How this instrument works
Two sides sharing a vertex, and the angle between them — that is the raw information the formula needs. A = a·b·sinθ turns a pair of lengths and an angle into an area without asking for the height first, because the height is buried inside the sine term: drop a perpendicular from the far end of side b down to the line containing side a, and that perpendicular's length is exactly b·sinθ. Multiply by a and the identity is just base times height, dressed in trigonometry.
The formula survives a slide, and that is not an accident. Euclid's Elements proves, in Book I Proposition 35, that any two parallelograms sharing a base and squeezed between the same pair of parallel lines have identical area — cut the triangle overhanging one end, slide it across, and it exactly refills the gap at the other end, turning the shape into a rectangle of the same base and height without changing so much as one square unit. That is why shearing a parallelogram, sliding its top edge sideways while the height stays fixed, never changes its area, even though every side length except the base is changing as it slides.
Side lengths alone do not pin an area down. Hold a = 5 and b = 7 fixed and swing the angle between them from a hairline crack near 0° up to a square 90°, and the area climbs from almost nothing to a maximum of exactly 35 — the full product a·b, reached only when the sides are perpendicular. Past 90° the area falls again, and it does so symmetrically: sin(θ) and sin(180° − θ) are equal, so either of a parallelogram's two distinct interior angles, plugged into the same formula, returns the identical area.
- Enter one side's length into Side a — any unit, as long as Side b is measured in the same one.
- Enter the adjacent side's length into Side b, measured from the same shared vertex as Side a.
- Enter the angle enclosed between those two sides into Included angle, choosing degrees, radians, or turns from its unit menu.
- Read Area for the result, a·b·sinθ worked to full precision — try 60°, π/3 radians, and 1/6 turn in turn and watch the same Area come back each time.
Worked example — sides of 5 and 7 meeting at 60°
Take a parallelogram-shaped garden bed with one side a = 5 m and the adjacent side b = 7 m, meeting at an included angle of 60°. The height implied by that angle is b·sinθ = 7 × sin(60°) = 7 × 0.8660254 = 6.062178 m, and the area follows as A = a × h = 5 × 6.062178 = 30.310889 m² — matching a·b·sinθ = 5 × 7 × sin(60°) = 30.310889 exactly.
Square the same two sides up to a 90° corner instead — same 5 m and 7 m sides, included angle 90° — and Area jumps to exactly 35 m², the full a·b product, because sin(90°) = 1 and the parallelogram has become a rectangle. Every other angle between 0° and 180° gives something between those two figures: 30.31 m² at 60° sits comfortably inside that range, closer to the maximum than to either flattened extreme.
Questions
What is the formula for the area of a parallelogram?
A = a·b·sinθ, where a and b are two adjacent sides sharing a vertex and θ is the angle enclosed between them. It reduces to the familiar base times height once you notice that b·sinθ is exactly the height measured perpendicular to side a — the sine term supplies the height without you having to measure it directly.
Why does the formula use sinθ instead of the angle itself?
Because sinθ converts the slanted side b into the vertical height relative to base a. Drop a perpendicular from the far end of side b down to the line containing a, and that perpendicular has length b·sinθ by definition of sine in a right triangle. Multiply by base a and the parallelogram's true base-times-height area appears.
Do the two side lengths alone determine the area?
No — the angle between them matters just as much. Sides of 5 and 7 can enclose anywhere from almost 0 m² at a near-flat angle up to a maximum of 35 m² at 90°, all with the identical two side lengths. Area is not a property of the sides in isolation; it depends on how they are hinged together.
Does it matter which of the two interior angles I enter?
No. A parallelogram has two distinct interior angles at adjacent vertices, and they always sum to 180°. Because sin(θ) equals sin(180° − θ), plugging in either one returns the exact same area — a convenient symmetry that means you never have to worry about picking the wrong corner.
What mistake do people make computing a parallelogram's area?
Multiplying the two sides directly, a×b, and forgetting the sine term entirely. That shortcut is only correct for the one special case where θ = 90° and the parallelogram happens to be a rectangle; for any other angle it silently overstates the true area, since sinθ is less than 1 everywhere except that right angle.
What happens to the area as the angle shrinks toward 0°?
It shrinks toward zero right along with it. As θ approaches 0° (or 180°), the two sides swing to lie almost on top of the same line, the parallelogram flattens into a sliver, and sinθ itself approaches 0 — so a·b·sinθ vanishes even though a and b stay exactly the same length.