1. Enter the three side lengths.
2. Read the area, perimeter and semi-perimeter.
3. Check the three interior angles and the classification (right, acute or obtuse).
4. If the sides cannot form a triangle, read the reason rather than a result.

How it works

Heron's formula: area from three sides alone

Heron's formula solves "three sides known, find the area": compute the semi-perimeter s = (a + b + c) / 2, then area = √(s(s−a)(s−b)(s−c)). For 3-4-5, s = 6 and area = √(6 × 3 × 2 × 1) = √36 = 6, matching base×height÷2; for scalene triangles like 7-8-9 it is practically the only direct method.

Interior angles via the law of cosines

Interior angles come from the law of cosines: cos A = (b² + c² − a²) / (2bc), then take the inverse cosine. 3-4-5 yields 36.87°, 53.13° and 90°, summing to 180°. The law generalizes the Pythagorean theorem — when A is the right angle, a² = b² + c² and the formula degenerates into it automatically.

Can these three sides form a triangle?

Triangle inequality decides validity: the sum of any two sides must exceed the third (1, 1, 5 fails because 1 + 1 < 5). Equivalently, the difference of two sides must be smaller than the third. The tool rejects invalid inputs and classifies the triangle: compare the square of the longest side with the sum of squares of the other two — equal means right, smaller acute, larger obtuse.

Calculation basis: area uses Heron's formula (s = (a+b+c)/2, area = √(s(s−a)(s−b)(s−c))); interior angles come from the law of cosines, cos A = (b²+c²−a²)/(2bc); triangle validity is checked with the triangle inequality; and the type follows from comparing the square of the longest side with the sum of the squares of the other two.

Code example

function triangle(a, b, c) {
  if (a + b <= c || b + c <= a || a + c <= b) throw new Error("not a valid triangle");
  const p = (a + b + c) / 2;                       // semi-perimeter
  const area = Math.sqrt(p * (p - a) * (p - b) * (p - c));
  const ang = (x, y, z) =>
    Math.acos((y * y + z * z - x * x) / (2 * y * z)) * 180 / Math.PI;
  return { area, angles: [ang(a, b, c), ang(b, a, c), ang(c, a, b)] };
}

triangle(3, 4, 5);   // area 6, angles [90, 53.13, 36.87]

import math

def triangle(a, b, c):
    assert a + b > c and b + c > a and a + c > b, "not a valid triangle"
    p = (a + b + c) / 2
    area = math.sqrt(p * (p - a) * (p - b) * (p - c))
    ang = lambda x, y, z: math.degrees(
        math.acos((y * y + z * z - x * x) / (2 * y * z)))
    return area, [ang(a, b, c), ang(b, a, c), ang(c, a, b)]

triangle(3, 4, 5)   # (6.0, [90.0, 53.13, 36.87])

FAQ