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ABCD is a rectangular sheet of paper that has been folded so that corner B is matched with point B' on edge AD. The crease is EF, where E is on AB and F is on CD. The dimensions AE=8, BE=12, and CF=2 are given. Find the perimeter of recangle ABCD.

 

 Jul 22, 2022

Best Answer 

 #1
avatar+2308 
+2

Note that\(BE = B'E = 12\)

 

By the Pythagorean Theorem, \(AB' = \sqrt{B'E^2 - AE^2} = \sqrt{144 - 64} = \sqrt{80} = 4 \sqrt 5\)

 

Also, note that by the Pythagorean Theorem, \(B'F = DF^2 + B'D^2 = CF^2 + BC^2\)

 

SImplifying the equation gives us: \(4 + BC^2 = 324 + B'D^2\)

 

But, note that \({B'D} = {BC - AB'} = BC - 4 \sqrt 5\)

 

This means that our equation simplifies to \(4 + BC^2 = 324 + (BC - 4 \sqrt5)^2\)

 

Now, just solve for \(BC\), and I assume you can take it from here?

 Jul 22, 2022
 #1
avatar+2308 
+2
Best Answer

Note that\(BE = B'E = 12\)

 

By the Pythagorean Theorem, \(AB' = \sqrt{B'E^2 - AE^2} = \sqrt{144 - 64} = \sqrt{80} = 4 \sqrt 5\)

 

Also, note that by the Pythagorean Theorem, \(B'F = DF^2 + B'D^2 = CF^2 + BC^2\)

 

SImplifying the equation gives us: \(4 + BC^2 = 324 + B'D^2\)

 

But, note that \({B'D} = {BC - AB'} = BC - 4 \sqrt 5\)

 

This means that our equation simplifies to \(4 + BC^2 = 324 + (BC - 4 \sqrt5)^2\)

 

Now, just solve for \(BC\), and I assume you can take it from here?

BuilderBoi Jul 22, 2022
 #2
avatar+124524 
+1

Very , very nice BuilderBoi.....this one had me stumped   !!!!

 

{Drawing B'F  was the key .....very crafty !!! )

 

I'd give you 5 points if  I could !!!

 

 

cool cool cool

CPhill  Jul 22, 2022
edited by CPhill  Jul 22, 2022
 #3
avatar+2308 
+2

LOL, Thank you!!

BuilderBoi  Jul 23, 2022

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