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

Home > Shapes > Geometry
Determine Area (Posted on 2010-03-27) Difficulty: 2 of 5

Let ABHF be a trapezoid such that 

   1) sides AB and FH are parallel,
   2) angle BAF is 90 degrees, and
   3) side FH is larger than side AB.

Let G be a point on side FH such that line segment BG 
is perpendicular to FH. Let line segment CDE be parallel 
to AB with points C, D, and E on line segments AF, BG, 
and BH respectively.

Let x = [ABDC], y = [CDGF], z = [BDE], and w = [DEHG].

Determine w in terms of x, y, and z.

Note: [PQ...UV] denotes the area of polygon PQ...UV.

See The Solution Submitted by Bractals    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution | Comment 1 of 2

Consider the parallel sides to be horizontal, for the discussion of height and base in the triangles below.

As the location of line CDE varies, the area of triangle BDE increases in proportion as the square of its height BD, as the base, DE is increasing in the same proportion. In the same way, consider the area of triangle BGH, which is z+w:

(z+w)/z = (GB/DB)^2

But on the rectangle side of the diagram:

(y+x)/x = GB/DB

Thus

(z+w)/z = ((y+x)/x)^2

z+w = z * ((y+x)/x)^2

w =  z * (((y+x)/x)^2 - 1)

w = z * ((y^2+x^2+2*x*y)/x^2 - 1)

w = z * ((y^2+2*x*y)/x^2)


  Posted by Charlie on 2010-03-27 16:00:37
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 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information