+0

# Rectangles

0
227
3

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

#1
+2667
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?

Jul 22, 2022

#1
+2667
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
#2
+129270
+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 !!!

CPhill  Jul 22, 2022
edited by CPhill  Jul 22, 2022
#3
+2667
0

LOL, Thank you!!

BuilderBoi  Jul 23, 2022