Five points A, B, C, D, and O lie on a flat field. A is directly north of O, B is directly west of O, C is directly south of O, and D is directly east of O. The distance between C and D is 140 m. A hot-air balloon is positioned in the air at H directly above O. The balloon is held in place by four ropes HA, HB, HC, and HD. Rope HC has length 150 m and rope HD has length 130 m.

To reduce the total length of rope used, rope HC and rope HD are to be replaced by a single rope HP where P is a point on the straight line between C and D. (The balloon remains at the same position H above O as described above.) Determine the greatest length of rope that can be saved.

Guest Jun 30, 2018

#1**-1 **

let oh = z and od = x and oc = y

z^2 + x^2 = 130^2

z^2 + y^2 = 150^2

x^2 + y^2 = 140^2

by pythagorean theorem

we can use system of equation rules to find x,y,z

x^2 - y^2 = 130^2 - 150^2

2x^2 = 130^2 - 150^2 +140^2

x= sqrt7000

z^2 + 7000 = 130^2

z = sqrt9900

7000 + y^2 = 140^2

y = sqrt12600

op can now be figured out using x or y

y^2 + 70^2 = (op)^2 or

12600 + 70^2 = 17500

op= sqrt17500

now we find z^2 + 17500 = (hp)^2 hp being the new rope

9900 + 17500 = 27400

the new rope is sqrt27400 or about 165.5 rounded.

the question wants to know how much can be saved.

so 130+150 is 280

280 - 165.5 is 114.5M rope saved

Guest Jul 1, 2018