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Garden Path (Posted on 2006-05-10) Difficulty: 3 of 5
A gardener has a rectangular garden 11m (AD) by 29m (AB) and wants to install a diagonal path exactly 1m wide. The edges of the path are ED and BF as shown in the diagram.
|            /  /|
|          /  /  |
|        /  /    |
|      /  /      |
|    /  /        |
|  /  /          |
D   F            C
Find the exact area of the path. (Note: EB is not the width of the path.)

See The Solution Submitted by Jer    
Rating: 3.6667 (3 votes)

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Solution solution | Comment 4 of 11 |

If x is the distance from the upper right corner to where the upper edge of the path intersects the top of the rectangle, then, construct a perpendicular from this intersection point to the lower edge of the path. The hypotenuse of the small triangle formed is x, and the leg across the path is length 1.  The small triangle is similar to the larger right triangle formed at the upper left, with legs 29-x and 11.  Therefore the smaller triangle's leg that lies along the lower edge of the path has length (29-x)/11, as the other leg has length 1.

By the Pythagorean theorem,

1 + ((29-x)/11)^2 = x^2

Eventually we get

(1-1/121)*x^2 + (58/121)*x - 962/121 = 0

The quadratic formula gives x = 2.6.

The original rectangle has area 11*29= 319. If we slide the lower right triangle to the left 2.6 units so as to expunge the path, the rectangle that's left is 11*26.4 = 290.4.  That leaves the difference, which is the area of the path, of 319 - 290.4 = 28.6.  Actually, we could simply have muliplied the x value of 2.6 by 11 to get this, i.e., 28.6 m^2.

Edited on May 10, 2006, 2:41 pm
  Posted by Charlie on 2006-05-10 14:39:42

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