# Thread:

1. ## Pythag Question

Hi,
This question is an extract from a book which i took a photo of
as I couldn't be bothered replicating it on the computer

http://i116.photobucket.com/albums/o...chet/Maths.jpg

Currently baffled as to what the answer is.
A Nudge in the right direction will be appreciated.

2. Originally Posted by immunofort
Hi,
This question is an extract from a book which i took a photo of
as I couldn't be bothered replicating it on the computer

http://i116.photobucket.com/albums/o...chet/Maths.jpg

Currently baffled as to what the answer is.
A Nudge in the right direction will be appreciated.
First find the hypotenuse of $\triangle ABC$. Then use that hypotenuse value to help you determine the leg $\overline{CD}$ in $\triangle ADC$.

Does that give you enough info to continue on with the problem?

3. ## errrr?

No Sorry, didn't help me much.
I know how to create the long version of the equation which has 2 squareroots but i cant to be able to figure the simplified version.
More Help would be appreciated

4. Hello, immunofort!

Chris is absolutely correct . . . and he did give you a nudge, right?
Code:
```    A *
      ** * 2
      * *   *
    x *  *     * D
      *   *   *
      *    * *
    B * * * *
         1   C```

Use Pythagorus in $\Delta ABC$
$AC^2 \:=\:AB^2 + BC^2\quad\Rightarrow\quad AC^2 \:=\:x^2+1\quad\Rightarrow\quad AC \:=\:\sqrt{x^2+1}$

Use Pythagorus in $\Delta ADC$
$CD^2 + AD^2 \:=\:AC^2\quad\Rightarrow\quad CD^2 + 2^2 \:=\:(\sqrt{x^2+1})^2 \quad\Rightarrow\quad CD^2 + 4 \:=\:x^2+1$

Therefore: . $CD^2 \:=\:x^2-3 \quad\Rightarrow\quad CD \:=\:\sqrt{x^2-3}$
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