Volume, Area Curves, lines   Imaginary math   Trigonometry   Calculus   Math ```Circle A = πr² length of chord L = 2√(r²-d²) r = radius d = perpendicular distance from chord to center length of chord = L, r = radius L = r crd θ where crd θ = 2 sin (θ/2) L = 2r sin (θ/2) solve for θ sin (θ/2) = (L/2r) θ = 2 arcsin (L/2r) Area of a segment A = (r²/2)[(π/180)θ – sinθ] equation of circle with center at a,b and radius r (x–a)² + (y–b)² = r² center at the origin x² + y² = r² triangle with inscribed circle radius = (1/s)√[s(s–a)(s–b)(s–c)] where s = (a+b+c)/2 a,b,c are the sides of the triangle Ellipse A = πab where a and b are one-half of the ellipse's major and minor axes, ie, the lengths of the semi-major and semi-minor axes. Ellipse whose axis correspond to the x and y axis has the equation (x/a)² + (y/b)² = 1 a & b are distance from origin to curve along x or y axis. Ellipse centered at x₁ and y₁, the equation is ((x – x₁)/a)² + ((y – y₁)/b)² = 1 a & b are distance from center to curve along x or y axis. Ellipse is not rotated. Sphere V = ⁴/₃πr³ A = 4πr² Volume of a partial sphere (cap) V = (πh²/3)(2r–h) V = (πh/6)(3a²+h²) where r is the radius of the sphere, h is the height of the cap, a is the radius of the cap. Cylinder V = πr²h A = 2πr² + 2πrh = 2πr(r+h) but skipping the ends, A = 2πrh Triangle equalaterial triangle A = (s²√3)/4) equalaterial triangle inscribed in circle, radius R A = (3√3/4)R² Isosceles triangle A = (b/4)√(4a²–b²) = (a²/2)sinθ a = length of equal sides, b the other θ = angle between equal sides any triangle A = ½bh b is base, h is height A = √[s(s–a)(s–b)(s–c)] where s = (a+b+c)/2 a,b,c are the sides of the triangle A = ½ab sin θ where θ is the angle between a and b a,b are two sides of the triangle Pythagorean theorem applies to only right triangles, where c is the side opposite the right angle, and a and b are the otehr two sides c = √(a² + b²) Cube Volume = s³ SA = 6s² Pyramid The volume of a pyramid is V = (1/3)Bh where B is the area of the base and h the height from the base to the apex. This works for any location of the apex, provided that h is measured as the perpendicular distance from the plane which contains the base. Surface area square base pyramid A = Aʙ + ps/2 Aʙ is area of base p is the perimeter of base s is slant height regular tetrahedron made up of 4 equilateral triangles A = s²√3 V = s³/(6√2) Cone surface area A = πr² + πrs = πr(r+s) where s is slant height volume V = ⅓πr²h, h is height of cone (not slant) h² + r² = s² A = πr² + πr√(h² + r²) Truncated Cone volume of truncated cone = (πh/3)(r² + R² + rR) r and R are the upper and lower radii Torus volume of a torus is 2π²Rr² where R is the distance from the center of the tube to the center of the torus, r is the radius of the tube. A conical frustum is a frustum created by slicing the top off a cone (with the cut made parallel to the base). For a right circular cone, let h be the height (not slant) and R₁ and R₂ the base and top radii. Then V = (1/3)πh(R₁² + R₁R₂ + R₂²) regular polygon with n sides area = s²(n/4)(cot(π/n)) where s is length of one side Pentagon (5), with sides all equal, Area is A = (s²/4)√(25+10√5) = 1.720s² where s is length of one side Σ interior angles 540º Hexagon (6), with sides all equal, Area = (3√3/2)(s²) = 2.598s² where s is length of one side Σ interior angles 720º Octagon (8), with sides all equal, Area = 2(√2–1)s² ≈ 4.828s² where s is length of one side Sum of the interior angles of a convex polygon with n sides 180(n–2) Square, Rectangle, Quadrilateral square A = s² rectangle A = L•W cube V = s³ A = 6s² cuboid (rectangular prism) V = L•W•H A = 2LW + 2LH + 2WH Simple Quadrilateral Σ interior angles 360º trapezium (UK) or trapezoid (US) had 2 parallel sides. isosceles trapezium/trapezoid is symmetrical around the center rectangle, sum of interior angles is 360º Rhombus A rhombus (equilateral quadrilateral) is a parallelogram with 4 equal sides. copposite angles are equal parallelogram A = sh = pq/2 where h is the height, s is a side and p,q are the diagonals. sum of interior angles is 360º Trapezoid A = (h/2)(b₁+b₂) h is the height b₁,b₂ are the lengths of the two parallel sides Parallelogram A = hb h is the height b is the base Parabola y = a(x–h)² + k vertex is at (h, k) focus relative to vertex is 1/4a If the chord has length b, and is perpendicular to the parabola's axis of symmetry, and if the perpendicular distance from the parabola's vertex to the chord is h, the parallelogram is a rectangle, with sides of b and h. 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