Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2014-01-12 17:41:44

thedarktiger
Member
Registered: 2014-01-10
Posts: 75

pentagon problem

In pentagon ABCDE, BC=CD=DE=2 units,  <E is a right angle and m<B = m<C = m<D = 135 degrees.  The length of segment AE can be expressed in simplest radical form as a+2sqrt(b) units.  What is the value of a+b?

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 wrongdown:(:D


Good. You can read.

Offline

#2 2014-01-12 19:58:12

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,469

Re: pentagon problem

hi thedarktiger

Hopefully the right diagram below.

I agree with everything up to the last calculation but simplify BE

Bob

View Image: thedarktiger3.gif

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

Offline

#3 2014-01-12 20:09:18

thedarktiger
Member
Registered: 2014-01-10
Posts: 75

Re: pentagon problem

nice! thats correct! I wish the thing I'm doing had solutions big_smile


Good. You can read.

Offline

#4 2014-01-12 20:13:06

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,469

Re: pentagon problem

hi

Is this from "Compuhigh" ?

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

Offline

#5 2014-01-13 17:25:46

phrontister
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 3,853

Re: pentagon problem

Hi,

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.

View Image: Geogebra.jpg View Image: Word.jpg

Last edited by phrontister (2014-01-14 19:28:50)


"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

Offline

Board footer

Powered by FluxBB