Mensuration of Plane Figures:
SQUARE:
x
Area : A = x
x
2
Perimeter : P = 4 x
RECTANGLE:
Area : A = xy
x
y
Perimeter : P = 2 x + 2 y
Mensuration of Plane Figures:
RIGHT TRIANGLES:
h
1
Area : A= bh
2
Pythagorean Theorem
c =b +h
2
b
2
2
Perimeter
P =b+h+c
Mensuration of Plane Figures:
OBLIQUE TRIANGLES:
a
h
b
c
1
Area = base * altitude
2
1
A = bh
2
A = s ( s − a)( s − b)( s − c)
( a + b + c)
where s =
2
Area of Triangles
x
Area :
θ
1
A = xy sin 
2
y
Area :
θ
y
β
1 2 sin  sin 
A= y
2 sin ( +  )
Mensuration of Plane Figures:
PARALLELOGRAM:
h
b
TRAPEZOID:
a
h
b
Area = base * altitude
A = bh
1
Area = (  bases ) * altitude
2
1
Area = ( a + b ) h
2
Mensuration of Plane Figures:
CIRCLE:
D
Circumference = 2 *(radius) =  *(diameter )
C = 2 R =  D
Area =  *(radius) =
2
A = *R =
2

4
D2

4
(diameter ) 2
Mensuration of Plane Figures:
SECTOR OF A CIRCLE:
Length of arc, L A
R
LA
LA =  R

1
Area = (radius ) * length of arc
2
1
A = LA ( R )
2
1 2
A= R 
2

D
Mensuration of Plane Figures:
SEGMENT OF A CIRCLE:
Areasegment = Areasec tor − Areatriangle
LA
θ
R
R
1
1 2
A = RLA − R sin 
2
2
1
1 2
A = R ( R ) − R sin 
2
2
R2
A=
( − sin  )
2
Mensuration of Plane Figures:
ELLIPSE:
b
a
Area =  ab
PARABOLIC SEGMENT:
h
w
2
Area = wh
3
Triangle and Circle (Inscribed Triangle)
Theorem: The inscribed angle is onehalf the central angle of an arc
a
R
B
b/2
b
c
b/2
abc
4R
AT = area of triangle
AT =
a
R
c
1
ac sin A
2
b
sinA =
2R
1  b  abc
AT = ac 
=
2  2R  4R
AT =
B
b
R = radius of circunscribed circle
a,b,and c are sides of triangle
Triangle Circumscribed About a circle
a
B
c
b
r
1
1
1
AT = ra + rb + rc
2
2
2
 a+b+c 
AT = r 

2


AT = rS
a+b+c
where S =
2
a
b
R
c
Triangle and escribed circle
AT =
1
R (a + b − c)
2
Tangent and Secant Lines
s t
=
t p
t
s
sp = t 2
p
p
x
y
Bisecting Chords
s
xy = ps
Tangent and Secant Lines
2 A = 180 − C
B
A
C
A
1
A = 90 − C
2
A + B = 90
1
90 − C + B = 90
2
1
B= C
2
Cirles, segments,
and sector
◼
Find the area of the circle
that can be inscribed in a
circular sector having a
central angle of 700 and a
radius of 10 cm.
A piece of land is bounded by an
arc of a circular lake of radius
10m and two lines AB and AC
which are tangent to the lake. If
the length of AB is 30 m, find the
area of the lot.
C
700
A
B
◼
Find the area of the circle
that can be inscribed in a
circular sector having a
central angle of 700 and a
radius of 10 cm.
A piece of land is bounded by an
arc of a circular lake of radius
10m and two lines AB and AC
which are tangent to the lake. If
the length of AB is 30 m, find the
area of the lot.
C
700
A
B
In the figure shown, the
length of AB=20 cm, the
length of AC = 18cm. If
the radius of the circle is
12 cm, find the angle
formed between AB and
AC.
The two chords AB and CD
intersect as shown. Find the area
of the smallest area. Knowing
that Ab is the diameter of the
circle. Distances are in meters
C
C
6
A
A
4
B
8
D
B
Solution
350
10-r
r
sin 35 =
10 − r
10sin 35 − r sin 35 = r
r
r
10sin 35
r=
(1 + sin 35)
r = 3.645cm
AO =  r 2
AO =  (3.645) 2
AO = 41.74cm 2
A piece of land is bounded by an arc of a circular lake of radius
10m and two lines AB and AC which are tangent to the lake. If the
length of AB is 30 m, find the area of the lot.
C
A
B
The two chords AB and CD intersect as shown. Find
the area of the smallest area. Knowing that Ab is the
diameter of the circle. Distances are in meters
C
6
A
4
8
D
B
Problem
◼
Given a triangle ABC, whose sides are AB
= 30 m; BC= 36 cm; and AC = 48 m. The
perpendicular bisectors of the sides
intersect at point P. How far is point P from
side BC?
Solution
30
Area of triangle
30 + 36 + 48
S=
2
S = 57
B
A
36
18
48
P
s − 30 = 27
s − 36 = 21
R
C
s − 48 = 9
A = s ( s − 30)( s − 36)( s − 48)
A = 57(27)(21)(9) = 539.32 m 2
Solution
Radius of circumscribed circle
B
A
30
36
48
P
D
18
R
C
abc
A=
4R
30(36)(48)
R=
= 24m
4(539.32)
PD = R 2 − 182
PD = 242 − 182 = 15.87m
Problem
◼
Two circles of radii 20 cm and 30 cm,
respectively intersect as shown. The angle
between the radii drawn to the point of
intersection is 860. Find the distance
between the centers of the circles.
Solution
20
860
30
x
x = 20 + 30 − 2(20)(30) cos86
2
2
2
x = 34.875cm
Problem
◼
A regular octagon is inscribed in a circle.
What is the radius of the circle if the length
of each side of the octagon is 12.6 inches?
Solution
θ
S
sin  =
2R
 360  12.6
sin 
=
 16  2 R
R = 16.46 in
Problem
◼
AB is a diameter of a circle. BC is a chord
10 cm long. CD is another chord. Angle
BDC=180. What is the area of a circle in sq
cm?
Solution 4 Area of circle
C
C
D
A
180
D
10
B
B
A
180
5
180
R
B
2
5
A=R
sin18 =
R
2
2
A
=

