#FIG 3.2 Landscape Center Inches Letter 100.00 Single -2 1200 2 2 1 0 1 0 7 100 0 -1 0.000 0 0 -1 0 0 2 1200 1200 1200 8400 2 1 0 1 0 7 100 0 -1 0.000 0 0 -1 0 0 2 8400 1200 1200 1200 2 1 0 1 0 7 100 0 -1 0.000 0 0 -1 0 0 2 8400 1200 8400 8400 2 2 0 1 0 7 100 0 -1 0.000 0 0 -1 0 0 5 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 2 2 0 1 0 7 100 0 -1 0.000 0 0 -1 0 0 5 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 2 1 0 1 0 7 100 0 -1 0.000 0 0 -1 0 0 2 1200 8400 8400 8400 2 1 0 1 0 7 100 0 -1 0.000 0 0 -1 0 0 2 8400 4800 1200 4800 2 1 0 1 0 7 100 0 -1 0.000 0 0 -1 0 0 2 8400 1200 6150 4800 2 1 0 1 0 7 100 0 -1 0.000 0 0 -1 0 0 2 6150 4800 8400 8400 2 1 0 1 0 7 100 0 -1 0.000 0 0 -1 0 0 2 1200 1200 3375 4800 2 1 0 1 0 7 100 0 -1 0.000 0 0 -1 0 0 2 3375 4800 1200 8400 2 1 0 1 0 7 100 0 -1 0.000 0 0 -1 0 0 2 8400 4800 10575 4800 2 1 0 1 0 7 100 0 -1 0.000 0 0 -1 0 0 2 8400 1200 10575 4800 4 0 0 100 0 0 12 0.0000 4 135 135 900 1200 A\001 4 0 0 100 0 0 12 0.0000 4 135 120 900 4800 B\001 4 0 0 100 0 0 12 0.0000 4 135 120 900 8400 C\001 4 0 0 100 0 0 12 0.0000 4 135 135 3525 4725 D\001 4 0 0 100 0 0 12 0.0000 4 135 120 6000 4725 E\001 4 0 0 100 0 0 12 0.0000 4 135 105 8550 1200 F\001 4 0 0 100 0 0 12 0.0000 4 135 135 8625 4875 G\001 4 0 0 100 0 0 12 0.0000 4 135 135 8625 8475 H\001 4 0 0 100 0 0 12 0.0000 4 180 6780 1725 9150 AB = BC = FG = GH = 5 miles, Angle ADC = Angle ADE = Angle CDE = 120 degrees.\001 4 0 0 100 0 0 12 0.0000 4 135 690 2025 2250 a slanted\001 4 0 0 100 0 0 12 0.0000 4 135 330 2100 2475 road\001 4 0 0 100 0 0 12 0.0000 4 135 1065 6675 2325 a slanted road\001 4 0 0 100 0 0 12 0.0000 4 135 1065 2250 6975 a slanted road\001 4 0 0 100 0 0 12 0.0000 4 135 1065 6150 7050 a slanted road\001 4 0 0 100 0 0 12 0.0000 4 180 1665 3900 4650 the left-to-right road \001 4 0 0 100 0 0 12 0.0000 4 180 5145 1650 9525 Angle BDA = Angle CDB = Angle FEG = Angle HEG = 60 degrees.\001 4 0 0 100 0 0 12 0.0000 4 180 6225 1575 9900 Angle BCD = Angle DAB = Anle EHG = Angle EFG = 30 degrees. BG = 10 miles\001 4 0 0 100 0 0 12 0.0000 4 180 6765 1575 10275 EFI is an equilateral triangle, i.e. all the sides are equal and all the angles are 60 degrees. \001 4 0 0 100 0 0 12 0.0000 4 180 5685 1500 10575 FG is its height as well as its bisector as well as its median, i.e. FE = 2EG.\001 4 0 0 100 0 0 12 0.0000 4 180 7125 1500 10800 By the Pythagoras theorem EG*EG + FG*FG = FE*FE = 4*EG*EG, so FG*FG = 3*EG*EG, \001 4 0 0 100 0 0 12 0.0000 4 180 5775 1500 11100 taking square roots of both sides, we get FG = sqrt(3)*EG = (sqrt(3)/2)*EF\001 4 0 0 100 0 0 12 0.0000 4 180 2955 1500 11775 Also EG = EF/2 = (1/sqrt(3))*5 miles. \001 4 0 0 100 0 0 12 0.0000 4 180 4095 1500 11475 and finally EF = (2/sqrt(3))*FG = (2/sqrt(3))*5 miles.\001 4 0 0 100 0 0 12 0.0000 4 180 4800 1500 12150 Now put it all together to calculate the total length of the roads.\001 4 0 0 100 0 0 12 0.0000 4 180 4455 1500 12525 Damt it, they don't teach math in this God forsaken coutry!\001 4 0 0 100 0 0 12 0.0000 4 135 60 10500 5025 I\001