Casey is in Florida when a Zombie Apocalypse suddenly breaks out in his area. He needs to help his family and friends cross a sinkhole to escape the marauding herd. Help Casey calculate the amount of rope he will need to zipline his family and friends to safety … don’t mess this up!

(a) Choose a point on the graph (using whole numbers) and use this point to form a triangle, making the red line one side of this triangle. Calculate the three midsegments of the triangle you created, these will stabilize the zipline. Show all of your calculations below and draw in your work in the image above.

(B) Calculate the perimeter of the triangle and the lengths of each midsegment. Add these together so that Casey knows just how much rope he will need to create this lifeline to safety. Show all of your work below. Only round off at the end of the problem. Round to the nearest hundredth digit.

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Attaching ropes to the midsegment of the triangle formed ensures that the

zipline can more safely cross the family and friends to safety.

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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