(16.18)
=
822.45
cm
R = 16.18cm
Problem
◼
Find the radius of the circumscribed circle of the triangle
whose given parts are A=560, B=240, and a=67 m.
B
c
A
a
b
C
abc 1
A=
= bc sin A
4R 2
a
67
R=
=
2sin A 2sin 56
R = 40.41m
Problem
◼ Find
the length of the median to the
longest side of the triangle whose
sides are 40, 50, and 70 cm.
Solution
a=40
M
b=50
c=70
1
2
2
2
M=
2 a + 2b − c
2
1
M=
2(40) 2 + 2(50) 2 − (70) 2
2
M = 28.72 cm
Problem
◼
The area of a rectangular lot is increased by
25 m2 if it is longer by 10 m and narrower by
2.5 m, but when it is shorter by 10 m and wider
by 5 m its area is reduced by 50 m2. Compute
the original area of the lot.
solution
A = LW  (1)
A + 25 = ( L + 10)(W − 2.5)
LW + 25 = LW − 2.5L + 10W − 25
10W − 2.5L = 50  (2)
A − 50 = ( L − 10)(W + 5)  (3)
LW − 50 = LW + 5L − 10W − 50
5L = 10W
L = 2W  (3)
L
W
L
Solution
W
10W − 2.5L = 50  (2)
L = 2W  (3)
10W − 2.5(2W ) = 50
W = 10;
L = 20
A = 20(10) = 200 m
2
The longer diagonal of the parallelogram has length 80.5 m and forms
the angels shown in the figure. Solve for X and Y
80.5
35012’
18024’
X
Y
35012’
80.5
18024’
Y
θ
 = 180 − 18024 '− 35012 '
 = 126024 '
X
X
Y
80.5
=
=
sin18024 ' sin 35012 ' sin126024 '
X = 31.57m
Y = 57.65m
Find the perimeter of the quadrilateral shown
75.8m
75.8m
430
720
92.6m
 = 180 − 72 − 43 = 650
x
y
16.8
=
=
sin 72 sin 43 sin 65
x = 17.63; y = 12.64
P = 92.6 + 17.63 + 75.8 + 12.64
P = 198.67m
θ
720
430
16.8m
Find the area of the
pentagon inscribed in a
circle of radius 10 cm
10
S
θ
10
360
=
= 720
5
1

