–Matt. 6:33
CHAPTER VI
AREA
COMPUTATION
METHODS OF DETERMINING AREA
1. AREA BY TRIANGLE
2. AREA BY COORDINATES
3. AREA
BY
DMD
(DOUBLE
MERIDIAN
DISTANCE)
4. AREA
BY
DPD
(DOUBLE
PARALLEL
DISTANCE)
5. AREA BY OFFSETS FROM A STRAIGHT LINE
5.1. TRAPEZOIDAL RULE
5.2. SIMPSON’S ONE-THIRD RULE
6. AREA BY PLANIMETER
7. AREA BY GRAPHICAL METHOD
8. AREA BY COORDINATE SQUARES
1. AREA BY TRIANGLES
1.1. Known Base and Altitude
B
A = ½ bh
A
Where:
A = Area of the tract of
land in sq. m. or has.
b = base of the triangle
h = altitude of the triangle
h
b
C
1. AREA BY TRIANGLES
1.2. Two sides and included
angle measured
B
a
A = ½ ab sinα
C
α
Where:
A = Area of the tract of
land in sq. m. or has.
a, b = sides of the triangle (m.)
α = angle between sides a and b
b
A
1. AREA BY TRIANGLES
1.3. Three sides measured
B
A = 𝐬(𝐬 − 𝐚)(𝐬 − 𝐛)(𝐬 − 𝐜)
c
a
A
Where:
b
A = Area of the tract of
land in sq. m. or has.
a, b, & c = sides of the triangle (m.)
s = semi-perimeter = ½ (a + b + c)
C
Example Problems
1. A surveyor sets up a transit at P which
is located on the middle portion of a
four-sided tract of land and reads
directions and measures distances as
given below to the four corners. Find
the area of the tract of land in
hectares.
Line
Bearing
Distance, m.
PA
N 42⁰ W
420.55
PB
N 39⁰ E
535.75
PC
S 73⁰ E
497.25
PD
S 61⁰ W
595.60
Solution:
Total Area = A1 + A2 + A3 + A4
θ1 = 42⁰ + 39⁰ = 81⁰; θ2 = 180⁰ - 39⁰ - 73⁰ = 68⁰
θ3 = 73⁰ + 61⁰ = 134⁰;
θ4 = 180⁰ - 42⁰ - 61⁰ = 77⁰
A
A1
420.55 m.
595.60 m.
N
39⁰
42⁰
535.75 m.
P
A4
B
θ1
θ4
61⁰
73⁰
θ2
Line
PA
PB
PC
PD
A2
497.25 m.
Bearing
N 42⁰ W
N 39⁰ E
S 73⁰ E
S 61⁰ W
θ3
D
A3
C
A1 = ½ (420.55) (535.75) sin 81⁰ = 111,267.863 sq. m.
A2 = ½ (535.75) (497.25) sin 68⁰ = 123,501.672 sq. m.
A3 = ½ (497.25) (595.60) sin 134⁰ = 106,520.593 sq. m.
A4 = ½ (595.60) (420.55) sin 77⁰ = 122,029.902 sq. m.
Total Area = 111,267.863 + 123,501.672 + 106,520.593 + 122,029.902
Total Area = 463,320.030 sq. m. =
46.332 Has.
Distance,m.
420.55
535.75
497.25
595.60
2. AREA BY COORDINATES
Y – Axis (North – South Line)
A
X1
A’
B’
X2
X4
D’
N
B
CORNER
D
Y1
Y4
C’
Y2
X3
A
B
C
D
COORDINATES
X
Y
x1
y1
x2
y2
x3
y3
x4
y4
C
Y3
ORIGIN
A=
𝟏
𝟐
X – Axis (East – West Line)
𝟏 𝒙𝟏 𝒙𝟐 𝒙𝟑 𝒙𝟒 𝒙𝟏
Area of ABCD =
𝟐 𝒚𝟏 𝒚𝟐 𝒚𝟑 𝒚𝟒 𝒚𝟏
𝒙𝟐 𝒚𝟏 + 𝒙𝟑 𝒚𝟐 + 𝒙𝟒 𝒚𝟑 + 𝒙𝟏 𝒚𝟒 − 𝒙𝟏 𝒚𝟐 + 𝒙𝟐 𝒚𝟑 + 𝒙𝟑 𝒚𝟒 + 𝒙𝟒 𝒚𝟏
Example Problems:
1.
