How this instrument works
A semicircle is what a circle becomes when a single straight diameter cuts it in two. The enclosed area, A = πr² ⁄ 2, is not a separate rule to memorize — it falls straight out of the full circle's πr² by symmetry: reflect one half of the disk across that diameter and it lands exactly on the other half, so the two pieces are congruent mirror images, and each must hold precisely half the enclosed area, whatever the radius happens to be.
The boundary itself stitches together two different kinds of edge — a straight diameter and a curved arc sweeping exactly 180° — but the enclosed area only answers for the surface swept between them, which is why the formula keeps the same r² term the full circle uses. A protractor's flat baseline, a lunette window set above a doorway, and the curved transition of a skate ramp are all instances of the identical half-disk, scaled up or down by whichever radius each one happens to use.
Shrink the radius toward zero and the shape collapses to a single point, area included: the arithmetic lands on exactly 0, with nothing left over to round away. And unlike a pie slice cut at some arbitrary angle, a semicircle's share of its parent circle never drifts: the cut is always a straight diameter, always 180°, so the ratio stays locked at exactly one half for every radius, from a fraction of a millimetre to a planet's equator.
- Type the radius of the parent circle into the Radius field — the straight-line distance out from the center to the curved arc, not a length taken along the flat diameter edge.
- Area updates immediately to A = πr² ⁄ 2, carried to eight decimal places.
- Starting from a known diameter rather than a radius? Halve it first — Radius wants r, not d — then read Area as usual.
- Sanity-check the result by doubling it in your head: twice whatever Area shows should land close to the parent circle's full πr².
- A radius of zero is accepted and returns an area of exactly zero, the shape's single degenerate point; negative values aren't, since a radius only ever measures a distance.
Worked example — a radius-4 fanlight window
A bricklayer sets a semicircular fanlight above a doorway, the curved brick arch spanning a radius of exactly 4 feet from the springing line to the crown. The glazed area to order is A = π × 4² ⁄ 2 = π × 16 ⁄ 2 = 8π = 25.132741228718345 square feet — half of the 16π ≈ 50.265 a full round window that size would need, since only the upper half of the circle is glazed at all.
Scale the same design up to a radius of 8 feet for a grander entrance, and the glazed area doesn't just double to match it. It becomes π × 8² ⁄ 2 = 32π = 100.53096491487338 square feet, four times the smaller window's 25.132741228718345, because area tracks the square of the radius while the radius itself only doubled. It's an easy point to miss when a client asks for a 'slightly bigger' arch and expects the glass bill to grow by roughly the same slight amount.
Questions
What formula gives a semicircle's area?
A = πr² ⁄ 2, where r is the radius of the circle the semicircle came from. It is exactly half of the familiar A = πr², since a semicircle is exactly half a disk. A radius of 4 gives A = π × 16 ⁄ 2 = 8π = 25.132741228718345, the figure this sheet's own worked example returns.
How does a semicircle's area compare to a quarter circle's?
A semicircle is bounded by one straight diameter, a single cut through the center, and keeps half the parent disk: πr² ⁄ 2. A quarter circle needs two perpendicular radii, a second cut, and keeps only a quarter: πr² ⁄ 4. For a matching radius, a semicircle therefore always covers exactly twice the area a quarter circle cut from the same circle would.
Why is the semicircle's share of the circle always exactly one half?
Because the cutting line is always a diameter, and reflecting one side of that line across it maps perfectly onto the other side — the two halves are mirror images, congruent by that reflection alone, so each carries exactly half the disk's area regardless of how large or small the radius is. No other angle of cut gives such a clean, radius-independent fraction.
What mistake do people most often make with this formula?
Forgetting the divide-by-two and reporting the full circle's πr² instead, which overstates the true area by a factor of exactly 2. A close second is mixing up area with perimeter: a semicircle's boundary length is πr + 2r (the arc plus the straight diameter edge), a completely different expression that does not share a simple ratio with the area.
How do I recover the radius from a known semicircle area?
Solve the equation for r instead: r = √(2A ⁄ π). Given the area 25.132741228718345 square units from this sheet's own example, r = √(2 × 25.132741228718345 ⁄ π) = √16 = 4 — the exact radius the example started from, with no rounding beyond the digits of π itself.
Can this be worked from a diameter rather than a radius?
Yes — substitute r = d ⁄ 2 into A = πr² ⁄ 2 and it simplifies to A = πd² ⁄ 8. A doorway arch measured at an 8-foot span, for instance, gives the same 8π ≈ 25.13 square feet as a radius of 4, since the two measurements describe the identical curve.