During a gale a maypole was broken in such a manner that it struck the ground at a distance twenty feet from the base of the pole.
It was repaired and later broke a second time at a point five feet lower. This time it struck the ground thirty-five feet from the base.
What was the original height of the pole?
H = Original height of the pole (say)
B= Height of the pole after the first gale.
Then, H-B is the hypotenuse of the broken pole after the first gale.
Also, the height of the pole after the second gale =B-5
and, H-B+5 is the hypotenuse of the broken pole after the second gale.
Then, we must have:
(H-B)^2 = B^2+20^2 ........(i)
(H-B+5)^2 = (B-5)^2+35^2 ..........(ii)
So, subtracting (i) from (ii), we have:
10(H-B)+25 = 25-10B+35^2-20^2
=> 10H-10B +25= 25-10B+35^2-20^2
=> 10H= 55*15 =825
=> H = 82.5
Consequently, the required original height of the maypole was 82.5 feet.
Puzzle Solution