Given
in
the
accompanying
tabulation are the observed data for
a closed traverse representing the
boundary of a parcel of land. Adjust
the traverse using transit rule. If
coordinates of A are xA = 6,000.00,
yA = 7,000.00, determine the lot area
in hectares using coordinate method.
Traverse Data
LINE
DISTANCE, m.
AZIMUTH
AB
495.85
185° 25’
BC
847.62
226° 08’
CD
855.45
292° 21’
DE
1,020.87
347° 30’
EF
1,117.26
83° 52’
FA
660.08
124° 49’
Solution:
Line
Distance, m.
Azimuth Bearing
AB
495.85
185° 25’
N05° 25’E +493.636 + 46.807
BC
847.62
226° 08’
N46° 08’E +587.386 +611.095
CD
855.45
292° 21’
S67° 39’E -325.296
+791.187
DE 1,020.87
347° 30’
S12° 30’E -996.671
+220.957
EF 1,117.26
83° 52’
S83° 52’W -119.371 -1,110.865
FA
124° 49’
N55° 11’W +376.874
660.08
σ = 4,997.13 m.
Latitude Departure
σ=
σ=
-541.915
+ 16.558 + 17.266
2,899.234 3,322.826
Solution:
CL = + 16.558
CD = + 17.266
LEC =
(𝐶𝐿 )2 +(𝐶𝐷 )2
= (16.558)2 +(17.266)2
LEC = 23.922 m.
Tan θ =
θ=
− (𝐶𝐷 )
− (𝐶𝐿 )
Tan-1
− (17.266)
(
)
− (16.558)
= S 46⁰ 11’ 57” W
D = 4,997.13 m.
𝐿𝐸𝐶
RP =
𝐷
=
23.922
4997.13
=
𝟏
𝟐𝟎𝟎
Solution:
Correction for latitude,cl =
𝑳𝒂𝒕
493.636
CL (σ ) = +16.558 (2,899.234
)
𝑳𝒂𝒕
Correction for latitude, cl = 2.819
Correction for departure,
cd =
𝑫𝒆𝒑
𝟒𝟔.𝟖𝟎𝟕
CD (
) = +17.266 (𝟑,𝟑𝟐𝟐.𝟖𝟐𝟔)
σ 𝑫𝒆𝒑
Correction for departure, cd = 0.243
Adjusted latitude, Lat’ = Lat ± cl = +493.636 – 2.819
Adjusted latitude, Lat’ =
+490.817
Adjusted departure, Dep’ = Dep ± cl = +46.807 – 0.243
Adjusted departure, Dep’ =
+46.564
Solution:
Line Distance
Correction
Correction
Adjusted
for Latitude for Departure Latitude
Adjusted
Departure
AB
495.85
-2.819
- 0.243
+ 490.817 + 46.564
BC
847.62
-3.355
- 3.175
+ 584.031 + 607.920
CD
855.45
-1.858
- 4.111
- 327.154 + 787.076
DE
1,020.87
-5.692
- 1.148
- 1,002.363 + 219.809
EF
1,117.26
-0.682
- 5.773
- 120.053 -1,116.638
FA
660.08
-2.152
- 2.816
+374.722 - 544.731
σ = 4,997.13 -16.558
-17.266
0
0
Solution:
292° 21’
C
855.45 m.
847.62 m.
347⁰ 30’
D
N
226⁰ 08’
B
495.85 m.
1,020.87 m.
185⁰ 25’
A
660.80 m.
124⁰ 49’
E
1,117.26 m.
F
83⁰ 52’
Solution:
Line
Adjusted
Adjusted
Latitude
Departure
Sta.
