All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
A Maypole Problem (Posted on 2006-05-01) Difficulty: 3 of 5
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?

See The Solution Submitted by Jer    
Rating: 3.3333 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Mayday! Mayday! | Comment 2 of 8 |
The original height of the pole is about 82.5 feet.

Each time the pole breaks, it forms a right triangle with the ground. For the first triangle, call the side segment x and the hypotenuse segment y. For the second triangle, the side is x-5 and the hypotenuse is y+5.

Applying Pythagorean theorem (a^2 + b^2 = c^2, where c is the hypotenuse) to the two triangles yields the following equations:

400 + x^2 = y^2
1225 + x^2 - 10x +25 = y^2 +10y +25

Substituting the y's in the second equation and simplifying leaves:

1225 + x^2 - 10x +25 = 400 + x^2 + 10*SQRT(400 + x^2) + 25
825 - 10x = 10*SQRT(400 + x^2)

Square the equation:

680625 - 16500x + 100x^2 = 40000 + 100x^2
16500x = 640625
x = 38.82575757...

Putting x back into the first equation (400 + x^2 = y^2) gives:

y = 43.67424242...

And x + y = 82.5 feet.

A quick check with the law of sines shows that these values work for both triangles. That's a big pole!


  Posted by Jyqm on 2006-05-01 14:15:59
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (23)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information