A pair of telephone poles d metres apart is supported by two cables which run from the top of each pole to the bottom of the other. The poles are 4 m and 6 m tall. Determine the height above the ground of the point, T , where the two cables intersect. What happens to this height as d increases?

Let the height of the poles QP = a and RS = b.

Let the distance PO = c. Let h be the height above ground of the point T.

Since ∆QPR and ∆TOR are similar , so a : d = h : (d – c) , or (d – c ) = dh / a.

Similarly , ∆SRP and ∆TOP are similar , so b : d = h : c . So , c = dh / b.

Eliminating c from the above two equations we get d = dh (1/a + 1/b) , which gives h = ab / (a+b)

If a = 4 m and b = 6 m then we get h = 12 / 5 m. Thus height of point T is 12/5 m , independent of d.

Let the height of the poles QP = a and RS = b.

Let the distance PO = c. Let h be the height above ground of the point T.

Since ∆QPR and ∆TOR are similar , so a : d = h : (d – c) , or (d – c ) = dh / a.

Similarly , ∆SRP and ∆TOP are similar , so b : d = h : c . So , c = dh / b.

Eliminating c from the above two equations we get d = dh (1/a + 1/b) , which gives h = ab / (a+b)

If a = 4 m and b = 6 m then we get h = 12 / 5 m. Thus height of point T is 12/5 m , independent of d.

**This can be seen by dragging points P or Q and see the path taken by point T**.
This is absolutely correct. Just a slightly more elegant solution, using the same principle, might look like this:

ReplyDeleteLet PO=x, OR=y. So, d=x+y, but it's not important.

From similarity of one pair of triangles:

h/a=y/(x+y)

From similarity of another pair

h/b=x/(x+y)

Add these two together and you get:

h(1/a + 1/b) = (x+y)/(x+y)

That is,

h(a+b)/ab = 1

and

h = ab/(a+b)

Thank you Zor Shekhtman

ReplyDelete