X
Y
Coordinates Coordinates
AB
+ 490.817
+ 46.564
A
6,000.000
7,000.000
BC
+ 584.031
+ 607.920
B
6,046.564
7,490.817
CD
- 327.154
+ 787.076
C
6,654.484
8,074.848
DE
- 1,002.363 + 219.809
D
7,441.560
7,747.694
EF
- 120.053
-1,116.638
E
7,661.369
6,745.331
FA
+374.722
- 544.731
F
6,544.731
6,625.278
0
0
A
6,000.000
7,000.000
σ=
Solution:
Area =
𝟏
𝟐
A =
𝟏
ሾ
𝟐
A =
𝟏
ൣ൛
𝟐
𝒙 𝑨 𝒙𝑩 𝒙𝑪 𝒙𝑫 𝒙𝑬 𝒙𝑭 𝒙𝑨
𝒚𝑨 𝒚𝑩 𝒚𝑪 𝒚𝑫 𝒚𝑬 𝒚𝑭 𝒚𝑨
𝒙𝑩 𝒚𝑨 + 𝒙𝑪 𝒚𝑩 + 𝒙𝑫 𝒚𝑪 + 𝒙𝑬 𝒚𝑫 + 𝒙𝑭 𝒚𝑬 + 𝒙𝑨 𝒚𝑭 −
𝒙𝑨 𝒚𝑩 + 𝒙𝑩 𝒚𝑪 + 𝒙𝑪 𝒚𝑫 + 𝒙𝑫 𝒚𝑬 + 𝒙𝑬 𝒚𝑭 + 𝒙𝑭 𝒚𝑨 ሿ
6,046.564 7,000.000 + 6,654.484 7,490.817 +
7,441.560 8,074.848 + 7,661.369 7,747.694 +
6,544.731 6,745.331 + 6,000.000 6,625.278 ൟ −
൛ 6,000.000 7,490.817 + 6,046.564 8,074.848 +
6,654.484 7,747.694 + 7,441.560 6,745.331 +
7,661.369 6,625.278 + 6,544.731 7,000.000 ൟ൧
A = 1,712,214.233 sq. m. (1 Ha./10,000 m2)
A = 171.22 Has.
3. AREA BY DMD (Double Meridian Distance)
B
B’
F’
E’
t
s
F
I
E
J
L
G’
H’
D’
C
K
C’
A
G
H
M
D
The Meridian Distance of a
line is defined as the shortest
distance from the midpoint of
the line to the Reference
Meridian.
DMD of AB = 2 (EE’)
= EE’ + EI
= Dep. Of AB
DMD of BC = 2 (FF’)
= 2(EE’) + 2(EI) + 2(IF)
= DMDAB+ DepAB+ DepBC
DMD of CD = 2 (GG’)
= 2(FF’) + 2(JC) - 2(CK)
= DMDBC+ DepBC - DepCD
Reference Meridian (North – South Line)
3. AREA BY DMD (Double Meridian Distance)
Based on these illustrations, following are the rules in
computing the DMD for each course of a traverse:
1. The DMD of the first course is equal to the
departure of the course.
2. The DMD of any course is equal to the DMD of the
preceding course, plus the departure of the
preceding course, plus the departure of the course
itself.
3. The DMD of the last course is numerically equal to
the departure of that course, but with the opposite
sign.
Double Area = DMD (Adjusted Latitude)
AREA = ½ (σ 𝑫𝑨)
Example Problems:
1.
Given
in
the
accompanying
tabulation are the observed data for
a closed traverse representing the
boundary of a parcel of land. Adjust
the traverse using transit rule and
determine the lot area in hectares
using DMD method.
Traverse Data
LINE
DISTANCE, m.
AZIMUTH
AB
495.85
185° 25’
BC
847.62
226° 08’
CD
855.45
292° 21’
DE
1,020.87
347° 30’
EF
1,117.26
83° 52’
FA
660.08
124° 49’
Solution:
Line
Distance, m.
