Prepared by: RTFVerterra
Plane and Solid Geometry Formulas
ASIAN
DEVELOPMENT
FOUNDATION
COLLEGE
Given four sides a, b, c, d, and sum of
two opposite angles:
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 
A=
Tacloban City
s=
a b c d
2
=
A C
B D
or =
2
2
A = ½ ab sin B + ½ cd sin D

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 = ab sin A
C
A
B
a
Given three angles A, B, and C and one
side a:
a 2 sinB sinC
A=
2 sin A
The area under this condition can also be
solved by finding one side using sine law and
h
b
c
d1
d2
D
Area =
d
A
A + C = 180°
B + D = 180°
Area = ½ C r
POLYGONS
d
Area, A = a2
Perimeter, P = 4a
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.
4
a
a
Diagonal, d = a 2
General quadrilateral
C
b

d1
2
d2
d
Given diagonals d1 and d2 and included
angle :
A = ½ d1 d2 sin 
5
3
c
D
A
4
3
B
2
1
1
5
=
=
=
=
=
=
=
Circle circumscribed about a quadrilateral
A circle is
circumscribed
about a
quadrilateral if it
passes through
the vertices of
the quadrilateral.
b
d
(ab cd)(ac bd)(ad bc)
r
180
C

2
r 
360
r


r
triangle
quadrangle or quadrilateral
pentagon
hexagon
heptagon or septagon
octagon
nonagon
C
r
2
3

r
a
Aquad
s
c
d
;
s = ½(a + b + c + d)
abcd
SOLID GEOMETRY
= 360 - 
r  r
POLYHEDRONS
A polyhedron is a closed solid whose faces are
polygons.
bh
h
Ellipse
b
Area = a b
Perimeter, P
P = 2
r=
Aquad =
O
Area = Asector + Atriangle
Area = ½ r2 r + ½ r2 sin 
Area = ½ r2 (r + sin )
Area =
A circle is
inscribed in a
quadrilateral
if it is tangent
to the three
sides of the
quadrilateral.
O
r = angle in radians
6
(s a)(s b)(s c)(s d)
Circle incribed in a quadrilateral
b
Area = Asector – Atriangle
Area = ½ r2 r – ½ r2 sin 
Area = ½ r2 (r – sin )
r
6
r c
a
D
Segment of a circle
Parabolic segment
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
b
ra = A T ; rc = A T ; rb = A T
s a
s c
s b
s = ½(a + b + c + d)
r
Note: 1 radian is the angle  such that C = r.
Square
c
a
4Aquad
Sector of a circle
Area = ½ r2 radians =
C
b
Circles escribed about a triangle
(Excircles)
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 =
Arc C = r radians =
r
r
A
r=
Area, A = r2 =  D2
4
a b c d
2
“For any cyclic quadrilateral, the product of the
diagonals equals the sum of the products of the
opposite sides”
d1 d2 = ac + bd
a2 b2
Perimeter, P = n x
n 2
Interior angle = n 180°
Circumference = 2r = D
(s a)(s b)(s c)(s d)
a
r
ra
Circle
Ptolemy’s theorem
Perimeter, P = 2(a + b)
c
Exterior angle = 360° / n
a
a circle.
AT
s
s = ½(a + b + c)
ra
= 360° / n
Area, A = ½ R2 sin n = ½ x r n
C
b
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
x
x = side
= 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
Cyclic Quadrilateral
A cyclic
quadrilateral is a
B
quadrilateral
whose vertices
lie on the
circumference of
s=
b
c
ra
x
a b
A=
h
2
Rectangle
a
abc
4A T
circle
Apothem
x
A = a2 sin A
Trapezoid
apply the formula for two sides and included
angle.
d
r=
r
a
b
 
r
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.
Circumcenter
of the triangle
Inscribed
R
R
  
A = ½ d1 d2
Given two sides a and b and included
angle :
A = ½ ab sin 
a
x
x
A
a
Given diagonals d1 and d2:
A = ½ bh
Diagonal, d =
d2
90°
Given base b and altitude h
Area, A = ab
D
d1
Given three sides a, b, and c: (Hero’s
Formula)
A = s(s a)(s b)(s c)
A circle is circumscribed about a triangle if it
passes through the vertices of the triangle.
r=
Regular polygons
Given two sides a and b and one angle A:
Rhombus

s=
b
D
a
Given diagonals d1 and d2 and included
angle :
A = ½ d1 d2 sin 
b

d2
A
B
C
C
d1
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
Parallelogram

