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

Home > Shapes
Gold-in pyramid (Posted on 2020-10-06) Difficulty: 3 of 5
A right pyramid has a unit square base ABCD and vertex V. Its height is 1 unit.

Points E and F are on CV and DV respectively such that ABEF is a plane section that splits the pyramid into two pieces of equal volume.

Find the length EF.

See The Solution Submitted by Jer    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
some thoughts | Comment 1 of 7
Since ABEF are in a plane, E and F must be the same height (H) above the base. Horizontal slices at height h of the lower shape (call it a wedge) will be trapezoids.  Preparing expressions for the lengths and separation of the parallel lines of each trapezoid (one found on the CDV face and the other with endpoints on the BCV and the ADV faces) as functions of h will give the trapezoidal area at each height. The areas can be integrated over h from 0 to H to get the wedge volume. Setting the result to 1/2 the volume of the pyramid will yield H and then EF.  

Edited on October 7, 2020, 4:51 am
  Posted by Steven Lord on 2020-10-06 22:49:58

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 (0)
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