How this calculator works
A semicircle is exactly half of a circle, cut along a diameter. Its curved edge is half the circle circumference, so the arc length is pi times the radius. Its area is half the circle area, so it is pi times the radius squared divided by two. The perimeter is not half the circumference: a semicircle also has the straight diameter as its base, so the full boundary is the arc plus twice the radius. This calculator can start from the radius, the diameter, the semicircle area or the full perimeter, and recovers the radius before reporting every property.
Formula
Semicircle area A = πr²/2. Arc length = πr. Perimeter P = πr + 2r = r(π + 2). The diameter is 2r. Reversing the area formula gives r = sqrt(2A/π), and reversing the perimeter formula gives r = P/(π + 2).
Worked example
A semicircle with radius 5 m has a diameter of 10 m, an arc of 5π ≈ 15.708 m, a perimeter of 5π + 10 ≈ 25.708 m, and an area of 25π/2 ≈ 39.270 m². A diameter of 6 cm gives a radius of 3 cm, an area of 4.5π ≈ 14.137 cm², and a perimeter of 3π + 6 ≈ 15.425 cm.
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Frequently asked questions
Why is the semicircle perimeter not half of 2πr?
Half of the circumference is only the curved part. A semicircle also has a straight diameter closing the shape, so its boundary is the arc πr plus the diameter 2r. Leaving out the diameter would measure only the curved edge.
How do I find the radius from the area?
Double the area, divide by pi, then take the square root. Doubling restores the full circle area and undoing πr² gives the radius. For an area of 39.27 m², doubling gives 78.54, dividing by pi gives 25, and the square root is 5 m.
How do I find the radius from the perimeter?
Divide the perimeter by pi plus two. The perimeter is r(π + 2), so dividing by that constant isolates the radius. This works only because the perimeter includes the diameter and the arc.
Is a semicircle the same as a half circle?
Yes. The shape is half of a circle, bounded by one diameter and one half of the circumference. Its straight edge is the diameter, and the centre of the original circle lies at the midpoint of that edge.