This is what I did:
Triangles ABG and EBG are congruent, from which...
AE = 2GE = 2BF = 2(2+√2) = 4+2√2.
The length of segment AE can be expressed in simplest radical form as a+2√b units.
a = 2BC = 2x2 = 4, and b = CD = 2......from which CF = √2 (Pythagoras) and hence CF also = √b.
∴ a+b = 4+2 = 6
Proof:
AE = a+2√b = 4+2√2.
The first image is with Geogebra and the second is with Word. I was just fiddling around, comparing ease of use, speed, accuracy, presentation etc. Because I use Word a fair bit it didn't take all that long there, but with Geogebra it took about 1/2 Word's time...once I'd worked out some moves.
]]>Is this from "Compuhigh" ?
Bob
]]>Hopefully the right diagram below.
I agree with everything up to the last calculation but simplify BE
Bob
]]>This is actually a lot of 45 45 90 triangles!
a+2*sqrt(b)
A xxxxxxxxxxxxxxxxxxxxxx E
x x
x x 2
x x D
n x 2 x x
x x x sqrt(2)
B xxxxxxxxxxxxxxx
C sqrt(2)
2
Then I drew BE and got that BE = sqrt(2(2+sqrt(2))^2)=sqrt(2)*(2+sqrt(2))
Then <ABE is 45 so ABE is 45 45 90 so AB = BE so AE = sqrt(2*(sqrt(2)*(2+sqrt(2)))^2) = 4*(2+sqrt(2)) = 8 + 4*sqrt(2) = 8 + 2*sqrt(8)
so the answers 8+8=16
whats wrong I got it wrong:(:D