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- Aug 13th 2008, 09:47 PMhotgal24SoYouThinkYouAreAGenius?
- Aug 14th 2008, 06:21 AMticbolHow about stubborn?
Okay, let us call AB =x; AD = y; CE = u and CF = v

In triangle ABE:

(1/2)(x)(y-u) = 4

y-u = 8/x

u = y -(8/x) ---------**

In triangle ADF:

(1/2)(y)(x-v) = 5

x-v = 10/y

v = x -(10/y) --------**

In triangle CEF:

(1/2)(u)(v) = 3

u*v = 6

Substitutions,

[y -(8/x)]*[x -(10/y)] = 6

xy -10 -8 +80/(xy) = 6

xy +80/(xy) = 24

Multiply both sides by xy,

(xy)^2 +80 = 24(xy)

(xy)^2 -24(xy) +80 = 0

Factoring that,

(xy -20)(xy -4) = 0

So,

xy = 20 or 4

xy cannot be 4, so xy = 20.

Therefore, in rectangle ABCD,

Let K = area of triangle AEF,

x*y = 3 +4 +5 +K

K = 20 -3-4-5 = 8 -----------answer.