Circle Square Footage Calculator

A Circular Border: Which Area Are You Buying For?

A round patio has an outside diameter of 20 ft and a uniform 1-ft border inside its edge. The inner surface and the border need different material quantities. Subtract the border on both sides, not just once.

1. Inner diameter: 20 − 2 × 1 = 18 ft.
2. Inner surface: π × 18² ÷ 4 ≈ 254.47 sq ft.
3. Border only: π × (20² − 18²) ÷ 4 ≈ 59.69 sq ft.

For outside diameter D and border width w, use border area = π × [D² − (D − 2w)²] ÷ 4. This assumes concentric circles, uniform width, matching units, and 0 ≤ 2w < D. An off-center or uneven border needs separate measurements.

To check with the calculator above, enter radius 10 ft, note Total Area, then enter 9 ft. Subtracting the displayed results gives 314.16 − 254.47 = 59.69 sq ft here; those displays are rounded. For an unrounded calculation, use π × (10² − 9²) and round the final answer. Apply any cutting allowance to the material you are ordering, not automatically to both surfaces. For a concrete volume, also supply thickness in the area-to-volume calculator.

What If Your Diameter Measurement Is Approximate?

Suppose you measure 20 ft across and judge that the actual diameter could be up to 2 inches smaller or larger. Two decimal places on a calculator do not make that tape measurement exact.

With diameter D and your chosen error bound e in the same units, the limits are π(D − e)²/4 and π(D + e)²/4, where 0 ≤ e < D. This is a measurement-bound example, not a statistical confidence interval or a recommended waste percentage. It assumes a true circle; measure across the center in more than one direction to check that assumption.

These worked examples are derived from the circle area formula. Calculations keep full precision until the displayed result is rounded.

Worked Examples: A Chord, a Diameter Error, and a Ring Band

Three measurements the tool above cannot take directly. Each is computed with the formula the calculator uses (π × r², full precision inside, rounding only on the display), then checked by typing the resulting radius into the widget. This tool has no chord, arc-height, border-width or thickness input, so every example below stops at a radius you can enter — and it returns an area, not a material order.

1. When the center is unreachable: a chord and the arc height

A round bed with a tree in the middle cannot be taped center-to-edge. What you can measure is a straight line across the curved edge — the chord c — and the bulge of the arc above the midpoint of that line, height h. The Math Doctors page on circular segments opens with exactly that problem: “Recent questions have dealt with calculations of various parts of a segment of a circle (chord, arc, sagitta, etc.). How, for example, can you find the arc length if you know the chord length and the height?” It defines h as “the distance from the midpoint of the chord to the midpoint of the arc”.

The radius then follows from r = (c² + 4h²) ÷ (8h). For a 10 ft chord with a 2 ft arc height:

• Radius: (10² + 4 × 2²) ÷ (8 × 2) = 116 ÷ 16 = 7.25 ft, so the recovered diameter is 14.5 ft.
• Full circle: enter radius 7.25 above and the tool displays 165.13 sq ft, which is 15.34 m².
• Back-check the geometry: the half-chord that this radius and height imply is √(2rh − h²) = √(2 × 7.25 × 2 − 2²) = 5.00 ft, exactly half of the 10 ft you measured.
• The crescent is not the circle. The segment “bounded by an arc of the circle and the chord subtending it” needs the central angle as well: θ = 2 × arcsin(c ÷ 2r) = 1.52203 rad and K = r²(θ − sin θ) ÷ 2 = 13.75 sq ft — 8.33% of 165.13 sq ft.

The same reference lists the seven quantities of a segment and states that “No one of these seven values is enough to determine the other six. Any two of them, however, are sufficient to determine the remaining five.” Chord plus height is one of those pairs; the radius it returns is what this calculator wants.

A flatter arc is far less forgiving, because the formula divides by h. A 12 ft chord with a 1.5 ft height gives radius 12.75 ft and 510.71 sq ft (its own segment is 12.15 sq ft, only 2.38% of the circle). Moving just the height reading by a tenth of a foot moves the area by more than 50 sq ft in either direction:

Take the height to the nearest fraction you can defend, then measure a second chord somewhere else. Two chords that recover different radii mean the edge is not one circle — no formula can fix a shape that is not round.

2. How a diameter error scales into the area

Area grows with the square of the diameter, so a measurement error never carries over one-for-one. Start from a pad believed to be 20 ft across (314.16 sq ft) and shift the diameter itself:

The rule is (1 + e)² − 1: 5% too big gives 1.05² − 1 = 10.25% more area, while 5% too small gives 0.95² − 1 = −9.75%. The directions are not symmetric — at 4 in the over-measure costs 10.73 cu ft while the equal under-measure saves 10.21 cu ft. Half an inch either side of 20 ft is 2.5% each way: 19.5 ft gives 298.65 sq ft (−4.94%) and 20.5 ft gives 330.06 sq ft (+5.06%).

What to do with the number once you have it is a documented question, not a style choice. NIST's guide for the use of the SI states that “Proper conversion procedure requires rounding this converted numerical value to the number of significant digits that is consistent with the maximum possible rounding error of the unconverted numerical value”, and adds that “one must often choose between discarding information or providing unwarranted information”. A diameter taped to the nearest half inch therefore does not justify a two-decimal area, even though the widget displays two decimals.

3. A ring band: its area, its edging length, its concrete

Take the border example above at a 20 ft outside diameter and a 2 ft band, so the inside diameter is 20 − 2 × 2 = 16 ft. One shape produces three different order quantities, and ordering the wrong one is the usual mistake:

The identity that ties them together is band area = width × centerline length, so 113.10 ÷ 2 = 56.55 lin ft. Dividing the area you already have by the width gives the linear run of anything placed through the middle of the band — a landscape-fabric strip, one row of pavers, an irrigation line — while the two edges together are 2 × 56.55 = 113.10 lin ft. Both follow from π × (R² − r²) = (R − r) × 2π × (R + r) ÷ 2; no measurement is duplicated.

Treat the matching digits as a coincidence of this width. At 2 ft the band area (113.10 sq ft) equals the two-edge length (113.10 lin ft); at a 1 ft band the area is 59.69 sq ft while the two edges are 119.38 lin ft. And the centerline at that 1 ft width is 59.69 lin ft, equal to the area again. Squares and linear feet are different units — use the width to decide which one you are buying.

To check any of this with the widget: enter radius 10 and then radius 8, and subtract the two displayed areas to get 113.10 sq ft. Thickness does not belong on this page — supply it in the area-to-volume calculator.