2
Ap = ( 5 )  (10) sin  
2

1

Ap = ( 5 )  (10) 2 sin 72  = 237.76 sq.cm.
2

The width of a river was determined by establishing a base line AB of
length 200m along the right bank of the river. A point C on the opposite
bank is then sighted from the end points of the baseline. The following
data were obtained. Calculate the width of the river.
C
710 24’
A
w
200m
540 30’
B
0
0
1
1
2 sin(71 24 ')sin(54 30')
AT = (200)h = ( 200 )
2
2
sin(71024 '+ 54030')
h = 190.51 m
C
Problem 9
Another solution
h
tan 71 24 ' =
x
h
0
tan 54 30' =
y
x + y = 200
h
710 24’
0
A
x
h
h
+
= 200
0
0
tan 71 24 ' tan 54 30'
h = 190.51m
200m
540 30’
y
B
RLF
VOLUMES
Volume calculated by the formula
V = Area of Base x altitude
V = Ah
A
A
A
h
Lateral Area
L. A. = Perimeter x altitude
L. A. = Ph
Cone and Pyramid
1
1
V =  Area of Base x altitude  = Ah
3
3
h
h
L
L
A
A
Lateral Area
1
1
L. A. = Perimeter x slant height = PL
2
2
Frustum of Cone and Pyramid
b
h
V =  B + b + Bb 
6
Frustum of cone
V=
h
6
h
L
B
 R + r + Rr 
Lateral Area
1
L. A. = ( P1 + P2 ) L
2
L is slant height
b
h
L
B
SPHERE
Formulas:
The area of the surface of a sphere
is equal to the area of four of its
great circles.
S = 4 r
2
The volume of a sphere is equal to 4π/3
times the cube of its radius
4 3
V = r
3
SPHERICAL ZONE
Formulas:
The area of the zone is equal to the product of its altitude
and the circumference of a great circle of the sphere.
Z = 2 Rh
h
R
SPHERICAL SEGMENT: Area and Volume
Area of Zone; Z = 2 Rh
Volume segment of two bases
1
V =  h ( 3a 2 + 3b 2 + h 2 )
6
Volume segment of one base
1
V =  h 2 ( 3R − h )
3
h
a
b
h
R
SPHERICAL SECTOR
❑ a spherical sector
having only one conical
surface is called a
spherical cone.
The volume of a
spherical sector is equal
to one-third of the
product of the area of the
zone which forms the
base, and the radius of
the sphere
1
V = ZR
3
Prismoidal Formula-For general
Prismatoid
Ay = ay 3 + by 2 + cy + d
A1
Am
L/2
L
y
A2
Volume
L
V =  A1 + 4 Am + A2 
6
A1andA 2 are end areas
A m = area at mid-height
Problem:
A spherical ball of
radius 3 in. is placed
into a conical vessel
of depth 8 in. and
radius of base 6 in.
What is the area of
the portion of the
sphere which lies
above the circle of
contact with the cone?
h
t 6
=
3 10
t = 1.8
6
h
t
3
8
h = 3 + t = 3 + 1.8
h = 4.8
A = 2 Rh
A = 2 ( 3)( 4.8 )
A = 90.48 sq.in
Problem
◼
Two balls, one 20 cm in diameter and the other 12 cm in
diameter, are placed in a cylindrical container 26 cm in
diameter. Find the volume of water poured in the container
to just submerged the two balls.
26
Solution
x = 26 − 10 − 6 = 10
6
z = 16 − 10 = 12.49
2
z
16
2
10
h = 10 + 12.49 + 6 = 28.49
6
V = Vc − (Vs1 + Vs2 )
V=
 (26) 2 (28.49)  4
4
V = 10032.6 cm 3
x
10
4
3
−   (6) +  (10) 
3
3

3
Two pipes of the same diameter intersect
each other at right angles. If the radius of
each pipe is 2m, find the common volume of
intersection.
Prismatoid
2m
L
VT =  A1 + 4 Am + A2 
6
4
VT = 0 + 4(4)(4) + 0
6
VT = 42.67
Voct
42.67
=
= 5.33 cu.m.
8
Problem
◼
The upper base of a frustum of a regular
triangular pyramid is an equilateral triangle
with and edge of 3m. The volume of the
frustum is 84.44 cu.m. and its altitude is 5
m. What is the lower base edge in meter?
Solution
3m
3m
3m
x
x
x
32 sin 60
A1 =
= 3.897
2
2
x
sin 60
h=5m
A2 =
= 0.433x 2
2
h
V=
A1 + A2 + A1 A2
3
5
84.44 = 3.897 + A2 + 3.897 A2
3
A2 = 35.08 = 0.433 x 2
(
)
(
x = 9m
)
SPHERICAL ZONE
Find the area illuminated by a candle h feet from the
surface of the ball r feet in radius. How much surface is
illuminated when a candle is 10 ft away from a ball 5 ft in
radius?
r
h
r
SPHERICAL SEGMENT
The center of each of two spheres of radius R lies in the
surface of the other sphere. Find the volume common to
the two spheres
Theorems of Pappus
Volume
V =  Ad
Volume
h
V =  Ad
h
R
R
2
 1  R   R h
V = 2  Rh   =
3
 2  3 
x2 y 2
The ellipse 2 + 2 = 1 is revolved
a
b
about the line x + 2 y - 2a = 0.
Determine the volume of the solid of
a
revolution.
b
V = A(2 )d
d =
(1)(0) + (2)(0) − 2a
12 + 22
2a 2 5
=
=
a
5
5
2 5  4
2 2
V =  ab ( 2 ) 
a =
5 a b
 5  5