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

Home > Shapes
RMS Resolution (Posted on 2014-06-17) Difficulty: 2 of 5
Two given line segments have the lengths a and b.
Using only a compass and an unmarked straightedge construct a line segment whose length is equal to the root-mean-square of a and b.

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Slightly different approach Comment 2 of 2 |
First construct a rectangle with sides of length a and b.
Then construct a circle whose diameter is a diagonal, PQ, of the rectangle.
Now draw the perpendicular bisector of that diagonal to cross the circle at R.
The distance |PR| is the RMS value of a and b.

Proof

|PQ| = sqrt(a2 + b2) is the hypotenuse of the right isosceles triangle PQR,

so |PR| = |PQ|/sqrt(2) = sqrt((a2 + b2)/2).



  Posted by Harry on 2014-06-18 13:36:48
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (9)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information