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.