Attaching ropes to the midsegment of the triangle formed ensures that the
zipline can more safely cross the family and friends to safety.
- The amount of rope needed to create a lifeline to safety is approximately 46.56
Reasons:
The coordinates of the endpoints of the redline = A(4, 3), and B(-8, -2)
The point selected, P = (4, -3)
[tex]\displaystyle Mid \ point \ of AB, \ D = \left(\frac{4 + (-8)}{2}, \ \frac{3 + (-2)}{2} \right) = \left(-2, \, 0.5\right)[/tex]
[tex]\displaystyle Mid \ point \ of AP, \ E = \left(\frac{4 +4}{2}, \ \frac{3 + (-3}{2} \right) = \left(4, \ 0 \right)[/tex]
[tex]\displaystyle Mid \ point \ of\, BP, \ F = \left(\frac{-8 +4}{2}, \ \frac{-2+ (-3}{2} \right) = \left(-2, \ -2.5\right)[/tex]
The lengths of the midsegments are found as follows
Length of DE = √((4 - (-2))² + (0 - (0.5))²) = 0.5·√(145)
Length of DF = √(((-2) - (-2))² + ((-2.5) - (0.5))²) = 3
Length of EF = √(((-2) - 4)² + ((-2.5) -0)²) = 6.5
(b) The length of side AB = √((4 - (-8))² + (3 - (-2))²) = 13
The length of side BP = √((4 - (-8))² + (-3 - (-2))²) = √(145)
The length of side AP = √((4 - 4)² + (-3 - 3)²) = 6
The perimeter of the triangle ΔABP = 13 + 6 + √(145) = 19 + √(145)
The amount of rope needed to create a lifeline to safety is therefore;
19 + √(145) + 0.5·√(145) + 3 + 6.5 = 28.5 + 1.5·√(145) ≈ 46.56
The amount of rope needed to create a lifeline to safety = 46.56
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