Geometry calculator with a scale drawing and working

Three independent ones, and at least one must be a length. Three angles fix only the shape: triangles with angles 30°, 60° and 90° come in any size. Dependent data do not help either: once α and β are known, γ is known too (γ = 180° − α − β). When the data are not enough, the solver says how many values are missing and suggests fitting ones — for a and b, for example, another side, an angle, a height or the area.

The angle β comes from the law of sines: sin β = b·sin α / a. Two angles share one sine — β and 180° − β. If the side a opposite the given angle is shorter than b but longer than b·sin α, both work, and there are two different triangles, a narrow one and a wide one. If a ≥ b, the obtuse option makes the angles add up to more than 180°, so there is one solution. If a < b·sin α, the side cannot reach the other arm of the angle, and there is no triangle.

Compare the square of the longest side with the sum of the squares of the other two. If c² < a² + b², the triangle is acute; if they are equal, it is right (the converse of the Pythagorean theorem); if c² is larger, it is obtuse. This follows from the law of cosines, c² = a² + b² − 2ab·cos γ: the sign of the cosine decides whether γ is acute or obtuse. For 5, 6, 7: 49 < 25 + 36, so the triangle is acute.

25π is the exact area of a circle of radius 5, while 78.5 is a rounding, and which one depends on the value of π used: π ≈ 3.14 gives 78.5, π ≈ 3.1416 gives 78.54. Teachers usually ask for the answer "in terms of π" or with π ≈ 3.14; both are in the setting "The number π". The exact form is kept through the working until the last step, so rounding errors do not pile up.

With the shoelace formula: A = ½·|x₁y₂ − x₂y₁ + x₂y₃ − x₃y₂ + … + xₙy₁ − x₁yₙ|. The vertices must be listed in the order you go around the polygon, clockwise or counterclockwise; otherwise the sides cross and the formula gives a different area. In the Points mode you can enter up to eight vertices: the tool checks that the sides do not cross and writes out the sum.

Dashed lines show hidden edges — those covered by the faces of the solid. In class solids are drawn in oblique projection: the front face without distortion, and the edges going back at 45° and half as long. Such a drawing shows which face is in front and which is behind, so diagonals and heights do not get mixed up. Here the hidden edges are found from the directions of the faces, so the dashes match the textbook.

Because a construction has to work for ANY given segments, not for measured numbers. The straightedge only draws a line through two points; lengths are carried by the compass, which keeps its width exactly. A measured 4.3 cm is already rounded, and a point placed by the markings is off by a fraction of a millimeter — the construction then works "by eye", not by proof. That is why "Your turn" here moves the givens slightly: a correct construction follows them, an eyeballed one stays put.

60°: draw a circle centered at the vertex through any point A on one side, then an arc of the same radius centered at A — the arcs cross at the point B for which triangle OAB is equilateral, so the angle AOB is 60°. 30° is the bisector of 60°, 90° is the perpendicular to a line through a point on it, and 45° is the bisector of 90°. The 60° angle, the bisector and the perpendicular are in the list of constructions and play step by step.