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  is equal to
360°.
= 360°
Divide the area into two triangles
B
PLANE AREAS
=
=
=
=
=
Sum of interior angles
The sum of interior angles of a polygon
of n sides is:
Sum, = (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
PRISM
a2 b2
2
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 = 2rh
2
h
Volume =
(3r h)
3
Directrix
b
a
a
Spherical segment
of two bases
h
r
As = 2rh
Ab
Volume = Ab h
Volume =
Volume = Ab h
d1
b
a
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
d2
d1
a
r


Ab
1
r
Vwedge
n
Similar to prisms, pyramids are classified
according to their bases.
Vertex
Ab = area of the base
h = altit ude,
perpendicular
distance fr om
o
the vertex t
the base
A2

L
C
R
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.
Spherical segment of one base
h
r
As = L 2 R
Second proposition of Pappus
h
r
1
m
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 R

2
SIMILAR SOLIDS
Two solids are similar if any two corresponding
cubes are similar.
First proposition of Pappus
3
3 A = 4r2
Surface area,
s
r
4A  A
x2
x2
x1
x2
x1

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.
A
Axis of
rotation
cg
SPHERE
A 1A 2

L
The prismoidal rule gives precise values of
volume for regular solid such as pyramids,
cones, frustums of pyramids or cones, spheres,
and prismoids.
r
x1
r
L/2
6
x1

A2
L/2
Volume =
E = spherical excess of the polygon
E = sum of angles – (n – 2)180°
r 3E
Volume =
540
L = slant height = h2 (R r)2
h 2 2
Volume =
R r Rr
3
Lateral area = (R + r) L
Volume =
3 
Am
B
SOLID OF REVOLUTION
4
3/2

L
r
R
A1
CYLINDERS
c
D

r
PRISMOIDAL RULE
C
A
3

h
h
3
b
Spherical pyramid
r
h A A
 1
2

r
R = lower base radius
A2
r2h
2







r  
4r  r 2 2


AL =
h

2  
3h2 

4

   
r E
180
E = sum of angles – (n – 2)180°
Frustum of right circular cone
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
1
Volume =
Area =
h
A A A
1 A 2
2  1
h
A h
h
A1
Frustum of pyramid
Volume =
270
B
D
A1
h
1e
d

Volume =
ln
e
r 3 
2
Frustum of a cone
A1 = lower base area
A2 = upper base area
h = altitude
Ab
3
360
A
r 2 h2
L = slant height =
1 2
1
r h
Ab h =
Volume =
3
3
Lateral area, AL = r L
A pyramid is a polyhedron with a polygonal base
and triangular faces that meet at a common
point called the vertex.
1
Vwedge =
a
r
2
PARABOLOID OF REVOLUTION
90

4 3
3 r
h3
h1
h
r 
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
L
h
2
a b
2
4r
360
=
3
As = 2a2 +
2
=
4
Volume =
Spherical polygons
3


Wedge

Ab h
a2 b2 / a
e=
Oblate spheroid

A lune
h4
Volume =
A zone r = 2 r 2h
3
Prolate spheroid is formed by revolving the
ellipse about its minor (Z) axis. Thus from the
figure above, c = a, then,
Directrix
a
Right circular cone
r = base radius
h = altitude
PYRAMIDS
3
Spherical lune and wedge
h
AR = area of the right section
n = number of sides
Volume = AR
Prolate spheroid is formed by revolving the
ellipse about its major (X) axis. Thus from the
figure above, c = b, then,
4
2
Volume = 
ab
3
arcsine
As = 2b2 + 2ab
e
r
Lune
Generator
Volume =
h2
1
Volume =
Similar to pyramids, cones are classified
according to their bases.
Vertex
Ab = base area
h = altitude
Truncated prism
AR
b
point, the vertex, and moving along a fixed
curve, the directrix.
a2 b2 c 2
Space diagonal, d2 =
4
abc
3
Prolate spheroid
r
A cone is the surface generated by a straight
a 2 c 2
Face diagonal, d1 =
(3a 3b h )
r
CONE
Volume = Ab h = abc
Lateral area, AL = 2(ac + bc)
2
h
h
Lateral area, AL
AL = Base perimeter h
AL = 2 r h
c
2
r
Volume = Ab h = r2 h
d2
6
Volume =
2
Spherical cone or spherical sector
Right circular cylinder
Rectangular parallelepiped
h
Y
h
Ab
h
X
c
x2
For all similar solids:

2
x

=  1 
x 



As2
 2

As1

3
and
1
V
V2
 1 
= 
x 
x

 2 
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.