Can't figure this one out.

The answer must be in exact value.

*Picture has been fixed to add the missing side length

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- Apr 3rd 2010, 06:47 PMshawliFinding side "x" (triangles)
Can't figure this one out.

The answer must be in exact value.

*Picture has been fixed to add the missing side length - Apr 3rd 2010, 07:01 PMSudharaka
- Apr 3rd 2010, 07:02 PMProve It
- Apr 3rd 2010, 07:07 PMshawli
Sorry! I missed a label, but I've reposted the picture with the correction.

- Apr 3rd 2010, 07:11 PMSudharaka
- Apr 3rd 2010, 07:14 PMProve It
You're expected to be able to recognise triangles that are of dimensions $\displaystyle (1, 1, \sqrt{2})$ and $\displaystyle (1, \sqrt{3}, 2)$.

Looking at the large triangle, its dimensions are

$\displaystyle (x, \sqrt{3}\,x, 2x)$.

So the base of this shape is $\displaystyle \sqrt{3}\,x$.

Using that information, you should be able to see that

$\displaystyle \sqrt{3}\,x - x = 4$

$\displaystyle x(\sqrt{3} - 1) = 4$

$\displaystyle x = \frac{4}{\sqrt{3} - 1}$

$\displaystyle x = \frac{4(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}$

$\displaystyle x = \frac{4(\sqrt{3} + 1)}{3 - 1}$

$\displaystyle x = \frac{4(\sqrt{3} + 1)}{2}$

$\displaystyle x = 2(\sqrt{3} + 1)$

$\displaystyle x = 2\sqrt{3} + 2$.