Area Formulas for All Shapes — Complete Reference Guide
Area measures how much 2D space a shape occupies — the number of unit squares that fit inside it. Every shape has a formula derived from its geometry. Understanding these formulas (not just memorizing them) makes it far easier to recall and apply them correctly.
This guide covers area formulas for every common shape, explains the logic behind each formula, and includes a complete reference table. All formulas use consistent notation: b = base, h = height, r = radius, s = side length.
Formula
Triangle: A = ½bh | Circle: A = πr² | Rectangle: A = lw
Area Formula Reference Table
The following table lists area formulas for all common 2D shapes. Use this as a quick reference or study guide.
| Shape | Formula | Variables | Example |
|---|---|---|---|
| Square | A = s² | s = side length | s = 5 → A = 25 |
| Rectangle | A = l × w | l = length, w = width | 8 × 5 = 40 |
| Triangle | A = ½bh | b = base, h = perpendicular height | ½ × 10 × 6 = 30 |
| Triangle (Heron's) | A = √(s(s−a)(s−b)(s−c)) | s = (a+b+c)/2 | sides 3,4,5 → A = 6 |
| Triangle (SAS) | A = ½ab·sin(C) | a,b = sides, C = included angle | ½×4×5×sin(60°) ≈ 8.66 |
| Circle | A = πr² | r = radius | r=7 → A = 153.94 |
| Semicircle | A = πr²/2 | r = radius | r=4 → A = 25.13 |
| Ellipse | A = πab | a,b = semi-axes | a=5,b=3 → A = 47.12 |
| Parallelogram | A = bh | b = base, h = perpendicular height | 9 × 4 = 36 |
| Rhombus | A = (d₁ × d₂)/2 | d₁,d₂ = diagonals | (6×8)/2 = 24 |
| Trapezoid | A = ½(a+b)h | a,b = parallel sides, h = height | ½(6+10)×5 = 40 |
| Kite | A = (d₁ × d₂)/2 | d₁,d₂ = diagonals | (8×10)/2 = 40 |
| Sector | A = (θ/360)πr² | θ = angle in degrees, r = radius | (90/360)π×4² = 12.57 |
| Regular Hexagon | A = (3√3/2)s² | s = side length | s=4 → A = 41.57 |
| Regular Pentagon | A ≈ 1.720s² | s = side length | s=5 → A ≈ 43.01 |
| Regular n-gon | A = ns²/(4tan(π/n)) | n = sides, s = side length | n=6,s=4 → A = 41.57 |
Triangle Area — Why ½ × base × height?
The area of a triangle is A = ½ × b × h. The "half" comes from the fact that any triangle is exactly half of a parallelogram with the same base and height. To see this: take any triangle and flip a copy of it. The two triangles fit together to form a parallelogram with area b × h. Each triangle is therefore ½ × b × h.
Critical: h must be the perpendicular height — the vertical distance from the base to the opposite vertex. If you're given a slanted side instead of the height, you can find h using trigonometry: h = side × sin(angle).
For an equilateral triangle (all sides equal to s), the height is h = (√3/2)s, giving A = ½ × s × (√3/2)s = (√3/4)s².
Circle Area — Why πr²?
The formula A = πr² can be derived by imagining the circle cut into very thin concentric rings. Each ring at radius x has circumference 2πx and infinitesimal width dx. Summing (integrating) all rings from 0 to r: A = ∫₀ʳ 2πx dx = πr².
A more visual proof: cut the circle into many thin wedges (like pizza slices) and rearrange them into a shape that approaches a rectangle. The width approaches πr (half the circumference) and the height approaches r, giving area ≈ πr × r = πr². As the slices get thinner, this approximation becomes exact.
Derived quantities: circumference C = 2πr; diameter d = 2r; A = π(d/2)² = πd²/4; if you know C, then A = C²/(4π).
Sector Area — Fraction of the Full Circle
A sector with central angle θ (degrees) is a fraction θ/360 of the full circle. So its area is (θ/360) × πr². In radians, the formula simplifies to A = ½r²θ, because a full circle in radians is 2π, so the fraction is θ/(2π) and area = θ/(2π) × πr² = ½r²θ.
The arc length (curved boundary) follows the same fraction logic: L = (θ/360) × 2πr in degrees, or L = rθ in radians.
Trapezoid Area — Average Width Times Height
A trapezoid with parallel sides a and b and height h has area A = ½(a+b)h. This equals (average width) × height, where the average of the two parallel sides gives an "average width." It's a natural extension of the rectangle formula (where both parallel sides are equal).
Visually, you can cut a trapezoid diagonally and rearrange the pieces into a parallelogram with base (a+b) and half the height, confirming A = ½(a+b)h.
Regular Polygon Area — ns²/4tan(π/n)
A regular polygon with n sides of length s can be divided into n isosceles triangles meeting at the center. Each triangle has base s and height equal to the apothem a = s/(2tan(π/n)). Triangle area = ½sa. Total polygon area = n × ½sa = ½n × s × s/(2tan(π/n)) = ns²/(4tan(π/n)).
As n increases, the polygon increasingly resembles a circle. The formula approaches πr² (where r is the circumradius) as n → ∞. At n = 6 (hexagon), area ≈ 0.827πr², already quite close to the circle.