Azimuth Bearing
AB
495.85
185° 25’
N05° 25’E +493.636 + 46.807
BC
847.62
226° 08’
N46° 08’E +587.386 +611.095
CD
855.45
292° 21’
S67° 39’E -325.296
+791.187
DE 1,020.87
347° 30’
S12° 30’E -996.671
+220.957
EF 1,117.26
83° 52’
S83° 52’W -119.371 -1,110.865
FA
124° 49’
N55° 11’W +376.874
660.08
σ = 4,997.13 m.
Latitude Departure
σ=
σ=
-541.915
+ 16.558 + 17.266
2,899.234 3,322.826
Solution:
Correction for latitude,cl =
𝑳𝒂𝒕
493.636
CL (σ ) = +16.558 (2,899.234
)
𝑳𝒂𝒕
Correction for latitude, cl = 2.819
Correction for departure,
cd =
𝑫𝒆𝒑
𝟒𝟔.𝟖𝟎𝟕
CD (
) = +17.266 (𝟑,𝟑𝟐𝟐.𝟖𝟐𝟔)
σ 𝑫𝒆𝒑
Correction for departure, cd = 0.243
Adjusted latitude, Lat’ = Lat ± cl = +493.636 – 2.819
Adjusted latitude, Lat’ =
+490.817
Adjusted departure, Dep’ = Dep ± cl = +46.807 – 0.243
Adjusted departure, Dep’ =
+46.564
Solution:
Line Distance
Correction
Correction
Adjusted
for Latitude for Departure Latitude
Adjusted
Departure
AB
495.85
-2.819
- 0.243
+ 490.817 + 46.564
BC
847.62
-3.355
- 3.175
+ 584.031 + 607.920
CD
855.45
-1.858
- 4.111
- 327.154 + 787.076
DE
1,020.87
-5.692
- 1.148
- 1,002.363 + 219.809
EF
1,117.26
-0.682
- 5.773
- 120.053 -1,116.638
FA
660.08
-2.152
- 2.816
+374.722 - 544.731
σ = 4,997.13 -16.558
-17.266
0
0
Solution:
Line
Adjusted
Adjusted
Latitude
Departure
DMD
DA
AB
+ 490.817
+ 46.564
+
46.564
+ 22,854.403
BC
+ 584.031
+ 607.920
+ 701.048
+ 409,433.765
CD
- 327.154
+ 787.076
+ 2,096.044
- 685,729.179
DE
- 1,002.363 + 219.809
+ 3,102.929
-3,110,261.221
EF
- 120.053
-1,116.638
+ 2,206.100
- 264,848.923
FA
+374.722
- 544.731
+
+ 204,122.690
0
0
σ=
544.731
- 3,424,428.465
AREA = ½ (3,424,428.465) = 1,712,214.233 m2 (1 Ha./10,000 m2)
AREA = 171.221 Has.
Sketch:
292° 21’
C
855.45 m.
847.62 m.
347⁰ 30’
D
N
226⁰ 08’
B
495.85 m.
1,020.87 m.
185⁰ 25’
A
660.80 m.
124⁰ 49’
E
1,117.26 m.
F
83⁰ 52’
4. AREA BY DPD (Double Parallel Distance)
The Parallel Distance of a line is defined as the
shortest distance from the midpoint of the line to
the Reference Parallel or the east-west line.
Rules in computing the DPD for each course of a traverse:
1. The DPD of the first course is equal to the latitude of
the course.
2. The DPD of any course is equal to the DPD of the
preceding course, plus the latitude of the preceding
course, plus the latitude of the course itself.
3. The DPD of the last course is numerically equal to the
latitude of that course, but with the opposite sign.
Double Area = DPD (Adjusted Departure)
AREA = ½ (σ 𝑫𝑨)
Example Problem:
1.
Given
in
the
accompanying
tabulation are the observed data for
a closed traverse representing the
boundary of a parcel of land. Adjust
the traverse using transit rule and
determine the lot area in hectares
using DPD method.
