advertisement

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. (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= A = ½ ab sin B + ½ cd sin D Parallelogram D a Given diagonals d1 and d2 and included angle θ: A = ½ d1 × d2 × sin θ A = ab sin A C Rhombus D d1 A b d2 a 90° B Given base b and altitude h A = ½ bh Given three sides a, b, and c: (Hero’s Formula) Given three angles A, B, and C and one side a: a 2 sin B sin C A= 2 sin A Trapezoid h a+b h A= 2 b Cyclic Quadrilateral b d1 d2 Area = D a d A d a Perimeter, P = 4a a Diagonal, d = a 2 General quadrilateral b C B Arc C = r × θradians = πrθ 180° Area = ½ r2 θradians = πr 2 θ 360° d Given diagonals d1 and d2 and included angle θ: A = ½ d1 × d2 × sin θ θ r θ5 β5 θ2 θ6 θ1 β1 β6 Polygons are classified according to the number of sides. The following are some names of polygons. 3 sides 4 sides 5 sides 6 sides 7 sides 8 sides 9 sides = = = = = = = triangle quadrangle or quadrilateral pentagon hexagon heptagon or septagon octagon nonagon r θ θr = angle in radians b AT AT AT ra = ; rc = ; rb = s −b s−a s−c Circle circumscribed about a quadrilateral A circle is circumscribed about a quadrilateral if it passes through the vertices of the quadrilateral. b r c a d (ab + cd)(ac + bd)(ad + bc ) 4 A quad ( s − a)(s − b)(s − c )(s − d) 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. 2 bh 3 A quad s c d ; s = ½(a + b + c + d) abcd SOLID GEOMETRY α = 360 - θ r θ r POLYHEDRONS A polyhedron is a closed solid whose faces are polygons. h Ellipse b Area = π a b Perimeter, P P = 2π c a r Parabolic segment Area = 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. Aquad = O Area = Asector + Atriangle Area = ½ r2 αr + ½ r2 sin θ Area = ½ r2 (αr + sin θ) C b Circles escribed about a triangle (Excircles) r= C Area = Asector – Atriangle Area = ½ r2 θr – ½ r2 sin θ Area = ½ r2 (θr – sin θ) r r A O Segment of a circle β2 D C r a r 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 d1 d2 r Note: 1 radian is the angle θ such that C = r. θ3 c s = ½(a + b + c + d) POLYGONS c θ AT s s = ½(a + b + c) r= Aquad = Area = ½ C × r Area, A = a2 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 r= Sector of a circle Diagonal, d = a 2 + b 2 A Perimeter, P = n × x n−2 × 180° Interior angle = n Exterior angle = 360° / n Circumference = 2π r = πD π 2 Area, A = π r2 = D 4 ( s − a)( s − b)( s − c )( s − d) “For any cyclic quadrilateral, the product of the diagonals equals the sum of the products of the opposite sides” d1 × d2 = ac + bd Square abc 4A T ra Circle Ptolemy’s theorem Perimeter, P = 2(a + b) c b ra Area, A = ½ R2 sin θ × n = ½ x r × n c a+b+c+d s= 2 b r= r a ra θ = 360° / n C B ∠A + ∠C = 180° ∠B + ∠D = 180° Rectangle a Circumcenter of the triangle 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 a A cyclic quadrilateral is a quadrilateral whose vertices lie on the circumference of a circle. x x A = a2 sin A 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 A circle is circumscribed about a triangle if it passes through the vertices of the triangle. x Apothem x Given side a and one angle A: 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. a R R θ θ θ r θ θ A = ½ d1 × d2 s( s − a)(s − b)(s − c ) Area, A = ab 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 x A a Regular polygons Given diagonals d1 and d2: Given two sides a and b and included angle θ: A = ½ ab sin θ s= b Given two sides a and b and one angle A: θ A= d2 θ A B C Circle circumscribed about a triangle (Cicumcircle) 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 d1 PLANE AREAS h RADIUS OF CIRCLES AT = area of the triangle PLANE GEOMETRY a decagon undecagon dodecagon quindecagon hexadecagon Sum of exterior angles The sum of exterior angles β is equal to 360°. ∑β = 360° Divide the area into two triangles B c = = = = = Sum of interior angles The sum of interior angles θ of a polygon of n sides is: