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

Home > Shapes
Quadrilateral Area Query (Posted on 2016-04-10) Difficulty: 3 of 5
ABC is a triangle with AB = 12, AC = 14 and BC = 16.
D is the midpoint of AB.
The vertex C is joined with D forming the crease EF with E on AC and F on BC.
Find the area of the quadrilateral BDEF

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts re: Long Solution Comment 2 of 2 |
(In reply to Long Solution by Jer)

I agree, it looks nasty:

The height of triangle ABC is sqrt( 14^2-3.5^2) = 7sqrt15/2, so that triangle's area is 42*sqrt15/2, A. Already nasty.

Turning to the kite DFCE, the larger pair of triangles (b) is bigger than the smaller pair (s) by the proportion 41/35.

Then 2b+ discarded (d) = 2(35/41)b +kept (k) = A/2.

The area d is easy: sqrt(32/5^2- 8/5^2)*3 = 24 sqrt15/5.

Now we have 2b+24 sqrt15/5 = 21*sqrt15/2, and b is the again nasty 57 sqrt15/20

The entire kept area, K, is s+b+k = 1/2A-s+b, = 21*sqrt15/2+ b- 35/41b = 267 sqrt15/20 - 399sqrt15/164, or (2238 sqrt(15))/205.



  Posted by broll on 2016-04-11 05:15:12
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 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information