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

Home > Shapes
A different angle (Posted on 2003-10-10) Difficulty: 3 of 5
Look at this shape:

Assume AB = AF = FE = ED and BC = CD, and all the angles in the shape are 90 degrees.

Let A be the area of this shape (in cm^2), and P -- the perimeter of this shape (in cm).

If A - P is 60, what is the length of line AC?

See The Solution Submitted by Lewis    
Rating: 1.9167 (12 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
is this a bit too easy? | Comment 21 of 23 |

Let AB = n, then BC = 2n

Area = (2n)^2 - n^2 = 3n^2

P = 4(n) + 2 (2n) = 8n

A-P = 3n^2 - 8n = 60

3n^2 - 8n - 60 = 0

(n - 6) (3n + 10) = 0

n = 6 or - 3/10 (rejected)

AC^2 = 6(6) + (12)(12)

AC = 13.42

well

Edited on October 15, 2005, 4:19 am

Edited on October 15, 2005, 4:20 am
  Posted by Terence on 2005-10-15 04:18:32

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 (24)
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