Sum, Σθ = (n – 2) × 180° Given four sides a, b, c, d, and two opposite angles B and D: Part of: Plane and Solid Geometry by RTFVerterra © October 2003 Triangle 10 sides 11 sides 12 sides 15 sides 16 sides Given four sides a, b, c, d, and sum of two opposite angles: a2 + b2 2 PRISM a b b 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 Directrix 2 πh (3r − h) 3 Volume = b a a Spherical segment of two bases h r As = 2πrh Ab Volume = Ab × h c b a a2 + b2 + c 2 Cube (Regular hexahedron) Volume = Ab × h = a3 Lateral area, AL = 4a2 Total surface area AS = 6a2 Face diagonal a d1 = a 2 Space diagonal d2 = a 3 r Similar to pyramids, cones are classified according to their bases. Vertex Ab = base area h = altitude d1 Directrix a a h Ab Truncated prism 1 Ab × h 3 Volume = AR = area of the right section n = number of sides AR Volume = AR 2 r θ Similar to prisms, pyramids are classified according to their bases. Vertex Ab = area of the base h = altitude, perpendicular distance from the vertex to the base R = lower base radius r = upper base radius; 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. b C c D C r πr 3E Volume = 540° Axis of rotation h2 + (R − r ) 2 ) cg First proposition of Pappus 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 h r L L [A 1 + 4A m + A 2 ] 6 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 x2 x1 x1 R 4 3 πr 3 Surface area, As = 4πr2 L/2 L/2 x2 x1 x2 x1 SPHERE Volume = A2 Two solids are similar if any two corresponding sides or planes are proportional. All spheres, cubes are similar. SOLID OF REVOLUTION ( Am A1 Volume = r = radius of sphere E = spherical excess of the polygon E = sum of angles – (n – 2)180° L πh 2 2 R + r + Rr 3 Lateral area = π (R + r) L r 3/2 3⎤ ⎡ ⎞ 4πr ⎢⎛⎜ r 2 ⎛r⎞ 2⎟ h + − ⎜⎜ ⎟⎟ ⎥ 2 ⎢⎜ 4 ⎟ 2 3h ⎝ ⎠ ⎥⎥ ⎠ ⎣⎢⎝ ⎦ PRISMOIDAL RULE B r A1 CYLINDERS r A R h⎛ ⎜ A 1 + A 2 + A 1A 2 ⎞⎟ ⎠ 3⎝ 1 2 πr h 2 Volume = Spherical pyramid h = altitude h h r πr 2E 180° E = sum of angles – (n – 2)180° r A2 h Area = h Frustum of right circular cone Volume = πr 3 θ 270° d A2 L = slant height = πb 2 1 + e ln 1− e e PARABOLOID OF REVOLUTION B A h Volume = ⎛⎜ A 1 + A 2 + A 1A 2 ⎞⎟ ⎠ 3⎝ 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 πr 2 θ 90° Vwedge = D Frustum of pyramid As = 2πa2 + Alune = a A1 1 Ab × h 3 4 2 πa b 3 2 A1 = lower base area A2 = upper base area h = altitude Ab Volume = AL = Frustum of a cone h θ Wedge A lune 4πr 2 = θ 360° 4 3 πr Vwedge = 3 θ 360° Volume = A pyramid is a polyhedron with a polygonal base and triangular faces that meet at a common point called the vertex. Volume = r L = slant height = r + h 1 1 2 Ab × h = πr h 3 3 Lateral area, AL = π r L Σh n PYRAMIDS Volume = Prolate spheroid is formed by revolving the ellipse about its minor (Z) axis. Thus from the figure above, c = a, then, r h2 Oblate spheroid 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 L h h3 h1 a2 − b2 / a Spherical polygons Right circular cone r = base radius h = altitude h4 Prolate spheroid is formed by revolving the ellipse about its major (X) axis. Thus from the figure above, c = b, then, 4 Volume = πab 2 3 arcsin e As = 2πb2 + 2πab e e= Spherical lune and wedge Lune Generator d2 r 1 2 A zone r = πr 2 h 3 3 Volume = 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. a2 + c 2 Space diagonal, d2 = r CONE Volume = Ab × h = abc Lateral area, AL = 2(ac + bc) Total surface area, AS = 2(ab + bc + ac) Face diagonal, d1 = h h Lateral area, AL AL = Base perimeter × h AL = 2 π r h 4 πabc 3 Prolate spheroid r Volume = Ab × h = π r2 h d1 b Spherical cone or spherical sector Right circular cylinder Rectangular parallelepiped Volume = πh 2 (3a + 3b 2 + h2 ) Volume = 6 Volume = Ab × h d2 Y h Ab h X c r 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 x2 For all similar solids: ⎛x As1 = ⎜⎜ 1 As 2 ⎝ x2 ⎞ ⎟ ⎟ ⎠ 2 and V1 V2 ⎛x = ⎜⎜ 1 ⎝ x2 ⎞ ⎟ ⎟ ⎠ 3 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.