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.

Guest Jul 22, 2022

#1**0 **

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

#1**0 **

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