Solution:
Line
Adjusted
Adjusted
Latitude
Departure
DPD
DA
AB
+ 490.817
+ 46.564
+ 490.817
+ 22,854.403
BC
+ 584.031
+ 607.920
+ 1,565.665
+ 951,799.067
CD
- 327.154
+ 787.076
+ 1,822.542
+1,434,479.067
DE
- 1,002.363 + 219.809
+
493.025
+ 108,371.332
EF
- 120.053
-1,116.638
-
629.391
+ 702,801.908
FA
+374.722
- 544.731
-
374.722
+ 204,122.690
0
0
σ=
- 3,424,428.467
AREA = ½ (3,424,428.467) = 1,712,214.234 m2 (1 Ha./10,000 m2)
AREA = 171.221 Has.
5. AREA BY OFFSET FROM A STRAIGHT LINE
5.1. TRAPEZOIDAL RULE
Figure:
5. AREA BY OFFSET FROM A STRAIGHT LINE
5.1. TRAPEZOIDAL RULE
5.1.1. For offsets at regular interval (d)
A=d
𝒉𝟏 +𝒉𝒏
(
𝟐
+ 𝒉𝟐 + 𝒉𝟑 + 𝒉𝟒 + ……….. + 𝒉𝒏−𝟏 )
Where:
A = summation of the areas of the trapezoids comprising the total area
d = common spacing between offsets
n = number of offsets
h1 = end (first) offset
hn = end (last) offset
h2, h3, etc. = intermediate offsets
hn-1 = last intermediate offset
5.1.2. For offsets at irregular interval (solve area of individual trapezoids)
2A = d1(𝒉𝟏 + 𝒉𝟐 )+ d2(𝒉𝟐 + 𝒉𝟑 )+ d3(𝒉𝟑 + 𝒉𝟒 )+...+ dn(𝒉𝒏−𝟏 + 𝒉𝒏 )
5. AREA BY OFFSET FROM A STRAIGHT LINE
5.1. SIMPSON’S ONE-THIRD RULE
Figure:
5. AREA BY OFFSET FROM A STRAIGHT LINE
5.1. SIMPSON’S ONE-THIRD RULE
5.1.1. For offsets at regular interval (d)
𝒅
A = 𝟑 (𝒉𝟏 +𝒉𝒏 ) + 𝟐(𝒉𝟑 + 𝒉𝟓 + … + 𝒉𝒏−𝟐 ) + 𝟒(𝒉𝟐 + 𝒉𝟒 + … + 𝒉𝒏−𝟏 )
Where:
A = Area of the tract bounded by the curved boundary, the base line,
and the end offsets
d = common spacing between offsets
n = number of offsets
h1 = end (first) offset
hn = end (last) offset
h3, h5, etc. = odd-numbered intermediate offsets
h2, h4, etc. = even-numbered intermediate offsets
hn-1 = last even-numbered intermediate offset
hn-2 = last odd-numbered intermediate offset
Note: The rule is only applicable when there is an odd number of offsets and if
they are equally spaced.
Example Problem:
1. A series of perpendicular offsets
were taken from a transit line to an
irregular boundary. These offsets
were taken at 3 meters apart and
were measured in the following
order; 1.3, 2.6, 4.3, 5.8, 3.9, 2.6, 4.7,
5.3, 6.5, 4.0, 1.8, and 5.4 all in
meters. Find the area included
between the transit line, the curved
boundary, and the end offsets using
trapezoidal rule and simpson’s onethird rule.
