How this instrument works
Inscribe the largest possible square inside a circle and its four corners land exactly on the circle's edge. Because the square's diagonal is a straight line running from one corner to the opposite corner, and both of those corners sit on the circle, that diagonal has to pass through the center and stretch across two points a full diameter apart. A square's diagonal is always its side times √2 (drop the diagonal in and a right triangle appears, so the Pythagorean theorem gives side² + side² = diagonal² directly), so setting the diagonal equal to 2r and solving backward hands over side = r√2 in one step.
Notice what is missing from that formula: π. Despite living inside a circle, neither the square's side nor its area depends on the circle's most famous constant at all — side = r√2 and area = 2r² are pure algebra, nothing trigonometric hiding underneath. That makes for a neat comparison with the circle's own area, πr²: divide the square's area by the circle's and the r² cancels, leaving a plain 2/π ≈ 0.6366, so the largest square that fits inside any circle always covers about 63.7 percent of it, no matter how big that circle is.
The formula scales the way two multiplied lengths should: doubling the radius doubles the side and quadruples the area, since area is a squared quantity. At the floor, r = 0 collapses the circle to a single point and the inscribed square collapses with it — side and area both return zero, a correct, unremarkable answer rather than an error. And this sheet is really the mirror image of a circumscribed-circle problem worked for a four-sided polygon: there, a side length feeds in and a circumradius comes out; here, the radius feeds in and the side comes out — the same right triangle, read from the opposite direction.
- Enter your circle's measurement into the Circle radius field — any unit works, as long as you read the results in that same unit.
- Read Inscribed square: side length for the edge length of the largest square that fits inside that circle.
- Read Inscribed square: area for that square's area, already squared into the matching unit.
- Cross-check the two outputs by hand: squaring the side length should reproduce the area field, since area = side².
- To size a circle around a known square instead, divide that square's side by √2 to recover the radius that would fit it exactly.
Worked example — a circle of radius 5
Set Circle radius to 5. The largest square that fits has a diagonal forced to equal the circle's diameter, 2 × 5 = 10, so its side comes straight from side = r√2 = 5 × 1.4142135623730951 = 7.0710678118654755 — Inscribed square: side length reports that full figure rather than rounding it down to a tidy 7.07.
Area follows from the shortcut area = 2r² = 2 × 25 = 50 exactly, and Inscribed square: area reports 50.0 with nothing left over. Squaring the side the long way agrees almost perfectly too: 7.0710678118654755² comes back as 50.00000000000001, off from 50 only by the last sliver a computer's floating-point arithmetic ever leaves behind.
Questions
What is the formula for a square inscribed in a circle?
side = r√2 for the edge length and area = 2r² for the area, where r is the circle's radius. Both come from one fact: the largest square that fits has a diagonal exactly equal to the circle's diameter, 2r, because the diagonal is the only straight line inside the square long enough to reach two opposite points on the boundary.
Why doesn't π appear in the inscribed square's formula?
Because the derivation never measures around the circle's curve — it only uses the diagonal, a straight chord equal to the diameter, together with the Pythagorean relationship between a square's side and its diagonal. π belongs to formulas that measure the circle's own curved boundary or area, such as circumference (2πr) or circle area (πr²); this square's edges are all straight lines, so no curve length is ever involved.
How much of the circle does the largest inscribed square cover?
About 63.7 percent. Dividing the square's area, 2r², by the circle's area, πr², cancels the r² and leaves 2/π ≈ 0.6366 — a ratio that holds for every circle regardless of size, since both areas scale the same way with r².
What is the most common mistake when finding the inscribed square's side?
Setting the side equal to the diameter, 2r, instead of to the diagonal. The diameter is the diagonal, not the side — a square's own side is shorter than its diagonal by a factor of √2, so the correct step divides the diameter by √2 (equivalently, multiplies the radius by √2) rather than using 2r directly as the answer.
How does this relate to the circumradius of a square?
It's the same relationship read in reverse. A square of side s has circumradius R = s/√2, the general circumscribed-circle formula for a four-sided polygon; this calculator starts from that radius and solves back to side = R√2, the identical right triangle worked from the opposite direction.
Does the formula still work for a very small or zero radius?
Yes — the relationship is linear with no lower limit besides zero itself. A radius of 0 collapses the circle to a single point, and the inscribed square collapses with it: side = 0 × √2 = 0 and area = 2 × 0² = 0, a correct answer rather than an undefined case.