Plane and Solid Geometry Formulas Prepared by: RTFVerterra

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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.
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