Calculate the area of a sector of a circle (\(A = \frac{\theta}{360^\circ}\pi r^2 = \frac{1}{2}r^2\alpha = \frac{1}{2}rL\)), arc length (\(L = \frac{\pi r\theta}{180^\circ}\)), closed perimeter (\(P = L + 2r\)), chord length (\(c = 2r\sin(\theta/2)\)), circular segment area, and sector centroid distance across degrees and radians.
Select input combination to calculate sector area, perimeter, and arc geometry.
Radius: 10.000 cm • Angle: 90.000° (1.571 rad) • Arc: 15.708 cm
| 📏 Circle Radius (\(r\)) | 10.000 cm |
| 📏 Circle Diameter (\(d\)) | 20.000 cm |
| 📐 Central Angle (\(\theta\)) | 90.000° (1.571 rad) |
| 🔄 Curved Arc Length (\(L\)) | 15.708 cm |
| 📏 Closed Perimeter (\(P\)) | 35.708 cm (L + 2r) |
| 🔵 Sector Area (\(A_{\text{sector}}\)) | 78.540 cm² |
| 🍕 Circle Coverage Fraction | 25.000% (1/4 of circle) |
| 📐 Chord Length (\(c\)) | 14.142 cm (2r sin(θ/2)) |
| 📏 Sagitta Height (\(h\)) | 2.929 cm |
| 🔺 Triangle Area under Chord | 50.000 cm² (½r² sin θ) |
| 🌘 Circular Segment Area | 28.540 cm² (A_sec - A_tri) |
| 🎯 Centroid Distance (\(\bar{x}\)) | 6.002 cm (from center) |
Circular sectors (pie-shaped portions of a circle) are integral to civil highway curve design, agricultural center-pivot irrigation coverage, wind turbine swept areas, architectural arches, and culinary portion sizing. Our Sector Area Calculator solves critical real-world challenges:
Calculates exact field acreage watered by rotating sprinkler heads covering partial arcs (\(90^\circ\), \(180^\circ\), or \(270^\circ\)) using \(A = \left(\frac{\theta}{360^\circ}\right)\pi r^2\).
Computes arc length (\(L = \frac{\pi r\theta}{180^\circ}\)), chord length (\(c = 2r\sin(\theta/2)\)), and sightline sagitta clearances for horizontal road curves and railway tracks.
Decomposes any circular sector into its inner triangle area (\(\frac{1}{2}r^2\sin\theta\)) and outer circular segment region (\(A_{\text{seg}} = A_{\text{sec}} - A_{\text{tri}}\)).
Solves for radius (\(r\)), central angle (\(\theta\)), arc length (\(L\)), or sector area (\(A\)) from any pair of given dimensions with seamless degree-to-radian conversions.
Select from Radius & Angle, Radius & Arc Length, Arc & Angle, Area & Radius, or Area & Angle.
Toggle between Degrees (°) and Radians (rad), and enter your numerical values.
Inspect calculated sector area, arc length, perimeter, chord, and segment dimensions.
Click Copy Sector Telemetry Card to export formatted results for engineering, CAD, or homework.
$$A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2$$ $$L_{\text{arc}} = \frac{\theta}{360^\circ} \times 2\pi r = \frac{\pi r \theta}{180^\circ}$$ $$P_{\text{sector}} = L_{\text{arc}} + 2r = r\left(\frac{\pi \theta}{180^\circ} + 2\right)$$
$$A_{\text{sector}} = \frac{1}{2} r^2 \alpha = \frac{1}{2} r L_{\text{arc}}$$ $$L_{\text{arc}} = r \alpha, \quad P = r(\alpha + 2)$$ $$\text{Relation: } A = \frac{L_{\text{arc}}^2}{2\alpha}$$
$$c = 2r \sin\left(\frac{\theta}{2}\right), \quad h_{\text{sagitta}} = r\left(1 - \cos\left(\frac{\theta}{2}\right)\right)$$ $$A_{\text{triangle}} = \frac{1}{2}r^2 \sin(\theta)$$ $$A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}} = \frac{1}{2}r^2(\alpha - \sin\alpha)$$
$$\bar{x} = \frac{2r \sin(\alpha/2)}{3(\alpha/2)} = \frac{4r \sin(\theta/2)}{3\alpha_{\text{rad}}}$$ $$\text{Quadrant } (90^\circ): \bar{x} = \frac{4\sqrt{2}r}{3\pi} \approx 0.6002 \cdot r$$
Arc length (\(L\)) only measures the curved outer edge of the sector. A closed sector includes the two straight radii borders: \(P = L + 2r\).
