Thread: Pythagoras Question

Hi All, Happy New-Year 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 = x-5
Also, I can note that length CD= a
So,

[TEX](x-5)^2 = a^2 + 1^2[\TEX]
[TEX]a^2 = x^2 -10x - 24[\TEX]
[TEX]a^2 = (x-6)(x-4)[\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.

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].

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

-Dan