Land Surveying Assignment: Parcel Division Report 1. ObjectiveTo divide parcel \(ABCDE\) into
two equal areas (\(A_{1}=A_{2}\)) such that they share equal frontage on line \(AB\). 2.
Coordinates Table Point Easting (X)Northing
(Y)A10000.0010000.00B10244.0010000.00C10211.2010130.60D10193.8010258.90E9922.301
0114.203. Total Area Calculation (Shoelace Method)Using the formula
\(2A=|(x_{A}y_{B}+x_{B}y_{C}+x_{C}y_{D}+x_{D}y_{E}+x_{E}y_{A})-
(y_{A}x_{B}+y_{B}x_{C}+y_{C}x_{D}+y_{D}x_{E}+y_{E}x_{A})|\): \(\sum (x_{i}y_{i+1})=(10000\times
10000)+(10244\times 10130.6)+(10211.2\times 10258.9)+(10193.8\times
10114.2)+(9922.3\times 10000)\)\(\sum
(x_{i}y_{i+1})=100,000,000+103,777,866.4+104,755,689.68+103,102,131.96+99,223,000=\mathb
f{510,858,688.04}\)\(\sum (y_{i}x_{i+1})=(10000\times 10244)+(10000\times
10211.2)+(10130.6\times 10193.8)+(10258.9\times 9922.3)+(10114.2\times 10000)\)\(\sum
(y_{i}x_{i+1})=102,440,000+102,112,000+103,269,290.28+101,791,833.47+101,142,000=\mathbf
{510,755,123.75}\)Total Area = \(0.5\times |510,858,688.04-
510,755,123.75|=\mathbf{51,782.145}\,\mathbf{m}^{\mathbf{2}}\)Target Half Area =
\(\mathbf{25,891.07}\,\mathbf{m}^{\mathbf{2}}\) 4. Determination of Point XEqual frontage on
\(AB\) means \(X\) is the midpoint of
\(AB\): \(X_{E}=\frac{10000+10244}{2}=\mathbf{10122.00}\)\(X_{N}=\frac{10000+10000}{2}=\
mathbf{10000.00}\)5. Determination of Point YWe calculate the area of polygon \(XBCD\): Area
\(XBC\) = \(7,966.60\,m^{2}\)Area \(XCD\) = \(6,858.40\,m^{2}\)Total \(XBCD\) =
\(14,825.00\,m^{2}\)Remaining Area Required from \(XDE\) = \(25,891.07-
14,825.00=\mathbf{11,066.07}\,\mathbf{m}^{\mathbf{2}}\) Area of full triangle \(XDE\) =
\(29,955.95\,m^{2}\).Ratio \(k=\frac{11,066.07}{29,955.95}=\mathbf{0.36941}\) Final
Coordinates for Y (along line DE): \(Y_{E}=10193.80+0.36941(9922.30-
10193.80)=\mathbf{10093.50}\)\(Y_{N}=10258.90+0.36941(10114.20-
10258.90)=\mathbf{10205.45}\)Final Result: Point X: (10122.00, 10000.00)Point Y: (10093.50,
10205.45)
two equal areas (\(A_{1}=A_{2}\)) such that they share equal frontage on line \(AB\). 2.
Coordinates Table Point Easting (X)Northing
(Y)A10000.0010000.00B10244.0010000.00C10211.2010130.60D10193.8010258.90E9922.301
0114.203. Total Area Calculation (Shoelace Method)Using the formula
\(2A=|(x_{A}y_{B}+x_{B}y_{C}+x_{C}y_{D}+x_{D}y_{E}+x_{E}y_{A})-
(y_{A}x_{B}+y_{B}x_{C}+y_{C}x_{D}+y_{D}x_{E}+y_{E}x_{A})|\): \(\sum (x_{i}y_{i+1})=(10000\times
10000)+(10244\times 10130.6)+(10211.2\times 10258.9)+(10193.8\times
10114.2)+(9922.3\times 10000)\)\(\sum
(x_{i}y_{i+1})=100,000,000+103,777,866.4+104,755,689.68+103,102,131.96+99,223,000=\mathb
f{510,858,688.04}\)\(\sum (y_{i}x_{i+1})=(10000\times 10244)+(10000\times
10211.2)+(10130.6\times 10193.8)+(10258.9\times 9922.3)+(10114.2\times 10000)\)\(\sum
(y_{i}x_{i+1})=102,440,000+102,112,000+103,269,290.28+101,791,833.47+101,142,000=\mathbf
{510,755,123.75}\)Total Area = \(0.5\times |510,858,688.04-
510,755,123.75|=\mathbf{51,782.145}\,\mathbf{m}^{\mathbf{2}}\)Target Half Area =
\(\mathbf{25,891.07}\,\mathbf{m}^{\mathbf{2}}\) 4. Determination of Point XEqual frontage on
\(AB\) means \(X\) is the midpoint of
\(AB\): \(X_{E}=\frac{10000+10244}{2}=\mathbf{10122.00}\)\(X_{N}=\frac{10000+10000}{2}=\
mathbf{10000.00}\)5. Determination of Point YWe calculate the area of polygon \(XBCD\): Area
\(XBC\) = \(7,966.60\,m^{2}\)Area \(XCD\) = \(6,858.40\,m^{2}\)Total \(XBCD\) =
\(14,825.00\,m^{2}\)Remaining Area Required from \(XDE\) = \(25,891.07-
14,825.00=\mathbf{11,066.07}\,\mathbf{m}^{\mathbf{2}}\) Area of full triangle \(XDE\) =
\(29,955.95\,m^{2}\).Ratio \(k=\frac{11,066.07}{29,955.95}=\mathbf{0.36941}\) Final
Coordinates for Y (along line DE): \(Y_{E}=10193.80+0.36941(9922.30-
10193.80)=\mathbf{10093.50}\)\(Y_{N}=10258.90+0.36941(10114.20-
10258.90)=\mathbf{10205.45}\)Final Result: Point X: (10122.00, 10000.00)Point Y: (10093.50,
10205.45)
