Prepared by: RTFVerterra Plane and Solid Geometry Formulas ASIAN DEVELOPMENT FOUNDATION COLLEGE Tacloban City The content of this material is one of the intellectual properties of Engr. Romel Tarcelo F. Verterra of Asian Development Foundation College. Reproduction of this copyrighted material without consent of the author is punishable by law. Part of: Plane and Solid Geometry by RTFVerterra © October 2003 10 sides = decagon 11 sides = undecagon 12 sides = dodecagon 15 sides = quindecagon 16 sides = hexadecagon Given four sides a, b, c, d, and sum of two opposite angles: (s −a)(s −b)(s −c)(s −d) −abcdcos2 θ A= a+b+c+d 2 ∠A + ∠C ∠B + ∠D θ= or θ = 2 2 s= Given four sides a, b, c, d, and two opposite angles B and D: C Parallelogram B Number of diagonals, D The diagonal of a polygon is the line segment joining two non-adjacent sides. The number of diagonals is given by: n D = (n − 3) 2 C d2 θ b D a Given diagonals d1 and d2 and included angle θ: A = ½ d1 × d2 × sin θ Given two sides a and b and one angle A: B A = ab sin A Rhombus C D d1 θ A d2 a 90° b A a Given diagonals d and d : 1 Given three sides a, b, and c: (Hero’s Formula) Given three angles A, B, and C and one side a: 2 a sinBsinC A= 2 sin A 2 Trapezoid a h a+b A= h 2 Cyclic Quadrilateral A cyclic quadrilateral is a B quadrilateral whose vertices lie on the circumference of a circle. ∠A + ∠C = 180° ∠B + ∠D = 180° b b d1 d2 D a d A ra a+b+c+d 2 × 180° n Exterior angle = 360° / n a Diagonal, d = a 2 General quad rilateral b C B θ a β2 D d Given diagonals d1 and d2 and included angle θ: A = ½ d1 × d2 × sin θ AT A ; rb = T c s−c s−b Circle circumscribed about a quadrilateral A circle is circumscribed about a quadrilateral if it passes through the vertices of the quadrilateral. b rc a d (ab + cd)(ac + bd)(ad + bc) 4Aquad s = ½(a + b + c + d) r D C πrθ 180° πr2θ Area = ½ r2 θradians = 360° Area = ½ C × r Arc C = r × θradians = r θ r Circle incribed in a quadrilateral b A circle is inscribed in a quadrilateral r if it is tangent to the three a sides of the quadrilateral. O r= C Segment of a circle Area = Asector – Atriangle 2 2 Area = ½ r θr – ½ r sin θ Area = ½ r2 (θr – sin θ) r θ θr = angle in radians Area = Asector + Atriangle 2 2 Area = ½ r αr + ½ r sin θ 2 Area = ½ r (αr + sin θ) Aquad s Aquad = c d ; s = ½(a + b + c + d) abcd r SOLID GEOMETRY O α = 360 - θ r θ r d1 d2 A β5 ;r= s−a Aquad = (s − a)(s − b)(s − c)(s − d) Sector of a circle θ5 θ3 c AT r= Note: 1 radian is the angle θ such that C = r. Perimeter, P = 4a ra= Perimeter, P = n × x n−2 Interior angle = Circumference = 2π r = πD π Area, A = π r2 = D2 4 Area = (s − a)(s − b)(s − c)(s − d) There are two basic types of polygons, a convex and a concave polygon. A convex polygon is one in which no side, when extended, will pass inside the polygon, otherwise it called concave polygon. The following figure is a convex polygon. β4 θ4 β3 a c a b Circle Square Area, A = A circle is escribed about a triangle if it is tangent to one side and to the prolongation of the other two sides. A triangle has three escribed circles. ra POLYGONS d Circles escribed about a triangle (Excircles) x Area, A = ½ R2 sin θ × n = ½ x r × n c 2 2 Diagonal, d = a + b a2 C b ra θ = 360° / n C “For any cyclic quadrilateral, the product of the diagonals equals the sum of the products of the opposite sides” d1 × d2 = ac + bd Perimeter, P = 2(a + b) A Inscribed circle x = side θ = angle subtended by the side from the center R = radius of circumscribing circle r = radius of inscribed circle, also called the apothem n = number of sides Ptolemy’s theorem