Leave one field blank (or 0) to solve for it. If all three are filled, the hypotenuse field is ignored and recalculated.
s = (a+b+c)/2, area √[s(s−a)(s−b)(s−c)]. Circles, rectangles and solids use the usual πr², 2πr, l×w, and 4/3πr³ formulas. Slope is (y2−y1)/(x2−x1); distance is the Pythagorean gap. Units cancel as long as you stay consistent.
For a triangle defined by three side lengths, the calculator first checks the triangle inequality (that any two sides must sum to more than the third) before computing area with Heron's formula: s = (a+b+c)/2, then area = √[s(s−a)(s−b)(s−c)].
Circles, rectangles, and solid shapes use the standard formulas you'd find in any geometry textbook — πr² for circle area, 2πr for circumference, length × width for a rectangle, and 4/3πr³ for a sphere's volume, among others depending on the shape you select.
Slope between two points is (y2−y1)/(x2−x1), and distance between them uses the Pythagorean theorem to find the straight-line gap. Since these are all pure ratios and formulas, your units simply carry through — enter everything in the same unit and your result will be in that unit (or its square/cube, for area and volume) consistently.
It states that the sum of any two sides of a triangle must be greater than the third side. If your three entered lengths fail this check, they can't actually form a real triangle, so the calculator flags this rather than returning a meaningless area.
Using Heron's formula: first calculate the semi-perimeter s = (a+b+c)/2, then area = the square root of s(s-a)(s-b)(s-c) — this works for any valid triangle without needing to know any angles.
No, as long as you're consistent — enter all your lengths in the same unit, and your area or volume result will be in that unit squared or cubed. Mixing units (like feet for one side and metres for another) would give a meaningless result.
Yes — enter two (x,y) coordinate points and the calculator finds the slope between them and the straight-line distance using the Pythagorean theorem.