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Pythagoras Question
Hi All, Happy NewYear to everyone.
I've attached below a diagram of the question. My attempts at it have left me stumped...
I have to find the length AB and length AB= x
I started this off by finding length AC using pythagoras and got AC= 5
Then I deduced that since AB= x, then the small length BC can be written as BC = x5
Also, I can note that length CD= a
So,
[TEX](x5)^2 = a^2 + 1^2[\TEX]
[TEX]a^2 = x^2 10x  24[\TEX]
[TEX]a^2 = (x6)(x4)[\TEX]
Hence,
[TEX]a = \pm\sqrt{6} or \pm\sqrt{4}[\TEX]
[TEX] a= 2.45 or 2[\TEX]
So, if CD = 2.45 or 2, then using pthagoras again we can now > calculate
[TEX]BC = \sqrt{2.45^2 + 1^2} = \sqrt{7}[\TEX]
Which would result in length AB = 5 + \sqrt{7} = 7.65[\TEX]
Using CD = 2 with the same above steps gives;
[TEX]BC = \sqrt{5}[\TEX] and AB = 7.24
However the answer is;
[TEX]AB= 4\sqrt{2} or 5.657[\TEX]
Where have i gone wrong? Thanks for the help.

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Re: Pythagoras Question
I think the figure is causing some confusion. First, you have two pouints both lanbeled 'D', and second it's not clear what the 4 cm length applies to. I believe that a corrected version of this figure is per the attached  note that I renamed the right hand D' as D' and that the 4cm length applies to the distance from D to D'. Finding length AB is now a simple use of Pythagoras: its the hypotenuse given two two legs of a right riangle with height 3+1= 4 cm and width 4 cm. That gives you AB = sqrt(4^2 + 4^2) = 4 sqrt(2).
The error you made are in here:
OK.
This should be
OK
Hence,
How did you get this? It dies not come from the previous equation at all  what happened to the 'x' variable?
BTW, the error in your LaTex expressions is that you have to use [/tex] to close it, not [\tex].

Re: Pythagoras Question
Here's a shortcut: Have you noticed that the two triangles are similar?
Dan