Solution:
a. Trapezoidal rule
𝟏𝟐
A = d (𝒉𝟏+𝒉
+ 𝒉𝟐 + 𝒉𝟑 + 𝒉𝟒 + 𝒉𝟓 + 𝒉𝟔 + 𝒉𝟕 + 𝒉𝟖 + 𝒉𝟗 + 𝒉𝟏𝟎 + 𝒉𝟏𝟏 )
𝟐
A = 3(𝟏.𝟑+𝟓.𝟒
+ 𝟐. 𝟔 + 𝟒. 𝟑 + 𝟓. 𝟖 + 𝟑. 𝟗 + 𝟐. 𝟔 + 𝟒. 𝟕 + 𝟓. 𝟑 + 𝟔. 𝟓 + 𝟒. 𝟎 +
𝟐
𝟏. 𝟖)
A = 134.55 sq. m.
b. Simpson’s one-third rule
A=
A=
𝒅
ሾ(𝒉𝟏 +𝒉𝟏𝟏 ) + 𝟐(𝒉𝟑 + 𝒉𝟓 + 𝒉𝟕 + 𝒉𝟗 ) + 𝟒(𝒉𝟐 + 𝒉𝟒 + 𝒉𝟔 + 𝒉𝟖
𝟑
𝒅
𝒉𝟏𝟎 )ሿ + (𝒉𝟏𝟏 +𝒉𝟏𝟐 )
𝟐
𝟑
ሾ(𝟏. 𝟑 + 𝟏. 𝟖) + 𝟐(𝟒. 𝟑 + 𝟑. 𝟗 + 𝟒. 𝟕 + 𝟔. 𝟓) + 𝟒(𝟐. 𝟔 + 𝟓. 𝟖 +
𝟑
𝟑
𝟐. 𝟔 + 𝟓. 𝟑 + 𝟒. 𝟎)ሿ + (𝟏. 𝟖 + 𝟓. 𝟒)
𝟐
A=
133.90 sq. m.
+
6. AREA BY PLANIMETER
Planimeter is a mechanical device used for determining the
area of any shape of figure plotted to a known scale
Two Types of Planimeter
1. Mechanical Planimeter
2. Electronic Planimeter
6. AREA BY PLANIMETER
By ratio and proportion between a test figure and the
plotted (specific) figure:
𝑫𝒊𝒇𝒇𝒕
𝑨𝒕
As =
=
𝑫𝒊𝒇𝒇𝒔
𝑨𝒔
𝑫𝒊𝒇𝒇𝒔
(𝑨𝒕 )
𝑫𝒊𝒇𝒇𝒕
Where:
As = Area of a specific figure
Diffs = Difference between planimeter readings for a specific figure
(final minus initial planimeter reading)
At = Computed area of the test figure
Difft = Difference between planimeter readings for the test figure
(final minus initial planimeter reading)
Example Problem:
1.
To determine the area of a
watershed portrayed on a map with
scale 1:5,000, a 3 cm. by 3 cm.
square was drawn on a piece of
paper and traced by the planimeter
with a difference in reading of 590.
When the watershed plotted on the
map was traced by the planimeter,
the difference in planimeter reading
is 1,510. determine the area of the
watershed in hectares.
Solution:
Diffs = 1,510; Difft = 590; At = 9 cm2
𝑫𝒊𝒇𝒇𝒕
𝑨𝒕
As =
As =
=
𝑫𝒊𝒇𝒇𝒔
𝑨𝒔
𝑫𝒊𝒇𝒇𝒔
(𝑨𝒕 )
𝑫𝒊𝒇𝒇𝒕
𝟏,𝟓𝟏𝟎
𝟗
𝟓𝟗𝟎
As = 23.034
𝟓𝟎 𝒎. 2
2
cm (
)
As = 57,584.746
As = 5.758 Has.
𝟏 𝒄𝒎.
m2 (
𝟏 𝒉𝒂.
)
𝟐
𝟏𝟎,𝟎𝟎𝟎 𝒎.
7. AREA BY GRAPHICAL METHOD
The given polygon is converted into a triangle equivalent in
area to the original figure.
Given below is a hexagon (ABCDEFA) whose area is to be
determined by gradually eliminating three out of the six
corners so that it will be converted to a triangle of equal area.
D
E
C
F
A
B
7. AREA BY GRAPHICAL METHOD cont’d.
Designate AB which is the longest line as the Base Line.
First step is to eliminate corner F yet still maintain an
equivalent area.
Area of triangle AEF’ is still equal to Area of triangle AEF.
