Thread: Surds (length of a side of a rhombus)

1. Surds (length of a side of a rhombus)

I tried Pythagora's theorem and I think I'm close to the answer, but I'm stuck at this point:

Sorry for the images, Latex is kind of confusing for me :s

Any help would be greatly appreciated and thank you in advance.

2. Hello !

You are close to the answer. You have the same one, written in a different form, but you need some more work to get to their actual answer :

$BC^2 = 18 \frac{4}{5} + 4 \sqrt{5} - \frac{12}{\sqrt{5}}$

First, we need to put everything in absolute fraction form (otherwise things get confusing) :

$BC^2 = \frac{94}{5} + 4 \sqrt{5} - \frac{12}{\sqrt{5}}$

Note that you can multiply everything by $5$ :

$5 BC^2 = 94 + 20 \sqrt{5} - 12 \sqrt{5}$

Then you can factor a $2$ out :

$5 BC^2 = 2 \left (47 + 10 \sqrt{5} - 6 \sqrt{5} \right )$

Simplifying the surds :

$5 BC^2 = 2 \left (47 + 4 \sqrt{5} \right )$

And finally, dividing by $5$ :

$BC^2 = \frac{2}{5} \left (47 + 4 \sqrt{5} \right )$

Does it make sense ?

3. Thank you very much!
If I use my original answer $( BC^2 = 18\frac{4}{5} + 4\sqrt5 - \frac{12}{\sqrt5} )$ is it still considered to be a correct answer if I use it in, say, a test? Or would it be incorrect because it is not fully simplified?
Thanks.

4. Originally Posted by caramelcake
Thank you very much!
If I use my original answer $( BC^2 = 18\frac{4}{5} + 4\sqrt5 - \frac{12}{\sqrt5} )$ is it still considered to be a correct answer if I use it in, say, a test? Or would it be incorrect because it is not fully simplified?
Thanks.
I believe you might lose some points depending on the context, but it wouldn't be penalized that much unless the question specifically asks for full simplification/factorization. But if you can factorize, do it (here there were two identical surds which sort of looks bad)

5. OK, thanks a lot for your input!

6. It was my pleasure.