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.
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
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
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