D
E
F
C
h1
F’ A
B
7. AREA BY GRAPHICAL METHOD cont’d.
Next to be eliminated will be corner E and still maintaining an
equivalent area.
Area of triangle DE’F’ is still equal to Area of triangle DEF’.
D
E
h2
F
C
h1
F’ A
B
7. AREA BY GRAPHICAL METHOD cont’d.
Last to be eliminated will be corner C and still maintaining an
equivalent area.
Area of triangle BC’D is still equal to Area of triangle BCD.
Area of the original polygon is now equal to area of triangle
C’DE’.
ABCDEFA
=½(
)( )
D
E
h2
h3
F
h1
F’ A
B
C
7. AREA BY COORDINATE SQUARES
Is a graphical solution of determining a closed traverse area or
an irregularly bounded area and is suited only for rough
estimates.
It involves plotting the closed figure to scale and marking of
squares of unit area.
Problem exercises: Area by Triangles
1. In the quadrilateral ABCD the sides were
measured as follows: AB, 535.98 m.; BC,
351.60 m.; CD, 631.33 m.; and DA, 297.20 m. if
the diagonal BC measures 675.68 m.,
determine the area of the quadrilateral.
C
D
A
B
Problem exercises: Area by Coordinates
2. Given below is the tabulation of the adjusted latitudes
and departures of a closed traverse. Determine the area
in hectares by coordinate method assuming that the
origin (0.0 m., 0.0 m.) is at station A.
Line
Adjusted
Adjusted
Sta.
Latitude
Departure
AB
+ 552.45
+ 212.63
A
BC
+ 232.06
+ 396.50
B
CD
- 359.70
+ 283.15
C
DE
- 209.37
- 366.89
D
EA
- 215.44
- 525.39
E
A
X
Y
Coordinates Coordinates
Problem exercises: Area by DMD and DPD
3. For the given tabulation of the adjusted latitudes and
departures shown in the accompanying tabulation,
determine the area of the traverse using the double
meridian distance (DMD) and the double parallel
distance (DPD).
Line
AB
Adjusted Latitude
Adjusted Departure
+ 715.20
+ 1,200.53
BC
-
414.29
+
401.78
CD
- 1,735.58
-
419.88
DE
+ 200.70
- 1,606.43
EF
+ 617.59
-
456.66
FG
+ 849.91
+
202.83
GA
-
+
677.83
233.53
Problem exercises: Offset from a Straight line
4. In the accompanying sketch, it is desired to determine
the area of the sand bar by the indicated short offsets
which were measured on both sides of the baseline AB.
Using the trapezoidal rule, determine the area of the sand
bar. Also determine the same area by employing
Simpson’s one-third rule and note the discrepancy in the
area computed by the two methods.
Problem exercises: Irregularly spaced Offsets
5. Vertical measurements were taken at varying
distances along a section of a stream as
illustrated in the accompanying figure.
Determine the area (in sq. m.) of the crosssection.
Problem exercises: Irregularly spaced Offsets
6. In the accompanying sketch, determine the
area included between the baseline and the
irregular boundary.
Problem exercises: Area by Planimeter
7. A draftsman drew a 10 – cm. square on a
piece of paper and traced it with a planimeter
three times and obtained the following
readings: 1985, 3612, 5241, and 6869. Then
the planimeter was used to determine the
area of a ricefield marked off on a map. The
boundary of the ricefield was traced three
times and readings were recorded as follows:
0105, 2705, 5307, and 7911. If the scale of the
map is 1 cm. = 60 m., determine the area of
the ricefield in square meters.
Problem exercises: Graphical Method
8. Given below is a closed traverse data
representing the sides of a five-sided farm lot. Plot
the traverse using an appropriate scale and
determine the area of the lot in hectares using
graphical method. Use line 1-2 as the baseline.
Line Distance, m. Bearing
1-2
1,082.75
Due East
2-3
637.62
N 10° 35’ W
3-4
781.45
N 57° 39’ W
4-5
587.01
S 35° 52’ W
5-1
570.53
S 03° 52’ E