Area, A = ab abc 4A T Circle inscribed in a triangle (Incircle) A circle is inscribed in a triangle if it is tangent to the three sides of the triangle. B Incenter of the triangle A c r= T a s r r s = ½(a + b + c) r x A = a2 sin A s= b c b Given side a and one angle A: Rectangle a Apothem x The area under this condition can also be solved by finding one side using sine law and apply the formula for two sides and included angle. d x R R θ θ θ r θθ 2 1 a+b+c 2 The area under this condition can also be solved by finding one angle using cosine law and apply the formula for two sides and included angle. Polygons whose sides are equal are called equilateral polygons. Polygons with equal interior angles are called equiangular polygons. Polygons that are both equilateral and equiangular are called regular polygons. The area of a regular polygon can be found by considering one segment, which has the form of an isosceles triangle. Circumscribing x circle A= ½d ×d A = s(s − a)(s − b)(s − c) s= Regular polygons x B Given base b and altitude h A = ½ bh Given two sides a and b and included angle θ: A = ½ ab sin θ r= r a AT = area of the triangle PLANE AREAS h Circumcenter of the triangle A = ½ ab sin B + ½ cd sin D PLANE GEOMETRY a A circle is circumscribed about a triangle if it passes through the vertices of the triangle. Sum of exterior angles The sum of exterior angles β is equal to 360°. β = 360° Divide the area into two triangles d1 c Circle circumscribed about a triangle (Cicumcircle) Sum of interior angles The sum of interior angles θ of a polygon of n sides is: Sum, Σθ = (n – 2) × 180° A Triangle RADIUS OF CIRCLES POLYHEDRONS A polyhedron is a closed solid whose faces are polygons. Parabolic segment θ θ6 2 β1 θ1 β6 Polygons are classified according to the number of sides. The following are some names of polygons. 3 sides = triangle 4 sides = quadrangle or quadrilateral 5 sides = pentagon 6 sides = hexagon 7 sides = heptagon or septagon 8 sides = octagon 9 sides = nonagon Area = 2 bh 3 h Ellipse b Area = π a b Perimeter, P P = 2π a2 + b2 2 a b b PRISM a A prism is a polyhedron whose bases are equal polygons in parallel planes and whose sides are parallelograms. Prisms are classified according to their bases. Thus, a hexagonal prism is one whose base is a Prepared by: RTFVerterra Plane and Solid Geometry Formulas hexagon, and a regular hexagonal prism has a base of a regular hexagon. The axis of a prism is the line joining the centroids of the bases. A right prism is one whose axis is perpendicular to the base. The height “h” of a prism is the distance between the bases. Like prisms, cylinders are classified according to their bases. Fixed straight line ELLIPSOID Z Azone = 2πrh 2 πh (3r − h) Volume = 3 Directrix b a a Spherical segment of two bases h X c r h As = 2πrh Ab Volume = Volume = Ab × h 2 b a Volume = Ab × h = abc Lateral area, AL = 2(ac + bc) Total surface area, AS = 2(ab + bc + ac) 2 Space diagonal, d2 = 2 a +b +c Cube (Regular hexahedron) 3 Volume = Ab × h = a 2 Lateral area, AL = 4a Total surface area 2 AS = 6a Face diagonal a d1 = a 2 Space diagonal d=a 3 2 2 AR = area of the right section n = number of sides A×h b 3 h4 L 1 A2 b C c A +A+ 1 3 A1A 2 2 r h C r SOLID OF REVOLUTION R 2 h + (R − r) ( Volume = πh R 2 + r 2 + Rr 3 Lateral area = π (R + r) L Axis of rotation 2 cg ) SPHERE First proposition of Pappus h A1 A 2 € CYLINDERS A cylinder is the surface generated by a straight line intersecting and moving along a closed plane curve, the directrix, while remaining parallel to a fixed straight line that is not on or parallel