Using \(\frac{1}{2}r^2\theta\) with \(\theta\) in degrees gives wildly incorrect answers! The formula \(\frac{1}{2}r^2\alpha\) requires angle \(\alpha\) in radians. For degrees, you must use \(\frac{\theta}{360^\circ}\pi r^2\).
A circular sector is a pie slice connected to the circle center. A circular segment is only the outer slice between the chord and the arc. Segment Area = Sector Area − Triangle Area.
If asked for the area of a shaded major sector with acute angle \(\theta\), the major sector angle is \(360^\circ - \theta\). Always ensure you are calculating the intended shaded region.
A circular sector has a radius of \(r = 12.0\,\text{cm}\) and a central angle of \(\theta = 60^\circ\). Calculate its area, arc length, and closed perimeter.
Solution: Sector Area: \(A = \frac{60}{360} \times \pi \times 12^2 = \frac{1}{6} \times 144\pi = 24\pi \approx \mathbf{75.398\,\text{cm}^2}\). Arc Length: \(L = \frac{60}{360} \times 2\pi(12) = 4\pi \approx \mathbf{12.566\,\text{cm}}\). Closed Perimeter: \(P = 12.566 + 2(12) = \mathbf{36.566\,\text{cm}}\).
A circle of radius \(r = 8\,\text{m}\) has a shaded sector with area \(A = 50\,\text{m}^2\). What is the central angle in degrees and radians?
Solution: In radians: \(\alpha = \frac{2A}{r^2} = \frac{100}{64} = \mathbf{1.5625\,\text{rad}}\). In degrees: \(\theta = 1.5625 \times \frac{180^\circ}{\pi} \approx \mathbf{89.525^\circ}\).
Find the area of the circular segment formed by a \(90^\circ\) sector of radius \(r = 10\,\text{cm}\).
Solution: Sector Area: \(A_{\text{sector}} = \frac{90}{360}\pi(10^2) = 25\pi \approx \mathbf{78.540\,\text{cm}^2}\). Triangle Area: \(A_{\text{tri}} = \frac{1}{2}(10)(10)\sin(90^\circ) = \mathbf{50.000\,\text{cm}^2}\). Segment Area: \(A_{\text{seg}} = 78.540 - 50 = \mathbf{28.540\,\text{cm}^2}\).
Standard dimensions across canonical central angles for a circle of radius \(r = 10\,\text{cm}\).
| Central Angle (\(\theta\)) | Radians (\(\alpha\)) | Sector Area (\(r=10\)) | Arc Length (\(L\)) | Chord Length (\(c\)) | Circle Fraction |
|---|---|---|---|---|---|
| \(30^\circ\) (1/12th Circle) | \(0.524\,\text{rad}\) (\(\frac{\pi}{6}\)) | \(26.180\,\text{cm}^2\) | \(5.236\,\text{cm}\) | \(5.176\,\text{cm}\) | \(8.33\%\) |
| \(45^\circ\) (Octant / 1/8th) | \(0.785\,\text{rad}\) (\(\frac{\pi}{4}\)) | \(39.270\,\text{cm}^2\) | \(7.854\,\text{cm}\) | \(7.654\,\text{cm}\) | \(12.50\%\) |
| \(60^\circ\) (Sextant / 1/6th) | \(1.047\,\text{rad}\) (\(\frac{\pi}{3}\)) | \(52.360\,\text{cm}^2\) | \(10.472\,\text{cm}\) | \(10.000\,\text{cm}\) | \(16.67\%\) |
| \(90^\circ\) (Quadrant / 1/4th) | \(1.571\,\text{rad}\) (\(\frac{\pi}{2}\)) | \(78.540\,\text{cm}^2\) | \(15.708\,\text{cm}\) | \(14.142\,\text{cm}\) | \(25.00\%\) |
| \(180^\circ\) (Semicircle / 1/2) | \(3.142\,\text{rad}\) (\(\pi\)) | \(157.080\,\text{cm}^2\) | \(31.416\,\text{cm}\) | \(20.000\,\text{cm}\) | \(50.00\%\) |
| \(270^\circ\) (Major 3/4th) | \(4.712\,\text{rad}\) (\(\frac{3\pi}{2}\)) | \(235.619\,\text{cm}^2\) | \(47.124\,\text{cm}\) | \(14.142\,\text{cm}\) | \(75.00\%\) |
Authoritative answers to common questions about circular sector areas, arc lengths, perimeters, chord lengths, and segment calculations.