to the plane of the directrix. r The surface area generated by a surface of revolution equals the product of the length of the generating arc and the distance traveled by its centroid. As = L × 2 π R Second proposition of Pappus Spherical segment of one base h r r h r 6 [A 1+ 4A +mA ] 2 The prismoidal rule gives precise values of volume for regular solid such as pyramids, cones, frustums of pyramids or cones, spheres, and prismoids. SIMILAR SOLIDS Two solids are similar if any two corresponding sides or planes are proportional. All spheres, cubes are similar. x2 x1 x2 x1 x2 r x2 x1 R 4 Volume = πr 3 3 2 Surface area, As = 4πr L x1 A1 A1 + A 2 + Volume = r = radius of sphere E = spherical excess of the polygon E = sum of angles – (n – 2)180° 3 πr E Volume = 540° L h L/2 L/2 B A € r L = slant height = A2 L D A2 Am A1 πr E 180° E = sum of angles – (n – 2)180° Frustum of right circular cone R = lower base radius r = upper base radius; h = altitude A frustum of a pyramid is the volume included between the base and a cutting plane parallel to the base. A1 = lower base area A2 = upper base area h = altitude PRISMOIDAL RULE 2 h Frustum of pyramid 3 r Spherical pyramid h Ab × h 1 Volume = πr2h 2 3 /2 ϒ 3⁄ 4πr ′ r2 r ∞ 2 +h − AL = 2 €∞ 3h2 ′ 4 € ≤ ƒ Area = 3 Volume = r d A1 Ab h B D A1 = lower base area A2 = upper base area h = altitude h PARABOLOID OF REVOLUTION 3 A Frustum of a cone Volume = πb2 1+ e ln e 1− e Vwedge = πr θ 270° a Ab = area of the base h = altitude, perpendicular distance from the vertex to the base πa2b 2 L = slant height = r + h 1 1 Volume = Ab × h = π r2 h 3 3 Lateral area, AL = π r L Similar to prisms, pyramids are classified according to their bases. Vertex πr 2θ 90° Alune = A spherical polygon is a polygon on the surface of a sphere whose sides are arcs of great circles. n = number of sides; r = radius of sphere E = spherical excess h 2 A pyramid is a polyhedron with a polygonal base and triangular faces that meet at a common point called the vertex. As = 2πa + 2 r h2 4 Spherical polygons h3 Σh Volume = AR n PYRAMIDS Volume = 1 Volume = Wedge 4πr A lune = θ 360° 4 3 πr Vwedge 3 = 360° θ Right circular cone r = base radius h = altitude AR θ 2 Ab Volume = h1 r 3 Directrix Truncated prism 2 Oblate spheroid θ h a 2 Prolate spheroid is formed by revolving the ellipse about its minor (Z) axis. Thus from the figure above, c = a, then, Lune a Prolate spheroid is formed by revolving the ellipse about its major (X) axis. Thus from the figure above, c = b, then, 4 Volume = πab2 3 arcsine As = 2πb2 + 2πab e e=a −b /a Spherical lune and wedge r Generator d1 r 1 Volume = A r = 2 πr 2h zone 3 3 Similar to pyramids, cones are classified according to their bases. Vertex Ab = base area h = altitude d2 Prolate spheroid r A cone is the surface generated by a straight line, the generator, passing through a fixed point, the vertex, and moving along a fixed curve, the directrix. πabc 3 b r CONE 2 2 Face diagonal, d1 = a + c 2 h h Lateral area, AL AL = Base perimeter × h AL = 2 π r h c d2 2 r Volume = Ab × h = π r h d1 2 (3a + 3b + h ) Spherical cone or spherical sector Right circular cylinder 4 Volume = πh 6 Volume = Ab × h Rectangular parallelepiped Y h Ab The volume area generated by a solid of revolution equals the product of the generating area and the distance traveled by its centroid. Volume = A × 2 π R For all similar solids: As1 As2 = 1 x 2 x2 € V and 1 = V2 3 1 x x2 € Where As is the surface, total area, or any corresponding area. The dimension x may be the height, base diameter, diagonal, or any corresponding dimension.